2 Preliminaries

A set $K\subset \mathbb{R}^{2}$ is convex if it contains the complete segment joining every two points in the set. We shall consider non-empty compact convex sets. The support function of $K$ is defined as

$$\begin{eqnarray}p_{K}(u)=\sup \{\langle x,u\rangle :x\in K\}\quad \text{for}~u\in \mathbb{R}^{2}.\end{eqnarray}$$

For a unit vector $u$ , the number $p_{K}(u)$ is the signed distance of the support line to  $K$ with outer normal vector  $u$ from the origin. The distance is negative if and only if $u$ points into the open half-plane containing the origin (cf. [Reference Schneider10, 1.7.1]). We shall denote by  $p(\unicode[STIX]{x1D711})$ the $2\unicode[STIX]{x1D70B}$ -periodic function obtained by evaluating $p_{K}(u)$ on $u=(\cos \unicode[STIX]{x1D711},\sin \unicode[STIX]{x1D711})$ . Note that $\unicode[STIX]{x2202}K$ is the envelope of the one parametric family of lines given by

$$\begin{eqnarray}x\cos \unicode[STIX]{x1D711}+y\sin \unicode[STIX]{x1D711}=p(\unicode[STIX]{x1D711}).\end{eqnarray}$$

If the support function $p(\unicode[STIX]{x1D711})$ is differentiable, we can parametrize the boundary $\unicode[STIX]{x2202}K$ by

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D711})=p(\unicode[STIX]{x1D711})N(\unicode[STIX]{x1D711})+p^{\prime }(\unicode[STIX]{x1D711})N^{\prime }(\unicode[STIX]{x1D711}), & & \displaystyle \nonumber\end{eqnarray}$$

where $N(\unicode[STIX]{x1D711})=(\cos \unicode[STIX]{x1D711},\sin \unicode[STIX]{x1D711}).$ When $p$ is a ${\mathcal{C}}^{2}$ function, the radius of curvature $\unicode[STIX]{x1D70C}(\unicode[STIX]{x1D711})$ of $\unicode[STIX]{x2202}K$ at the point $\unicode[STIX]{x1D6FE}(\unicode[STIX]{x1D711})$ is given by $p(\unicode[STIX]{x1D711})+p^{\prime \prime }(\unicode[STIX]{x1D711})$ . Then, convexity is equivalent to $p(\unicode[STIX]{x1D711})+p^{\prime \prime }(\unicode[STIX]{x1D711})\geqslant 0$ . From now on, we will assume that $p$ is of class  $C^{2}$ .

It can be seen (cf. [Reference Santaló9, I.1.2]) that the length $L$ of $\unicode[STIX]{x2202}K$ and the area $F$ of $K$ are given in terms of the support function, respectively, by

(3) $$\begin{eqnarray}\displaystyle L=\int _{0}^{2\unicode[STIX]{x1D70B}}p\,d\unicode[STIX]{x1D711}\quad \text{and}\quad F=\frac{1}{2}\int _{0}^{2\unicode[STIX]{x1D70B}}(p^{2}-{p^{\prime }}^{2})\,d\unicode[STIX]{x1D711}. & & \displaystyle\end{eqnarray}$$

We will consider $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}(P)$ the visual angle of $\unicode[STIX]{x2202}K$ from an exterior point $P$ , that is, the angle between the tangents from $P$ to $\unicode[STIX]{x2202}K$ . For a function $f(\unicode[STIX]{x1D714})$ of the visual angle $\unicode[STIX]{x1D714}$ , we will deal with the integral of $f(\unicode[STIX]{x1D714})$ with respect to the area measure $dP$ .

Sets of constant width form a special class of convex sets; they are those whose orthogonal projection on any direction have the same length  $w$ . In terms of the support function  $p$ of $K$ , constant width means that $p(\unicode[STIX]{x1D711})+p(\unicode[STIX]{x1D711}+\unicode[STIX]{x1D70B})=w$ . Expanding $p$ in Fourier series

(4) $$\begin{eqnarray}\displaystyle p(\unicode[STIX]{x1D711})=a_{0}+\mathop{\sum }_{n=1}^{\infty }(a_{n}\cos n\unicode[STIX]{x1D711}+b_{n}\sin n\unicode[STIX]{x1D711}), & & \displaystyle\end{eqnarray}$$

it follows that

$$\begin{eqnarray}p(\unicode[STIX]{x1D711})+p(\unicode[STIX]{x1D711}+\unicode[STIX]{x1D70B})=2\mathop{\sum }_{n=0}^{\infty }(a_{2n}\cos 2n\unicode[STIX]{x1D711}+b_{2n}\sin 2n\unicode[STIX]{x1D711}),\end{eqnarray}$$

and therefore constant width is equivalent to $a_{n}=b_{n}=0$ for all even $n>0$ .

Given a compact convex set $K$ with support function $p(\unicode[STIX]{x1D711})$ , the Steiner point of  $K$ is defined by the vector-valued integral

$$\begin{eqnarray}s(K)=\frac{1}{\unicode[STIX]{x1D70B}}\int _{0}^{2\unicode[STIX]{x1D70B}}p(\unicode[STIX]{x1D711})N(\unicode[STIX]{x1D711})\,d\unicode[STIX]{x1D711}.\end{eqnarray}$$

This functional on the space of convex sets is additive with respect to the Minkowski sum. The Steiner point is rigid motion equivariant; this means that $s(gK)=gs(K)$ for every rigid motion  $g$ . We remark that $s(K)$ can be considered, in the ${\mathcal{C}}^{2}$ case, as the centroid with respect to the curvature measure in the boundary $\unicode[STIX]{x2202}K$ ; also it is known that $s(K)$ lies in the interior of $K$ (cf. [Reference Groemer5, p. 56]). In terms of the Fourier coefficients of $p(\unicode[STIX]{x1D711})$ given in (4), the Steiner point is

$$\begin{eqnarray}s(K)=(a_{1},b_{1}).\end{eqnarray}$$

The relation between the support function $p(\unicode[STIX]{x1D711})$ of a convex set $K$ and the support function $q(\unicode[STIX]{x1D711})$ of the same convex set but with respect to a new reference with origin at the point $(a,b)$ and axes parallel to the previous $x$ and $y$ axes, is given by

$$\begin{eqnarray}q(\unicode[STIX]{x1D711})=p(\unicode[STIX]{x1D711})-a\cos \unicode[STIX]{x1D711}-b\sin \unicode[STIX]{x1D711}.\end{eqnarray}$$

Hence, taking the Steiner point as a new origin,

$$\begin{eqnarray}q(\unicode[STIX]{x1D711})=a_{0}+\mathop{\sum }_{n\geqslant 2}(a_{n}\cos n\unicode[STIX]{x1D711}+b_{n}\sin n\unicode[STIX]{x1D711}).\end{eqnarray}$$

The associated pedal curve to $K$ will be the curve that in polar coordinates with respect to the Steiner point as origin is given by $r=p(\unicode[STIX]{x1D711})$ . In fact, it is the geometrical locus of the orthogonal projection of the center on the tangents to the curve. The area  $A$ enclosed by the pedal curve is

$$\begin{eqnarray}A=\frac{1}{2}\int _{0}^{2\unicode[STIX]{x1D70B}}p(\unicode[STIX]{x1D711})^{2}\,d\unicode[STIX]{x1D711}.\end{eqnarray}$$

3 First integral formula

Let $f(\unicode[STIX]{x1D714})$ be a function of the visual angle $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}(P)$ of a given compact convex set  $K$ from a point  $P$ outside $K$ . In this section, we will give a formula to compute the integral of $f(\unicode[STIX]{x1D714})$ with respect to the area measure  $dP$ , $\int _{P\notin K}f(\unicode[STIX]{x1D714})\,dP$ , in terms of the area of  $K$ , the length of the boundary of  $K$ , and the Fourier coefficients of the support function of  $K$ .

For each point $P\notin K$ , let $\unicode[STIX]{x1D711}$ be the angle at the origin formed by the normal to one of the tangents from $P$ to $\unicode[STIX]{x2202}K$ with the $x$ axis; the pair $(\unicode[STIX]{x1D711},\unicode[STIX]{x1D714})$ can be considered as a system of coordinates of $\mathbb{R}^{2}\setminus K$ .

Figure 1

The visual angle $\unicode[STIX]{x1D714}$ .

Denoting by $A$ , $A_{1}$ the contact points of the tangents from $P$ to $\unicode[STIX]{x2202}K$ , and by  $p=p(\unicode[STIX]{x1D711})$ the support function of $K$ with respect to an origin $O$ inside $K$ (see Figure 1), we have

$$\begin{eqnarray}\displaystyle A=(p\cos \unicode[STIX]{x1D711}-p^{\prime }\sin \unicode[STIX]{x1D711},p\sin \unicode[STIX]{x1D711}+p^{\prime }\cos \unicode[STIX]{x1D711}) & & \displaystyle \nonumber\end{eqnarray}$$

and hence, denoting

$$\begin{eqnarray}\unicode[STIX]{x1D711}_{1}=\unicode[STIX]{x1D70B}+\unicode[STIX]{x1D711}-\unicode[STIX]{x1D714}\end{eqnarray}$$

we get

$$\begin{eqnarray}\displaystyle A_{1}=(-p_{1}\cos (\unicode[STIX]{x1D711}-\unicode[STIX]{x1D714})+p_{1}^{\prime }\sin (\unicode[STIX]{x1D711}-\unicode[STIX]{x1D714}),-p_{1}\sin (\unicode[STIX]{x1D711}-\unicode[STIX]{x1D714})-p_{1}^{\prime }\cos (\unicode[STIX]{x1D711}-\unicode[STIX]{x1D714})), & & \displaystyle \nonumber\end{eqnarray}$$

where $p_{1}(\unicode[STIX]{x1D711})=p(\unicode[STIX]{x1D711}_{1})$ , $p_{1}^{\prime }(\unicode[STIX]{x1D711})=p^{\prime }(\unicode[STIX]{x1D711}_{1})$ .

The intersection point $P=(X,Y)$ of the tangent lines to $\unicode[STIX]{x2202}K$ at points  $A$ and  $A_{1}$ is given by

$$\begin{eqnarray}\displaystyle X & = & \displaystyle -\frac{1}{\sin \unicode[STIX]{x1D714}}(p\sin (\unicode[STIX]{x1D711}-\unicode[STIX]{x1D714})+p_{1}\sin \unicode[STIX]{x1D711}),\nonumber\\ \displaystyle Y & = & \displaystyle \frac{1}{\sin \unicode[STIX]{x1D714}}(p\cos (\unicode[STIX]{x1D711}-\unicode[STIX]{x1D714})+p_{1}\cos \unicode[STIX]{x1D711}).\nonumber\end{eqnarray}$$

From this, it is easy to see that the distances $T=PA$ and $T_{1}=PA_{1}$ are given by the positive quantities

(5) $$\begin{eqnarray}\displaystyle \begin{array}{@{}c@{}}\displaystyle T=\frac{1}{\sin \unicode[STIX]{x1D714}}(p\cos \unicode[STIX]{x1D714}-p^{\prime }\sin \unicode[STIX]{x1D714}+p_{1}),\\ \displaystyle T_{1}=\frac{1}{\sin \unicode[STIX]{x1D714}}(p_{1}\cos \unicode[STIX]{x1D714}+p_{1}^{\prime }\sin \unicode[STIX]{x1D714}+p),\end{array} & & \displaystyle\end{eqnarray}$$

due to the fact that the origin is inside $K$ .

The area element $dP$ of $\mathbb{R}^{2}\setminus K$ is

$$\begin{eqnarray}dP=dX\wedge dY=\bigg(\frac{\unicode[STIX]{x2202}X}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\frac{\unicode[STIX]{x2202}Y}{\unicode[STIX]{x2202}\unicode[STIX]{x1D711}}-\frac{\unicode[STIX]{x2202}X}{\unicode[STIX]{x2202}\unicode[STIX]{x1D711}}\frac{\unicode[STIX]{x2202}Y}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\bigg)\,d\unicode[STIX]{x1D711}\wedge d\unicode[STIX]{x1D714}.\end{eqnarray}$$

A straightforward computation shows that

$$\begin{eqnarray}\displaystyle dP=\frac{TT_{1}}{\sin \unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D711}\wedge d\unicode[STIX]{x1D714}. & & \displaystyle \nonumber\end{eqnarray}$$

This expression of the area element, introduced by Crofton in [Reference Crofton1], appears also in [Reference Masotti8] and [Reference Santaló9, I.2.2].

Hence, the integral on $\mathbb{R}^{2}\setminus K$ of a suitable function of the visual angle $f(\unicode[STIX]{x1D714})$ is given by

$$\begin{eqnarray}\int _{P\notin K}f(\unicode[STIX]{x1D714})\,dP=\int _{0}^{\unicode[STIX]{x1D70B}}\int _{0}^{2\unicode[STIX]{x1D70B}}\frac{f(\unicode[STIX]{x1D714})}{\sin \unicode[STIX]{x1D714}}TT_{1}\,d\unicode[STIX]{x1D711}\,d\unicode[STIX]{x1D714}=\int _{0}^{\unicode[STIX]{x1D70B}}\frac{f(\unicode[STIX]{x1D714})}{\sin \unicode[STIX]{x1D714}}\biggl(\int _{0}^{2\unicode[STIX]{x1D70B}}TT_{1}\,d\unicode[STIX]{x1D711}\biggr)\,d\unicode[STIX]{x1D714}.\end{eqnarray}$$

Now we will write the product $TT_{1}$ in terms of the Fourier coefficients of $p(\unicode[STIX]{x1D711})$ given in (4) and the Fourier coefficients of $p_{1}(\unicode[STIX]{x1D711})$ given by

$$\begin{eqnarray}p_{1}(\unicode[STIX]{x1D711})=a_{0}+\mathop{\sum }_{k>0}(A_{k}\cos k\unicode[STIX]{x1D711}+B_{k}\sin k\unicode[STIX]{x1D711}),\end{eqnarray}$$

which are related to the coefficients of $p(\unicode[STIX]{x1D711})$ by

$$\begin{eqnarray}\displaystyle A_{k} & = & \displaystyle (-1)^{k+1}(-a_{k}\cos k\unicode[STIX]{x1D714}+b_{k}\sin k\unicode[STIX]{x1D714}),\nonumber\\ \displaystyle B_{k} & = & \displaystyle (-1)^{k+1}(-a_{k}\sin k\unicode[STIX]{x1D714}-b_{k}\cos k\unicode[STIX]{x1D714}).\nonumber\end{eqnarray}$$

Substituting these Fourier series in (5), a straightforward but long calculation gives

$$\begin{eqnarray}\displaystyle \int _{0}^{2\unicode[STIX]{x1D70B}}TT_{1}\,d\unicode[STIX]{x1D711}=\frac{1}{\sin ^{2}\unicode[STIX]{x1D714}}\biggl(\frac{L^{2}}{2\unicode[STIX]{x1D70B}}(1+\cos \unicode[STIX]{x1D714})^{2}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k>0}c_{k}^{2}h_{k}(\unicode[STIX]{x1D714})\biggr), & & \displaystyle \nonumber\end{eqnarray}$$

where $c_{k}^{2}=a_{k}^{2}+b_{k}^{2}$ and

(6) $$\begin{eqnarray}\displaystyle h_{k}(\unicode[STIX]{x1D714}) & = & \displaystyle 2\cos \unicode[STIX]{x1D714}+(-1)^{k+1} (-\cos \,k\unicode[STIX]{x1D714}(1+\cos ^{2}\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle -\,2k\sin k\unicode[STIX]{x1D714}\sin \unicode[STIX]{x1D714}\cos \unicode[STIX]{x1D714}+k^{2}\cos k\unicode[STIX]{x1D714}\sin ^{2}\unicode[STIX]{x1D714}\,).\end{eqnarray}$$

These functions can also be written as

$$\begin{eqnarray}\displaystyle h_{k}(\unicode[STIX]{x1D714}) & = & \displaystyle \frac{(-1)^{k}}{4}[\,(k+1)^{2}\cos ((k-2)\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle +\,(k-1)^{2}\cos ((k+2)\unicode[STIX]{x1D714})-2(k^{2}-3)\cos (k\unicode[STIX]{x1D714})\,]+2\cos \unicode[STIX]{x1D714}.\nonumber\end{eqnarray}$$

Notice that $h_{1}\equiv 0$ , $h_{k}(0)=2(1+(-1)^{k})$ and $h_{k}(\unicode[STIX]{x1D714})=O((\unicode[STIX]{x1D714}-\unicode[STIX]{x1D70B})^{4})$ , as $\unicode[STIX]{x1D714}$ tends to  $\unicode[STIX]{x1D70B}$ . Hence we have obtained the following result.

Theorem 3.1. Let $K$ be a compact convex set with boundary of class $C^{2}$ and let $L$ be the length of $\unicode[STIX]{x2202}K$ . Let $c_{k}^{2}=a_{k}^{2}+b_{k}^{2}$ , where $a_{k}$ , $b_{k}$ are the Fourier coefficients of the support function of $K$ . Then, for every continuous function of the visual angle  $f(\unicode[STIX]{x1D714})$ on $[0,\unicode[STIX]{x1D70B}]$ such that $f(\unicode[STIX]{x1D714})=O(\unicode[STIX]{x1D714}^{3})$ , as $\unicode[STIX]{x1D714}$ tends to zero,

(7) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{P\notin K}f(\unicode[STIX]{x1D714})\,dP\nonumber\\ \displaystyle & & \displaystyle \quad =\bigg(\int _{0}^{\unicode[STIX]{x1D70B}}\frac{f(\unicode[STIX]{x1D714})(1+\cos \unicode[STIX]{x1D714})^{2}}{\sin ^{3}\unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}\bigg)\frac{L^{2}}{2\unicode[STIX]{x1D70B}}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\bigg(\int _{0}^{\unicode[STIX]{x1D70B}}\frac{f(\unicode[STIX]{x1D714})h_{k}(\unicode[STIX]{x1D714})}{\sin ^{3}\unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}\bigg)c_{k}^{2},\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $h_{k}$ , for $k\geqslant 2$ , are the universal functions given in (6).

As a first example, we can easily compute the integral in (1) to get

(8) $$\begin{eqnarray}\displaystyle \int _{P\notin K}\sin ^{3}\unicode[STIX]{x1D714}\,dP & = & \displaystyle \frac{L^{2}}{2\unicode[STIX]{x1D70B}}\int _{0}^{\unicode[STIX]{x1D70B}}(1+\cos \unicode[STIX]{x1D714})^{2}\,d\unicode[STIX]{x1D714}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}c_{k}^{2}\int _{0}^{\unicode[STIX]{x1D70B}}h_{k}(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & = & \displaystyle \frac{3}{4}L^{2}+\frac{9}{4}\unicode[STIX]{x1D70B}^{2}c_{2}^{2}.\end{eqnarray}$$

Notice that $\unicode[STIX]{x1D6FE}_{2}^{2}=9c_{k}^{2}$ (see the footnote on page 884). This formula shows that the quantity $c_{2}^{2}$ is invariant with respect to Euclidean motions of $K$ .

4 The area of level sets and second integral formula

As an application of Theorem 3.1, we will now compute the area $F(\unicode[STIX]{x1D714})$ enclosed by the locus $C_{\unicode[STIX]{x1D714}}$ of the points from which the convex set $K$ is viewed under the same angle $\unicode[STIX]{x1D714}$ . The corresponding formula (11) was first given by Hurwitz in [Reference Hurwitz6, p. 383] and we will use it later to obtain another version of formula (7).

Applying formula (7) with $f$ the characteristic function of the domain enclosed by the level set $C_{\unicode[STIX]{x1D714}}$ gives

(9) $$\begin{eqnarray}\displaystyle F(\unicode[STIX]{x1D714}) & = & \displaystyle F+\bigg(\int _{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}\frac{(1+\cos \unicode[STIX]{x1D70F})^{2}}{\sin ^{3}\unicode[STIX]{x1D70F}}\,d\unicode[STIX]{x1D70F}\bigg)\frac{L^{2}}{2\unicode[STIX]{x1D70B}}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\bigg(\int _{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}\frac{h_{k}(\unicode[STIX]{x1D70F})}{\sin ^{3}\unicode[STIX]{x1D70F}}\,d\unicode[STIX]{x1D70F}\bigg)c_{k}^{2}\nonumber\\ \displaystyle & = & \displaystyle F+\frac{L^{2}}{2\unicode[STIX]{x1D70B}}\bigg[-\frac{1}{2\sin ^{2}(\unicode[STIX]{x1D70F}/2)}\bigg]_{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\int _{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}\bigg(\frac{h_{k}(\unicode[STIX]{x1D70F})}{\sin ^{3}\unicode[STIX]{x1D70F}}\,d\unicode[STIX]{x1D70F}\bigg)c_{k}^{2}\nonumber\\ \displaystyle & = & \displaystyle F+\frac{L^{2}}{4\unicode[STIX]{x1D70B}}\cot ^{2}(\unicode[STIX]{x1D714}/2)+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\int _{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}\bigg(\frac{h_{k}(\unicode[STIX]{x1D70F})}{\sin ^{3}\unicode[STIX]{x1D70F}}\,d\unicode[STIX]{x1D70F}\bigg)c_{k}^{2}.\end{eqnarray}$$

Using (3) and the Fourier expansions of $p$ and $p^{\prime }$ , one gets (see, for instance, [Reference Cufí and Reventós2] or [Reference Groemer5, 4.2.2])

$$\begin{eqnarray}\displaystyle F=\frac{L^{2}}{4\unicode[STIX]{x1D70B}}-\frac{\unicode[STIX]{x1D70B}}{2}\mathop{\sum }_{k\geqslant 2}(k^{2}-1)c_{k}^{2} & & \displaystyle \nonumber\end{eqnarray}$$

and equation (9) can be written as

$$\begin{eqnarray}F(\unicode[STIX]{x1D714})=\frac{L^{2}}{4\unicode[STIX]{x1D70B}}\frac{1}{\sin ^{2}(\unicode[STIX]{x1D714}/2)}-\frac{\unicode[STIX]{x1D70B}}{2}\mathop{\sum }_{k\geqslant 2}(k^{2}-1)c_{k}^{2}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\int _{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}\bigg(\frac{h_{k}(\unicode[STIX]{x1D70F})}{\sin ^{3}\unicode[STIX]{x1D70F}}\,d\unicode[STIX]{x1D70F}\bigg)c_{k}^{2}.\end{eqnarray}$$

Equivalently,

$$\begin{eqnarray}\displaystyle F(\unicode[STIX]{x1D714})\sin ^{2}\unicode[STIX]{x1D714} & = & \displaystyle \frac{L^{2}}{2\unicode[STIX]{x1D70B}}(1+\cos \unicode[STIX]{x1D714})-\frac{\unicode[STIX]{x1D70B}}{2}\sin ^{2}\unicode[STIX]{x1D714}\mathop{\sum }_{k\geqslant 2}(k^{2}-1)c_{k}^{2}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D70B}\sin ^{2}\unicode[STIX]{x1D714}\mathop{\sum }_{k\geqslant 2}\int _{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}\bigg(\frac{h_{k}(\unicode[STIX]{x1D70F})}{\sin ^{3}\unicode[STIX]{x1D70F}}\,d\unicode[STIX]{x1D70F}\bigg)c_{k}^{2}.\nonumber\end{eqnarray}$$

Introducing the functions

(10) $$\begin{eqnarray}\displaystyle g_{k}(\unicode[STIX]{x1D714})=1+\frac{(-1)^{k}}{2}((k+1)\cos (k-1)\unicode[STIX]{x1D714}-(k-1)\cos (k+1)\unicode[STIX]{x1D714}), & & \displaystyle\end{eqnarray}$$

already considered by Hurwitz, and using that

$$\begin{eqnarray}\frac{h_{k}(\unicode[STIX]{x1D70F})}{\sin ^{3}\unicode[STIX]{x1D70F}}=-\bigg(\frac{g_{k}(\unicode[STIX]{x1D70F})}{\sin ^{2}\unicode[STIX]{x1D70F}}\bigg)^{\prime }\end{eqnarray}$$

we get the following proposition.

Proposition 4.1. Under the same hypothesis as that of Theorem 3.1, the area $F(\unicode[STIX]{x1D714})$ enclosed by the locus $C_{\unicode[STIX]{x1D714}}$ of the points from which the compact convex set $K$ is viewed under the same angle $\unicode[STIX]{x1D714}$ is given by

(11) $$\begin{eqnarray}\displaystyle F(\unicode[STIX]{x1D714})\sin ^{2}\unicode[STIX]{x1D714}=\frac{L^{2}}{2\unicode[STIX]{x1D70B}}(1+\cos \unicode[STIX]{x1D714})+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}c_{k}^{2}g_{k}(\unicode[STIX]{x1D714}), & & \displaystyle\end{eqnarray}$$

where the functions $g_{k}(\unicode[STIX]{x1D714})$ are defined in (10).

Notice that the asymptotic behavior of $F(\unicode[STIX]{x1D714})$ for $\unicode[STIX]{x1D714}$ near zero is given by

(12) $$\begin{eqnarray}\displaystyle \lim _{\unicode[STIX]{x1D714}\rightarrow 0}(F(\unicode[STIX]{x1D714})\sin ^{2}\unicode[STIX]{x1D714})=\frac{L^{2}}{\unicode[STIX]{x1D70B}}+2\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2,\,\text{even}}c_{k}^{2}, & & \displaystyle\end{eqnarray}$$

an equality that appears in [Reference Hurwitz6, p. 383]. In the special case of a compact set of constant width, it is

$$\begin{eqnarray}\lim _{\unicode[STIX]{x1D714}\rightarrow 0}(F(\unicode[STIX]{x1D714})\sin ^{2}\unicode[STIX]{x1D714})=\frac{L^{2}}{\unicode[STIX]{x1D70B}}.\end{eqnarray}$$

As $dP=P(\unicode[STIX]{x1D711},\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D711}\wedge d\unicode[STIX]{x1D714}$ , we can write

$$\begin{eqnarray}F(\unicode[STIX]{x1D714})=F+\int _{0}^{2\unicode[STIX]{x1D70B}}\int _{\unicode[STIX]{x1D714}}^{\unicode[STIX]{x1D70B}}P(\unicode[STIX]{x1D711},\unicode[STIX]{x1D70F})\,d\unicode[STIX]{x1D70F}\,d\unicode[STIX]{x1D711},\end{eqnarray}$$

so that $F^{\prime }(\unicode[STIX]{x1D714})=-\int _{0}^{2\unicode[STIX]{x1D70B}}P(\unicode[STIX]{x1D711},\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D711}$ . Therefore

$$\begin{eqnarray}\int _{P\notin K}f(\unicode[STIX]{x1D714})\,dP=-\int _{0}^{\unicode[STIX]{x1D70B}}f(\unicode[STIX]{x1D714})F^{\prime }(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}.\end{eqnarray}$$

Integrating by parts and using the functions $g_{k}$ given in (10), we obtain the following proposition.

Proposition 4.2. With the same hypothesis as that of Theorem 3.1,

$$\begin{eqnarray}\displaystyle \int _{P\notin K}f(\unicode[STIX]{x1D714})\,dP=-[f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})]_{0_{+}}^{\unicode[STIX]{x1D70B}_{-}}+\frac{L^{2}}{2\unicode[STIX]{x1D70B}}M(f)+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\unicode[STIX]{x1D6FD}_{k}(f)c_{k}^{2}, & & \displaystyle \nonumber\end{eqnarray}$$

where

(13) $$\begin{eqnarray}\displaystyle M(f)=\int _{0}^{\unicode[STIX]{x1D70B}}\frac{f^{\prime }(\unicode[STIX]{x1D714})}{1-\cos \unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}\quad \text{and}\quad \unicode[STIX]{x1D6FD}_{k}(f)=\int _{0}^{\unicode[STIX]{x1D70B}}\frac{f^{\prime }(\unicode[STIX]{x1D714})g_{k}(\unicode[STIX]{x1D714})}{\sin ^{2}\unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}. & & \displaystyle\end{eqnarray}$$

Notice that $M(f)$ and $\unicode[STIX]{x1D6FD}_{k}(f)$ depend only on the function $f$ and not on the shape of the convex set $K$ .

As an application of Proposition 4.2, we can easily prove Crofton’s formula

(14) $$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}-\sin \unicode[STIX]{x1D714})\,dP=-\unicode[STIX]{x1D70B}F+\frac{L^{2}}{2}. & & \displaystyle\end{eqnarray}$$

Indeed, $M(\unicode[STIX]{x1D714}-\sin \unicode[STIX]{x1D714})=\unicode[STIX]{x1D70B}$ and $\unicode[STIX]{x1D6FD}_{k}(\unicode[STIX]{x1D714}-\sin \unicode[STIX]{x1D714})=\int _{0}^{\unicode[STIX]{x1D70B}}g_{k}(x)/(1+\cos (x))\,dx=0$ , as can be easily seen by integrating by parts and using elementary trigonometric identities. Since

$$\begin{eqnarray}-\lim _{\unicode[STIX]{x1D714}\rightarrow \unicode[STIX]{x1D70B}}\,f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})+\lim _{\unicode[STIX]{x1D714}\rightarrow 0}f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})=-\unicode[STIX]{x1D70B}F,\end{eqnarray}$$

the formula follows.

We will now find another expression for the universal factors $g_{k}(\unicode[STIX]{x1D714})/\sin ^{2}(\unicode[STIX]{x1D714})$ appearing in the integral defining the coefficients $\unicode[STIX]{x1D6FD}_{k}(f)$ .

Lemma 4.3. The following identities hold.

$$\begin{eqnarray}\displaystyle \frac{g_{k}(\unicode[STIX]{x1D714})}{\sin ^{2}(\unicode[STIX]{x1D714})} & = & \displaystyle \frac{1}{1-\cos \unicode[STIX]{x1D714}}+2\mathop{\sum }_{j=1,\,\text{odd}}^{k-1}j\cos (j\unicode[STIX]{x1D714})\quad \text{for }k\text{ even},\nonumber\\ \displaystyle \frac{g_{k}(\unicode[STIX]{x1D714})}{\sin ^{2}(\unicode[STIX]{x1D714})} & = & \displaystyle -2\mathop{\sum }_{j=2,\,\text{even}}^{k-1}j\cos (j\unicode[STIX]{x1D714})\qquad \qquad ~\,\quad \text{for }k\text{ odd.}\nonumber\end{eqnarray}$$

Proof. From the expression of the conjugate Dirichlet kernel,

(15) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\sin (j\unicode[STIX]{x1D714})=\frac{1-\cos (k\unicode[STIX]{x1D714})}{2\sin \unicode[STIX]{x1D714}}\quad \text{for }k\text{ even}, & & \displaystyle\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{j=1,\,\text{even}}^{k-1}\sin (j\unicode[STIX]{x1D714})=\frac{\cos (\unicode[STIX]{x1D714})-\cos (k\unicode[STIX]{x1D714})}{2\sin \unicode[STIX]{x1D714}}\quad \text{for }k\text{ odd}. & & \displaystyle \nonumber\end{eqnarray}$$

Differentiating these formulas proves the lemma. ◻

From this Lemma and Proposition 4.2, we get the following proposition.

Proposition 4.4. With the same hypothesis as that of Theorem 3.1,

(16) $$\begin{eqnarray}\displaystyle \int _{P\notin K}f(\unicode[STIX]{x1D714})\,dP & = & \displaystyle -[f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})]_{0}^{\unicode[STIX]{x1D70B}}+\frac{L^{2}}{2\unicode[STIX]{x1D70B}}M(f)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2,\,\text{even}}\bigg(M(f)+2\mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}f^{\prime }(\unicode[STIX]{x1D714})j\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\bigg)c_{k}^{2}\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 3,\,\text{odd}}\bigg(-2\mathop{\sum }_{j=2,\,\text{even}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}f^{\prime }(\unicode[STIX]{x1D714})j\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\bigg)c_{k}^{2},\end{eqnarray}$$

where $M(f)$ is given in (13).

This is a useful formula because it does not involve auxiliary functions and the coefficients of the $c_{k}^{2}$ do not depend on the convex set.

5 Some applications of the integral formulas

5.1 Hurwitz functions

In formula (16), the individual Fourier coefficients, $a_{k},b_{k},$ of the support function of $K$ do not appear. Rather, it is the $c_{k}^{2}=a_{k}^{2}+b_{k}^{2}$ that seems to be of interest. So it will be important to see how these quantities depend on the geometry of $K$ . In fact, Hurwitz, in [Reference Hurwitz6, p. 392], found a formula relating the $c_{k}^{2}$ with the length of $\unicode[STIX]{x2202}K$ and the integral outside $K$ of an elementary function of the visual angle. By a direct application of formula (16), we can prove the following theorem.

Theorem 5.1 (Hurwitz, [Reference Hurwitz6]).

Let $K$ be a compact convex set with boundary of class  $C^{2}$ and length  $L$ . Let $c_{k}^{2}=a_{k}^{2}+b_{k}^{2}$ , where $a_{k},b_{k}$ are the Fourier coefficients of the support function of $K$ . For the functions $f_{m}(\unicode[STIX]{x1D714})$ given by

(17) $$\begin{eqnarray}\displaystyle f_{m}(\unicode[STIX]{x1D714})=-2\sin \unicode[STIX]{x1D714}+\frac{m+1}{m-1}\sin ((m-1)\unicode[STIX]{x1D714})-\frac{m-1}{m+1}\sin ((m+1)\unicode[STIX]{x1D714}), & & \displaystyle\end{eqnarray}$$

we haveFootnote ¹

(18) $$\begin{eqnarray}\displaystyle \int _{P\not \in K}f_{m}(\unicode[STIX]{x1D714})\,dP=L^{2}+(-1)^{m}\,\unicode[STIX]{x1D70B}^{2}(m^{2}-1)c_{m}^{2}\quad m\geqslant 2. & & \displaystyle\end{eqnarray}$$

Proof. In order to apply formula (16), we need to compute $[f_{m}(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})]_{0}^{\unicode[STIX]{x1D70B}}$ , $M(f_{m})$ and the integrals $\int _{0}^{\unicode[STIX]{x1D70B}}f_{m}^{\prime }(\unicode[STIX]{x1D714})\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}$ , for $j$ an integer.

First, we have $[f_{m}(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})]_{0}^{\unicode[STIX]{x1D70B}}=0$ since, by (12),

$$\begin{eqnarray}\lim _{\unicode[STIX]{x1D714}\rightarrow 0}f_{m}(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})=c\lim _{\unicode[STIX]{x1D714}\rightarrow 0}\frac{f_{m}(\unicode[STIX]{x1D714})}{\sin ^{2}\unicode[STIX]{x1D714}}=0.\end{eqnarray}$$

For $M(f_{m})$ , we need the equalities

$$\begin{eqnarray}\displaystyle f_{m}^{\prime }(\unicode[STIX]{x1D714})=2(1-\cos \unicode[STIX]{x1D714})\bigg(1+2\mathop{\sum }_{j=1}^{m-1}j\cos (j\unicode[STIX]{x1D714})+(m-1)\cos (m\unicode[STIX]{x1D714})\bigg)\quad m\geqslant 2, & & \displaystyle \nonumber\end{eqnarray}$$

which are obtained by direct computation using $2\cos (\unicode[STIX]{x1D714})\cos (j\unicode[STIX]{x1D714})=\cos (j+1)\unicode[STIX]{x1D714}+\cos (j-1)\unicode[STIX]{x1D714}.$ Then

$$\begin{eqnarray}\displaystyle M(f_{m}) & = & \displaystyle \int _{0}^{\unicode[STIX]{x1D70B}}\frac{f_{m}^{\prime }(\unicode[STIX]{x1D714})}{1-\cos \unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & = & \displaystyle 2\int _{0}^{\unicode[STIX]{x1D70B}}\biggl(1+2\mathop{\sum }_{j=1}^{m-1}j\cos (j\unicode[STIX]{x1D714})+(m-1)\cos (m\unicode[STIX]{x1D714})\biggr)\,d\unicode[STIX]{x1D714}=2\unicode[STIX]{x1D70B}.\nonumber\end{eqnarray}$$

Finally,

$$\begin{eqnarray}\displaystyle \int _{0}^{\unicode[STIX]{x1D70B}}f_{m}^{\prime }(\sin (j\unicode[STIX]{x1D714}))^{\prime }\,d\unicode[STIX]{x1D714} & = & \displaystyle -\int _{0}^{\unicode[STIX]{x1D70B}}f_{m}^{\prime \prime }\sin (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & = & \displaystyle -\int _{0}^{\unicode[STIX]{x1D70B}}(2\sin \unicode[STIX]{x1D714}+2(m^{2}-1)\cos (m\unicode[STIX]{x1D714})\sin \unicode[STIX]{x1D714})\sin (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FF}_{1,j}-2(m^{2}-1)\int _{0}^{\unicode[STIX]{x1D70B}}\sin \unicode[STIX]{x1D714}\cos (m\unicode[STIX]{x1D714})\sin (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FF}_{1,j}-2(m^{2}-1)\frac{\unicode[STIX]{x1D70B}}{4}(\unicode[STIX]{x1D6FF}_{j,m+1}-\unicode[STIX]{x1D6FF}_{j,m-1}),\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}_{ij}=0$ for $i\neq j$ and $\unicode[STIX]{x1D6FF}_{ii}=1$ .

Hence, if $k$ is even,

$$\begin{eqnarray}\displaystyle 2\mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}f_{k}^{\prime }(\sin (j\unicode[STIX]{x1D714}))^{\prime }\,d\unicode[STIX]{x1D714} & = & \displaystyle -2\unicode[STIX]{x1D70B}-(m^{2}-1)\unicode[STIX]{x1D70B}(-\unicode[STIX]{x1D6FF}_{k-1,m-1})\nonumber\\ \displaystyle & = & \displaystyle -2\unicode[STIX]{x1D70B}+(k^{2}-1)\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FF}_{k,m}\nonumber\end{eqnarray}$$

since

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{j+2,m+1}-\unicode[STIX]{x1D6FF}_{j,k-1}=0,\quad j=1,\ldots ,k-1.\end{eqnarray}$$

And, if $k$ is odd,

$$\begin{eqnarray}\displaystyle 2\mathop{\sum }_{j=2,\,\text{even}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}f_{m}^{\prime }(\sin (j\unicode[STIX]{x1D714}))^{\prime }\,d\unicode[STIX]{x1D714}=-(m^{2}-1)\unicode[STIX]{x1D70B}(-\unicode[STIX]{x1D6FF}_{k-1,m-1})=(m^{2}-1)\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FF}_{k,m}. & & \displaystyle \nonumber\end{eqnarray}$$

Substituting in (16) gives

$$\begin{eqnarray}\displaystyle \int _{P\notin K}f_{m}(\unicode[STIX]{x1D714})\,dP & = & \displaystyle L^{2}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2,\,\text{even}}c_{k}^{2}(2\unicode[STIX]{x1D70B}-2\unicode[STIX]{x1D70B}+(k^{2}-1)\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FF}_{k,m})\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 3,\,\text{odd}}c_{k}^{2}(-(k^{2}-1)\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FF}_{k,m})\nonumber\\ \displaystyle & = & \displaystyle L^{2}+(-1)^{m}\unicode[STIX]{x1D70B}^{2}c_{m}^{2}(m^{2}-1).\nonumber\end{eqnarray}$$

The proof is now complete. ◻

The theorem shows that the quantities $c_{m}^{2}$ are invariant with respect to Euclidean motions of $K$ .

Notice that the functions $f_{m}$ are related to the functions $g_{m}$ introduced in (10) by

$$\begin{eqnarray}g_{m}(\unicode[STIX]{x1D714})=1+\frac{(-1)^{m}}{2}(f_{m}^{\prime }(\unicode[STIX]{x1D714})+2\cos (\unicode[STIX]{x1D714})).\end{eqnarray}$$

5.2 Masotti integral formula

In [Reference Masotti7], Masotti gives without proof a Crofton-type formula evaluating $\int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP$ . Here, we will derive Masotti’s formula from (16).

Consider the function $f(\unicode[STIX]{x1D714})=\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714}$ , which clearly satisfies the hypothesis of Theorem 3.1, and let us compute $[f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})]_{0}^{\unicode[STIX]{x1D70B}}$ , $M(f)$ and the integrals $\int _{0}^{\unicode[STIX]{x1D70B}}f^{\prime }(\unicode[STIX]{x1D714})\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}$ , for $j$ an integer.

We have

$$\begin{eqnarray}\lim _{\unicode[STIX]{x1D714}\rightarrow \unicode[STIX]{x1D70B}}f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})=\unicode[STIX]{x1D70B}^{2}F\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \lim _{\unicode[STIX]{x1D714}\rightarrow 0}f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})=\lim _{\unicode[STIX]{x1D714}\rightarrow 0}\frac{\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714}}{\sin ^{2}\unicode[STIX]{x1D714}}\bigg(\frac{L^{2}}{2\unicode[STIX]{x1D70B}}(1+\cos \unicode[STIX]{x1D714})+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}c_{k}^{2}g_{k}(\unicode[STIX]{x1D714})\bigg)=0, & & \displaystyle \nonumber\end{eqnarray}$$

since the term inside the parentheses is bounded. Hence, $[f(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})]_{0}^{\unicode[STIX]{x1D70B}}=\unicode[STIX]{x1D70B}^{2}F.$

On the other hand,

$$\begin{eqnarray}\displaystyle M(f) & = & \displaystyle \int _{0}^{\unicode[STIX]{x1D70B}}\frac{f^{\prime }(\unicode[STIX]{x1D714})}{1-\cos \unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}=\int _{0}^{\unicode[STIX]{x1D70B}}\frac{2\unicode[STIX]{x1D714}-\sin (2\unicode[STIX]{x1D714})}{1-\cos \unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & = & \displaystyle [\sin ^{2}(\unicode[STIX]{x1D714}/2)-3\cos ^{2}(\unicode[STIX]{x1D714}/2)-2\unicode[STIX]{x1D714}\cot (\unicode[STIX]{x1D714}/2)]_{0}^{\unicode[STIX]{x1D70B}}=8.\nonumber\end{eqnarray}$$

Moreover, for $j\neq 2$ ,

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\unicode[STIX]{x1D70B}}f^{\prime }(\unicode[STIX]{x1D714})\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{0}^{\unicode[STIX]{x1D70B}}(2\unicode[STIX]{x1D714}-\sin (2\unicode[STIX]{x1D714}))\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & & \displaystyle \quad =\bigg[\frac{2}{j^{2}}(\cos (j\unicode[STIX]{x1D714})+j\unicode[STIX]{x1D714}\sin (j\unicode[STIX]{x1D714}))-\frac{\cos ((j-2)\unicode[STIX]{x1D714})}{2(j-2)}+\frac{\cos ((j+2)\unicode[STIX]{x1D714})}{2(j+2)}\bigg]_{0}^{\unicode[STIX]{x1D70B}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{8(1-(-1)^{j})}{j^{2}(j^{2}-4)},\nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\int _{0}^{\unicode[STIX]{x1D70B}}(2\unicode[STIX]{x1D714}-\sin (2\unicode[STIX]{x1D714}))\cos (2\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}=0.\end{eqnarray}$$

It follows that

$$\begin{eqnarray}\mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}f^{\prime }(\unicode[STIX]{x1D714})\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}=\mathop{\sum }_{j=1,\,\text{odd}}\frac{16}{j(j^{2}-4)}=\frac{4k^{2}}{1-k^{2}}.\end{eqnarray}$$

Summing up, we obtain the following theorem.

Theorem 5.2 (Masotti, [Reference Masotti7]).

Let $K$ be a compact convex set of area $F$ with boundary of class $C^{2}$ and length $L$ . Let $c_{k}^{2}=a_{k}^{2}+b_{k}^{2}$ , where $a_{k}$ , $b_{k}$ are the Fourier coefficients of the support function of $K$ . Then

(19) $$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP=-\unicode[STIX]{x1D70B}^{2}F+\frac{4L^{2}}{\unicode[STIX]{x1D70B}}+8\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2,\,\text{even}}\bigg(\frac{1}{1-k^{2}}\bigg)c_{k}^{2}. & & \displaystyle\end{eqnarray}$$

Moreover, the equality

$$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP=-\unicode[STIX]{x1D70B}^{2}F+\frac{4L^{2}}{\unicode[STIX]{x1D70B}} & & \displaystyle \nonumber\end{eqnarray}$$

holds if and only if the compact convex set $K$ has constant width.

In addition to the obvious inequality

$$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP\leqslant -\unicode[STIX]{x1D70B}^{2}F+\frac{4L^{2}}{\unicode[STIX]{x1D70B}} & & \displaystyle \nonumber\end{eqnarray}$$

that follows from (19), Santaló states, in [Reference Santaló9, I.4.5], the lower bound

(20) $$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP\geqslant (16-\unicode[STIX]{x1D70B}^{2})F, & & \displaystyle\end{eqnarray}$$

with equality only for circles. We will now improve this last inequality.

Theorem 5.3. Under the same hypothesis as that in Theorem 5.2,

(21) $$\begin{eqnarray}\displaystyle -\unicode[STIX]{x1D70B}^{2}F+\frac{4L^{2}}{\unicode[STIX]{x1D70B}}-\frac{4}{3}\biggl(H-\frac{L^{2}}{\unicode[STIX]{x1D70B}}\biggr)\leqslant \int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP\leqslant -\unicode[STIX]{x1D70B}^{2}F+\frac{4L^{2}}{\unicode[STIX]{x1D70B}}, & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $H=\lim _{\unicode[STIX]{x1D714}\rightarrow 0}F(\unicode[STIX]{x1D714})\sin ^{2}\unicode[STIX]{x1D714}$ is given in (12). Equality in the left-hand side holds if and only if $\unicode[STIX]{x2202}K$ is a circle or a curve parallel to an astroid.

Proof. For the left-hand side, just write

$$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP & {\geqslant} & \displaystyle -\unicode[STIX]{x1D70B}^{2}F+\frac{4L^{2}}{\unicode[STIX]{x1D70B}}-\frac{8\unicode[STIX]{x1D70B}}{3}\mathop{\sum }_{k\geqslant 2,\,\text{even}}c_{k}^{2}\nonumber\\ \displaystyle & = & \displaystyle -\unicode[STIX]{x1D70B}^{2}F+\frac{4L^{2}}{\unicode[STIX]{x1D70B}}-\frac{4}{3}\biggl(H-\frac{L^{2}}{\unicode[STIX]{x1D70B}}\biggr).\nonumber\end{eqnarray}$$

Equality in the left-hand side holds if and only if the support function of $K$ with respect to the Steiner point is of the form $p(\unicode[STIX]{x1D711})=a_{0}+a_{2}\cos (2\unicode[STIX]{x1D711})+b_{2}\sin (2\unicode[STIX]{x1D711})$ . This means that $\unicode[STIX]{x2202}K$ is a circle or a curve parallel to an astroid (see, for instance, [Reference Cufí and Reventós2]).◻

In terms of the area $A$ of the pedal curve of $K$ with respect to the Steiner point, we get the following corollary.

Corollary 5.1. Under the same hypothesis as that in Theorem 5.2,

$$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{2}-\sin ^{2}\unicode[STIX]{x1D714})\,dP\geqslant (16-\unicode[STIX]{x1D70B}^{2})F+\frac{32}{3}(A-F). & & \displaystyle \nonumber\end{eqnarray}$$

Equality holds if and only if $\unicode[STIX]{x2202}K$ is a circle or a curve parallel to an astroid.

Proof. This is a simple consequence of (21) and the two inequalities $H-L^{2}/\unicode[STIX]{x1D70B}\leqslant A-F$ and $\unicode[STIX]{x1D6E5}\geqslant 3\unicode[STIX]{x1D70B}(A-F)$ (see [Reference Cufí and Reventós2]).◻

The above argument shows that (21) improves Santaló’s inequality (20).

6 Integral powers of the sine of the visual angle

In (8), we have seen that

$$\begin{eqnarray}\int _{P\notin K}\sin ^{3}\unicode[STIX]{x1D714}\,dP=\frac{3}{4}L^{2}+\frac{9}{4}\unicode[STIX]{x1D70B}^{2}c_{2}^{2}.\end{eqnarray}$$

In this section, we compute the integral of $\sin ^{m}(\unicode[STIX]{x1D714})$ for integer values of $m$ greater than three. We have that $[\sin ^{m}(\unicode[STIX]{x1D714})F(\unicode[STIX]{x1D714})]_{0}^{\unicode[STIX]{x1D70B}}=0,$ so

$$\begin{eqnarray}\int _{P\notin K}\sin ^{m}(\unicode[STIX]{x1D714})\,dP=M(\sin ^{m}(\unicode[STIX]{x1D714}))\frac{L^{2}}{2\unicode[STIX]{x1D70B}}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\unicode[STIX]{x1D6FD}_{k}(\sin ^{m}\unicode[STIX]{x1D714})c_{k}^{2},\end{eqnarray}$$

where $M$ and $\unicode[STIX]{x1D6FD}_{k}$ are as given in (13).

As, for $j$ even, $\cos (\unicode[STIX]{x1D714})\cos (j\unicode[STIX]{x1D714})$ is an odd function with respect $\unicode[STIX]{x1D70B}/2$ , it can be seen from (16) that $\unicode[STIX]{x1D6FD}_{k}(\sin ^{m}\unicode[STIX]{x1D714})=0$ for every $k$ odd.

When $K$ is a convex set of constant width, $c_{k}=0$ for every even value of $k$ , and we get

$$\begin{eqnarray}\int _{P\notin K}\sin ^{m}(\unicode[STIX]{x1D714})\,dP=M(\sin ^{m}(\unicode[STIX]{x1D714}))\frac{L^{2}}{2\unicode[STIX]{x1D70B}}.\end{eqnarray}$$

Since $M(\sin ^{m}\unicode[STIX]{x1D714})$ does not depend on the convex set $K$ , we can compute this constant by applying the above formula to the unit circle centered at the origin. If $r$ is the distance to the origin of the point $P$ , we have $\sin (\unicode[STIX]{x1D714})=2\sin (\unicode[STIX]{x1D714}/2)\cos (\unicode[STIX]{x1D714}/2)=2\sqrt{r^{2}-1}/r^{2},$ and so

$$\begin{eqnarray}\displaystyle \int _{P\not \in K}\sin ^{m}(\unicode[STIX]{x1D714})\,dP & = & \displaystyle \int _{0}^{2\unicode[STIX]{x1D70B}}\!\!\int _{1}^{\infty }\sin ^{m}(\unicode[STIX]{x1D714})r\,dr\,d\unicode[STIX]{x1D703}=2\unicode[STIX]{x1D70B}\,2^{m}\int _{1}^{\infty }\bigg(\frac{\sqrt{r^{2}-1}}{r^{2}}\bigg)^{m}r\,dr\nonumber\\ \displaystyle & = & \displaystyle 2^{m}\unicode[STIX]{x1D70B}B\bigg(\frac{m}{2}+1,\frac{m}{2}-1\bigg),\nonumber\end{eqnarray}$$

where $B(x,y)$ is the beta function. Hence

$$\begin{eqnarray}M(\sin ^{m}\unicode[STIX]{x1D714})=2^{m-1}B\bigg(\frac{m}{2}+1,\frac{m}{2}-1\bigg)=2^{m-1}\frac{\unicode[STIX]{x1D6E4}(m/2+1)\unicode[STIX]{x1D6E4}(m/2-1)}{(m-1)!}.\end{eqnarray}$$

Using the relation

$$\begin{eqnarray}\unicode[STIX]{x1D6E4}(z)\unicode[STIX]{x1D6E4}\big(z+{\textstyle \frac{1}{2}}\big)=2^{1-2z}\sqrt{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6E4}(2z)\end{eqnarray}$$

we obtain

$$\begin{eqnarray}\unicode[STIX]{x1D6E4}\biggl(\frac{m}{2}+1\biggr)\unicode[STIX]{x1D6E4}\biggl(\frac{m}{2}-1\biggr)=\frac{\unicode[STIX]{x1D70B}m!(m-1)!}{2^{2m-2}(m-2)\unicode[STIX]{x1D6E4}((m+1)/2)^{2}},\end{eqnarray}$$

and hence

(22) $$\begin{eqnarray}\displaystyle M(\sin ^{m}(\unicode[STIX]{x1D714}))=\frac{\unicode[STIX]{x1D70B}\,m!}{2^{m-1}(m-2)\unicode[STIX]{x1D6E4}((m+1)/2)^{2}}. & & \displaystyle\end{eqnarray}$$

Notice that $M(\sin ^{m}\unicode[STIX]{x1D714})$ decreases with $m$ , its maximum value $3\unicode[STIX]{x1D70B}/2$ is attained for $m=3$ and it behaves as $1/\sqrt{m}$ when $m$ tends to infinity.

We have proved the following proposition.

Proposition 6.1. Let $K$ be a compact convex of constant width with boundary of class $C^{2}$ and length $L$ . Then

$$\begin{eqnarray}\displaystyle \int _{P\notin K}\sin ^{m}\unicode[STIX]{x1D714}\,dP=\frac{\unicode[STIX]{x1D70B}\,m!}{2^{m-1}(m-2)\unicode[STIX]{x1D6E4}((m+1)/2)^{2}}\;\frac{L^{2}}{2\unicode[STIX]{x1D70B}}. & & \displaystyle \nonumber\end{eqnarray}$$

For general convex sets, we have the following theorem.

Theorem 6.1. Let $K$ be a compact convex set with boundary of class $C^{2}$ and length $L$ . Write $c_{k}^{2}=a_{k}^{2}+b_{k}^{2}$ , where $a_{k}$ , $b_{k}$ are the Fourier coefficients of the support function of $K$ . Then

(23) $$\begin{eqnarray}\displaystyle \int _{P\notin K}\sin ^{m}\unicode[STIX]{x1D714}\,dP & = & \displaystyle M(\sin ^{m}\unicode[STIX]{x1D714})\frac{L^{2}}{2\unicode[STIX]{x1D70B}}+\frac{m!\unicode[STIX]{x1D70B}^{2}}{2^{m-1}(m-2)}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{k\geqslant 2,\,\text{even}}\frac{(-1)^{k/2+1}(k^{2}-1)}{\unicode[STIX]{x1D6E4}((m+1+k)/2)\unicode[STIX]{x1D6E4}((m+1-k)/2)}c_{k}^{2},\end{eqnarray}$$

where $M(\sin ^{m}(\unicode[STIX]{x1D714}))$ is given in (22). For $m$ odd, the index $k$ in the sum runs only from $2$ to $m-1$ .

Proof. From (16), it is clear that we need to compute

$$\begin{eqnarray}M(\sin ^{m}\unicode[STIX]{x1D714})+2\mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}(\sin ^{m}\unicode[STIX]{x1D714})^{\prime }j\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\quad \text{for }k\;\text{even}.\end{eqnarray}$$

Using (15) and integrating by parts gives

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}(\sin ^{m}\unicode[STIX]{x1D714})^{\prime }(\sin (j\unicode[STIX]{x1D714}))^{\prime }\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & & \displaystyle \quad =-\int _{0}^{\unicode[STIX]{x1D70B}}(m(m-1)\sin ^{m-2}\unicode[STIX]{x1D714}-m^{2}\sin ^{m}\unicode[STIX]{x1D714})\frac{1-\cos (k\unicode[STIX]{x1D714})}{2\sin \unicode[STIX]{x1D714}}\,d\unicode[STIX]{x1D714}.\nonumber\end{eqnarray}$$

Denoting $I_{m,k}=\int _{0}^{\unicode[STIX]{x1D70B}}\sin ^{m}\unicode[STIX]{x1D714}\cos (k\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}$ ,

(24) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}(\sin ^{m}\unicode[STIX]{x1D714})^{\prime }(\sin (j\unicode[STIX]{x1D714}))^{\prime }\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & & \displaystyle \quad =-\frac{m(m-1)}{2}I_{m-3,0}+\frac{m^{2}}{2}I_{m-1,0}+\frac{m(m-1)}{2}I_{m-3,k}-\frac{m^{2}}{2}I_{m-1,k}.\end{eqnarray}$$

By induction on $m$ and using known relations of the gamma function, it can be seen that

$$\begin{eqnarray}I_{m,k}=(-1)^{k/2}\frac{2^{-m}m!\unicode[STIX]{x1D70B}}{\unicode[STIX]{x1D6E4}(1+(m-k)/2)\unicode[STIX]{x1D6E4}(1+(m+k)/2)},\end{eqnarray}$$

(see, for instance, [Reference Gradshteyn and Ryzhik4, p. 372]). Performing the operation on the right-hand side of (24) with these values of $I_{m,k}$ , we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}(\sin ^{m}\unicode[STIX]{x1D714})^{\prime }(\sin (j\unicode[STIX]{x1D714}))^{\prime }\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & & \displaystyle \quad =-\frac{1}{2}M(\sin ^{m}\unicode[STIX]{x1D714})+\frac{m!\unicode[STIX]{x1D70B}}{2^{m}(m-2)}\frac{(-1)^{k/2+1}(k^{2}-1)}{\unicode[STIX]{x1D6E4}((m+1+k)/2)\unicode[STIX]{x1D6E4}((m+1-k)/2)}.\nonumber\end{eqnarray}$$

From this, formula (23) follows.

When $m$ is odd and $k>m$ , we have that $m+1-k$ is an even non-positive integer and hence $\unicode[STIX]{x1D6E4}((m+1-k)/2)=\infty$ . From this remark, the last assertion of the theorem is proved.◻

For instance, for the special cases $m=3$ , $4$ and $5$ , we obtain the equalities

$$\begin{eqnarray}\displaystyle \int _{P\notin K}\sin ^{3}\unicode[STIX]{x1D714}\,dP & = & \displaystyle \frac{3}{4}L^{2}+\frac{9}{4}\unicode[STIX]{x1D70B}^{2}c_{2}^{2},\nonumber\\ \displaystyle \int _{P\notin K}\sin ^{4}\unicode[STIX]{x1D714}\,dP & = & \displaystyle \frac{4}{3\unicode[STIX]{x1D70B}}L^{2}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k=2,\text{even}}^{\infty }\frac{24}{9-k^{2}}\,c_{k}^{2},\nonumber\\ \displaystyle \int _{P\notin K}\sin ^{5}\unicode[STIX]{x1D714}\,dP & = & \displaystyle \frac{5}{16}L^{2}+\frac{5\unicode[STIX]{x1D70B}^{2}}{4}c_{2}^{2}-\frac{25\unicode[STIX]{x1D70B}^{2}}{16}c_{4}^{2}.\nonumber\end{eqnarray}$$

To finish, we make the following remarks

1. (a) If $m=2r$ , the coefficient of $c_{k}^{2}$ in (23) for $k\leqslant r$ is positive if and only if $k/2$ is odd. For $k>r$ , this coefficient is positive if and only if $r$ is odd.

2. (b) If $m=2r-1$ , the coefficient of $c_{k}^{2}$ vanishes for $k>m$ .

In [Reference Hurwitz6, p. 392], Hurwitz computed the integral of $\sin ^{3}(\unicode[STIX]{x1D714})$ and the integrals of the functions  $f_{m}(\unicode[STIX]{x1D714})$ given in (17) without any relationship between them. We will show now that the integrals of the powers of the sine of the visual angle are a linear combination of the integrals of the functions $f_{m}$ .

Proposition 6.2. For a compact convex set $K$ with boundary of class $C^{2}$ and  $m\geqslant 3$ ,

$$\begin{eqnarray}\displaystyle \int _{P\notin K}\sin ^{m}(\unicode[STIX]{x1D714})\,dP & = & \displaystyle \frac{m!}{2^{m-1}(m-2)}\mathop{\sum }_{p=1}^{\infty }\frac{(-1)^{p+1}}{\unicode[STIX]{x1D6E4}((m+1)/2+p)\unicode[STIX]{x1D6E4}((m+1)/2-p)}\nonumber\\ \displaystyle & & \displaystyle \cdot \int _{P\notin K}f_{2p}(\unicode[STIX]{x1D714})\,dP,\nonumber\end{eqnarray}$$

where the functions $f_{2p}(\unicode[STIX]{x1D714})$ are given in (17).

Proof. Substituting in (23) the value of $c_{k}^{2}$ given by (18) gives

(25) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{P\notin K}\sin ^{m}\unicode[STIX]{x1D714}\,dP\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{m!}{2^{m}(m-2)}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\bigg(\frac{1}{\unicode[STIX]{x1D6E4}((m+1)/2)^{2}}-2\mathop{\sum }_{p=1}^{\infty }\frac{(-1)^{p+1}}{\unicode[STIX]{x1D6E4}((m+1)/2+p)\unicode[STIX]{x1D6E4}((m+1)/2-p)}\bigg)L^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{m!}{2^{m-1}(m-2)}\mathop{\sum }_{p=1}^{\infty }\frac{(-1)^{p+1}}{\unicode[STIX]{x1D6E4}((m+1)/2+p)\unicode[STIX]{x1D6E4}((m+1)/2-p)}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \int _{P\notin K}f_{2p}(\unicode[STIX]{x1D714})\,dP.\end{eqnarray}$$

Using the standard notation for hypergeometric series,

$$\begin{eqnarray}\displaystyle & & \displaystyle 2\mathop{\sum }_{p=1}^{\infty }\frac{(-1)^{p+1}}{\unicode[STIX]{x1D6E4}((m+1)/2+p)\unicode[STIX]{x1D6E4}((m+1)/2-p)}\nonumber\\ \displaystyle & & \displaystyle \quad =2\frac{\text{}_{2}F_{1}((3-m)/2,1;(3+m)/2;1)}{\unicode[STIX]{x1D6E4}((m+1)/2+1)\unicode[STIX]{x1D6E4}((m+1)/2-1)},\nonumber\end{eqnarray}$$

and by the Gauss summation formula (see [Reference Gauss3, Vol. III, p. 147]) we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle 2\frac{\text{}_{2}F_{1}((3-m)/2,1;(3+m)/2;1)}{\unicode[STIX]{x1D6E4}((m+1)/2+1)\unicode[STIX]{x1D6E4}((m+1)/2-1)}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{2}{\unicode[STIX]{x1D6E4}((m+1)/2+1)\unicode[STIX]{x1D6E4}((m+1)/2-1)}\cdot \frac{\unicode[STIX]{x1D6E4}(m-1)\unicode[STIX]{x1D6E4}((m+3)/2)}{\unicode[STIX]{x1D6E4}(m)\unicode[STIX]{x1D6E4}((m+1)/2)}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{\unicode[STIX]{x1D6E4}((m+1)/2)^{2}}.\nonumber\end{eqnarray}$$

Hence the coefficient of $L^{2}$ in (25) vanishes.◻

7 Extension of the Crofton and Masotti formulas and related inequalities

In this section, we consider the integral

$$\begin{eqnarray}\int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP,\end{eqnarray}$$

where $\unicode[STIX]{x1D714}$ is the visual angle of the convex set $K$ from the point $P$ . For $m=1$ and $m=2$ , these are the integrals appearing in Crofton’s formula (14) and in the Masotti integral formula (19), respectively.

For the general case, we have, by Proposition 4.2,

(26) $$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP=-\unicode[STIX]{x1D70B}^{m}F+M_{m}\frac{L^{2}}{2\unicode[STIX]{x1D70B}}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k\geqslant 2}\unicode[STIX]{x1D6FD}_{k}c_{k}^{2}, & & \displaystyle\end{eqnarray}$$

where $M_{m}=M(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})$ and $\unicode[STIX]{x1D6FD}_{k}=\unicode[STIX]{x1D6FD}_{k}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})$ are given in (13). The quantities $M_{m}$ can be explicitly computed. In fact, $M(\sin ^{m}\unicode[STIX]{x1D714})$ is given in (22) and

$$\begin{eqnarray}M(\unicode[STIX]{x1D714}^{m})=2m(m-1)\unicode[STIX]{x1D70B}^{m-2}\bigg(\frac{1}{m-2}+\mathop{\sum }_{k=1}^{\infty }\frac{(-1)^{k}\unicode[STIX]{x1D70B}^{2k}B_{2k}}{(m-2+2k)(2k)!}\bigg),\end{eqnarray}$$

where $B_{2k}$ are the Bernoulli numbers (see [Reference Gradshteyn and Ryzhik4, p. 189]). As the parenthesized expression tends to zero as $1/m^{2}$ when $m$ tends to infinity, we see that $M(\unicode[STIX]{x1D714}^{m})$ behaves like $e^{m}$ and therefore $M_{m}$ grows exponentially with $m$ .

For $\unicode[STIX]{x1D6FD}_{k}$ , recall that, by Proposition 4.4,

(27) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FD}_{k}=\left\{\begin{array}{@{}ll@{}}\displaystyle M_{m}+2\mathop{\sum }_{j=1,\,\text{odd}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})^{\prime }j\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\quad & \text{for }k\;\text{even},\\ \displaystyle -2\mathop{\sum }_{j=1,\,\text{even}}^{k-1}\int _{0}^{\unicode[STIX]{x1D70B}}(\unicode[STIX]{x1D714}^{m})^{\prime }j\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\quad & \text{for }k\;\text{odd}.\end{array}\right. & & \displaystyle\end{eqnarray}$$

Although the integrals appearing in the expression of the $\unicode[STIX]{x1D6FD}_{k}$ can be explicitly computed, they are not easily handled.

For instance, in the case $m=3$ , it can be seen that

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{3}-\sin ^{3}\unicode[STIX]{x1D714})\,dP\nonumber\\ \displaystyle & & \displaystyle \quad =-\unicode[STIX]{x1D70B}^{3}F+\bigg(12\unicode[STIX]{x1D70B}\ln (2)-\frac{3\unicode[STIX]{x1D70B}}{2}\bigg)\frac{L^{2}}{2\unicode[STIX]{x1D70B}}+12\unicode[STIX]{x1D70B}^{2}\biggl(\ln (2)-\frac{19}{16}\biggr)c_{2}^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,6\unicode[STIX]{x1D70B}^{2}\mathop{\sum }_{k\geqslant 3}\bigg(\unicode[STIX]{x1D6F9}\bigg(\frac{k+1}{2}\bigg)+\unicode[STIX]{x1D6FE}\bigg)c_{k}^{2},\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D6F9}(x)$ is the digamma function, $\unicode[STIX]{x1D6F9}(x)=(\ln \unicode[STIX]{x1D6E4}(x))^{\prime }$ , and $\unicode[STIX]{x1D6FE}$ is the Euler–Mascheroni constant.

7.1 Upper bounds

We now obtain an upper bound for $\int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP$ . For $m=3$ , since $\unicode[STIX]{x1D6F9}(x)>0$ for $x\geqslant 2$ ,

(28) $$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{3}-\sin ^{3}\unicode[STIX]{x1D714})\,dP\leqslant -\unicode[STIX]{x1D70B}^{3}F+\biggl(12\unicode[STIX]{x1D70B}\ln (2)-\frac{3\unicode[STIX]{x1D70B}}{2}\biggr)\frac{L^{2}}{2\unicode[STIX]{x1D70B}}. & & \displaystyle\end{eqnarray}$$

For the general case, we obtain the following result.

Theorem 7.1. Let $K$ be a compact convex set with boundary of class  ${\mathcal{C}}^{2}$ , area  $F$ and length of the boundary $L$ , and let $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}(P)$ be the visual angle from the point  $P$ . Then

$$\begin{eqnarray}\int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP\leqslant -\unicode[STIX]{x1D70B}^{m}F+M_{m}\frac{L^{2}}{2\unicode[STIX]{x1D70B}}\quad \text{for }m\geqslant 1,\end{eqnarray}$$

where $M_{m}=\int _{0}^{\unicode[STIX]{x1D70B}}((\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})^{\prime }/(1-\cos \unicode[STIX]{x1D714}))\,d\unicode[STIX]{x1D714}$ . Equality holds only for circles.

Remark 7.1. Since $M_{1}=\unicode[STIX]{x1D70B}$ , $M_{2}=8$ and $M_{3}=(12\ln (2)-3\unicode[STIX]{x1D70B}/2)$ , this result agrees with Crofton’s formula (14) for $m=1$ , and it is a generalization of the upper bounds for Masotti’s integral given in (21) for $m=2$ and of the upper bound given in (28) for  $m=3$ .

Proof. We shall see that $\unicode[STIX]{x1D6FD}_{k}\leqslant 0$ , for $k\geqslant 2$ . Writing $V_{r,j}=\int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{r}\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}$ , the following recurrence formula can be checked:

$$\begin{eqnarray}V_{r,j}=\frac{r}{j^{2}}((-1)^{j}\unicode[STIX]{x1D70B}^{r-1}-(r-1)V_{r-2,j}).\end{eqnarray}$$

This gives $V_{r,j}\geqslant 0$ for $j$ even and $V_{r,j}\leqslant 0$ for $j$ odd. As, for $k$ odd, we have $\unicode[STIX]{x1D6FD}_{k}=-2m\sum _{j=2,\,\text{even}}^{k-1}jV_{m-1,j}$ , it follows that $\unicode[STIX]{x1D6FD}_{k}\leqslant 0$ for $k$ odd.

For $k$ even, by integrating by parts, we get

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{k}-\unicode[STIX]{x1D6FD}_{k+2}=2\int _{0}^{\unicode[STIX]{x1D70B}}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})^{\prime \prime }\sin ((k+1)\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}.\end{eqnarray}$$

It can be seen that, for $m\geqslant 4$ , the function $\unicode[STIX]{x1D713}(\unicode[STIX]{x1D714}):=(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})^{\prime \prime }$ is non-negative and increasing on the interval $[0,\unicode[STIX]{x1D70B}].$ Then, partitioning $[0,\unicode[STIX]{x1D70B}]$ by $t_{j}=j\unicode[STIX]{x1D70B}/(k+1)$ with $j=0,\ldots ,k+1$ , we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D713}(\unicode[STIX]{x1D714})\sin (k+1)\unicode[STIX]{x1D714}\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j=0}^{k}\int _{t_{j}}^{t_{j+1}}\unicode[STIX]{x1D713}(\unicode[STIX]{x1D714})\sin (k+1)\unicode[STIX]{x1D714}\,d\unicode[STIX]{x1D714}>\mathop{\sum }_{j=1}^{k}\int _{t_{j}}^{t_{j+1}}\unicode[STIX]{x1D713}(\unicode[STIX]{x1D714})\sin (k+1)\unicode[STIX]{x1D714}\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j=2,\,\text{even}}^{k}\bigg(\int _{t_{j}}^{t_{j+1}}\unicode[STIX]{x1D713}(\unicode[STIX]{x1D714})\sin (k+1)\unicode[STIX]{x1D714}\,d\unicode[STIX]{x1D714}-\int _{t_{j-1}}^{t_{j}}\unicode[STIX]{x1D713}(\unicode[STIX]{x1D714})|\sin (k+1)\unicode[STIX]{x1D714}|\,d\unicode[STIX]{x1D714}\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j=2,\,\text{even}}^{k}\int _{t_{j}}^{t_{j+1}}\bigg(\unicode[STIX]{x1D713}(\unicode[STIX]{x1D714})-\unicode[STIX]{x1D713}\biggl(\unicode[STIX]{x1D714}-\frac{\unicode[STIX]{x1D70B}}{k+1}\biggr)\bigg)\sin (k+1)\unicode[STIX]{x1D714}\,d\unicode[STIX]{x1D714}>0,\nonumber\end{eqnarray}$$

so that $\unicode[STIX]{x1D6FD}_{k}-\unicode[STIX]{x1D6FD}_{k+2}>0$ for all $k\geqslant 2$ . Thus, in order to see that $\unicode[STIX]{x1D6FD}_{k}\leqslant 0$ , it is enough to show that $\unicode[STIX]{x1D6FD}_{2}\leqslant 0$ , with

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{2}=\int _{0}^{\unicode[STIX]{x1D70B}}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})^{\prime }\bigg(\frac{1}{1-\cos \unicode[STIX]{x1D714}}+2\cos \unicode[STIX]{x1D714}\bigg)\,d\unicode[STIX]{x1D714}.\end{eqnarray}$$

For simplicity in the exposition, we write $h_{m}(\unicode[STIX]{x1D714})=\unicode[STIX]{x1D714}^{m-1}-\sin ^{m-1}\unicode[STIX]{x1D714}\cos \unicode[STIX]{x1D714}$ and $g(\unicode[STIX]{x1D714})=1/(1-\cos \unicode[STIX]{x1D714})+2\cos \unicode[STIX]{x1D714}.$ The function $g$ has exactly one root in the interval $[0,\unicode[STIX]{x1D70B}]$ , $\unicode[STIX]{x1D701}=\arccos ((1-\sqrt{3})/2).$

Notice that

$$\begin{eqnarray}h_{m}(\unicode[STIX]{x1D714})g(\unicode[STIX]{x1D714})=\unicode[STIX]{x1D714}^{m-1}g(\unicode[STIX]{x1D714})-\sin ^{m-1}\unicode[STIX]{x1D714}\cos \unicode[STIX]{x1D714}g(\unicode[STIX]{x1D714})\leqslant \unicode[STIX]{x1D714}^{m-1}g(\unicode[STIX]{x1D714})+1\end{eqnarray}$$

for $0\leqslant \unicode[STIX]{x1D714}\leqslant \unicode[STIX]{x1D70B}.$ Then, in order to see that $\unicode[STIX]{x1D6FD}_{2}<0$ , it suffices to prove that $\int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\leqslant -\unicode[STIX]{x1D70B}$ . First, we see that the sequence $\int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}$ decreases with $m$ , that is,

$$\begin{eqnarray}\int _{0}^{\unicode[STIX]{x1D70B}}(\unicode[STIX]{x1D714}^{m}-\unicode[STIX]{x1D714}^{m-1})g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}=\int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}(\unicode[STIX]{x1D714}-1)g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\leqslant 0.\end{eqnarray}$$

The integrand is positive if $\unicode[STIX]{x1D714}\in [1,\unicode[STIX]{x1D701}]$ and negative otherwise. Therefore, if we see that

$$\begin{eqnarray}\int _{1}^{\unicode[STIX]{x1D701}}\unicode[STIX]{x1D714}^{m}(\unicode[STIX]{x1D714}-1)g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}<\int _{\unicode[STIX]{x1D701}}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}(\unicode[STIX]{x1D714}-1)|g(\unicode[STIX]{x1D714})|\,d\unicode[STIX]{x1D714},\end{eqnarray}$$

the proof is complete.

On the one hand, $\int _{1}^{\unicode[STIX]{x1D701}}\unicode[STIX]{x1D714}^{m-1}(\unicode[STIX]{x1D714}-1)g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}<\unicode[STIX]{x1D701}^{m-1}(\unicode[STIX]{x1D701}-1)\int _{1}^{\unicode[STIX]{x1D701}}g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}$ . On the other hand,

$$\begin{eqnarray}\displaystyle \int _{\unicode[STIX]{x1D701}}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}(\unicode[STIX]{x1D714}-1)|g(\unicode[STIX]{x1D714})| & {>} & \displaystyle \int _{\unicode[STIX]{x1D701}+\unicode[STIX]{x1D700}}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}(\unicode[STIX]{x1D714}-1)|g(\unicode[STIX]{x1D714})|\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & {>} & \displaystyle (\unicode[STIX]{x1D701}+\unicode[STIX]{x1D716})^{m-1}(\unicode[STIX]{x1D701}+\unicode[STIX]{x1D716}-1)\int _{\unicode[STIX]{x1D701}+\unicode[STIX]{x1D716}}^{\unicode[STIX]{x1D70B}}|g(\unicode[STIX]{x1D714})|\,d\unicode[STIX]{x1D714}\nonumber\end{eqnarray}$$

for $0<\unicode[STIX]{x1D716}<\unicode[STIX]{x1D70B}-\unicode[STIX]{x1D701}.$ Considering $\unicode[STIX]{x1D716}=1$ and using the fact that the function $2\sin \unicode[STIX]{x1D714}-\cot (\unicode[STIX]{x1D714}/2)$ is an antiderivative of $g$ , a simple computation shows that

$$\begin{eqnarray}\unicode[STIX]{x1D701}^{m-1}(\unicode[STIX]{x1D701}-1)\int _{1}^{\unicode[STIX]{x1D701}}g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}<(\unicode[STIX]{x1D701}+1)^{m-1}\unicode[STIX]{x1D701}\int _{\unicode[STIX]{x1D701}+1}^{\unicode[STIX]{x1D70B}}|g(\unicode[STIX]{x1D714})|\,d\unicode[STIX]{x1D714}\end{eqnarray}$$

and the sequence $\int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}$ decreases with $m$ . Since $\int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{2}g(\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}<-\unicode[STIX]{x1D70B}$ , we have proved that $\unicode[STIX]{x1D6FD}_{k}\leqslant \unicode[STIX]{x1D6FD}_{2}<0$ . This gives us the upper bound

$$\begin{eqnarray}\int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP\leqslant -\unicode[STIX]{x1D70B}^{m}F+M_{m}\frac{L^{2}}{2\unicode[STIX]{x1D70B}}\quad \text{for }m\geqslant 4.\end{eqnarray}$$

As the cases $m=1,2,3$ are already known, this finishes the proof of the inequality in the theorem. Finally, since the $\unicode[STIX]{x1D6FD}_{k}$ are not zero, equality holds only when $c_{k}=0$ for $k\geqslant 2$ , that is, when $\unicode[STIX]{x2202}K$ is a circle.◻

7.2 Lower bounds

For the case of constant width, we have the following theorem.

Theorem 7.2. Let $K$ be a compact convex set of constant width, with boundary of class ${\mathcal{C}}^{2}$ , of area $F$ and length of the boundary $L$ , and let $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}(P)$ be the visual angle from the point $P$ . Then

$$\begin{eqnarray}\int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP\geqslant -\unicode[STIX]{x1D70B}^{m}F+M_{m}\frac{L^{2}}{2\unicode[STIX]{x1D70B}}-\frac{\unicode[STIX]{x1D70B}^{m-1}}{4}\bigg(1-\bigg(\frac{3}{4}\bigg)^{m}\bigg)\unicode[STIX]{x1D6E5}\geqslant 0,\end{eqnarray}$$

where $\unicode[STIX]{x1D6E5}=L^{2}-4\unicode[STIX]{x1D70B}F$ is the isoperimetric deficit. The first inequality becomes an equality only for circles.

Proof. In the constant width case, we have $c_{k}=0$ for $k$ even and the only contribution in $\unicode[STIX]{x1D6FD}_{k}$ comes from $\unicode[STIX]{x1D714}^{m}$ when $k$ is odd. Therefore, since

$$\begin{eqnarray}\int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP=-\unicode[STIX]{x1D70B}^{m}F+M_{m}\frac{L^{2}}{2\unicode[STIX]{x1D70B}}+\unicode[STIX]{x1D70B}\mathop{\sum }_{k>2,\,\text{odd}}\unicode[STIX]{x1D6FD}_{k}c_{k}^{2},\end{eqnarray}$$

an upper bound for the positive quantity $K_{m}=-\unicode[STIX]{x1D70B}\sum _{k>2,\,\text{odd}}\unicode[STIX]{x1D6FD}_{k}c_{k}^{2}$ will give a lower bound for $\int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP$ . Using (27),

$$\begin{eqnarray}K_{m}=2\unicode[STIX]{x1D70B}m\mathop{\sum }_{k>2,\,\text{odd}}\bigg(\mathop{\sum }_{j=2,\,\text{even}}^{k-1}j\int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\bigg)c_{k}^{2}.\end{eqnarray}$$

The following estimate holds.

$$\begin{eqnarray}\displaystyle \int _{0}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714} & {<} & \displaystyle \int _{\unicode[STIX]{x1D70B}-\unicode[STIX]{x1D70B}/2j}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}\cos (j\unicode[STIX]{x1D714})\,d\unicode[STIX]{x1D714}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \int _{\unicode[STIX]{x1D70B}-\unicode[STIX]{x1D70B}/2j}^{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D714}^{m-1}\,d\unicode[STIX]{x1D714}=\frac{\unicode[STIX]{x1D70B}^{m}}{m}\bigg(1-\bigg(\frac{3}{4}\bigg)^{m}\bigg).\nonumber\end{eqnarray}$$

Moreover, $\sum _{j=2,\,\text{even}}j=\frac{1}{4}(k^{2}-1),$ so that

(29) $$\begin{eqnarray}\displaystyle K_{m}\leqslant \frac{\unicode[STIX]{x1D70B}^{m+1}}{2}\bigg(1-\bigg(\frac{3}{4}\bigg)^{m}\bigg)\mathop{\sum }_{k>2,\,\text{odd}}(k^{2}-1)c_{k}^{2}. & & \displaystyle\end{eqnarray}$$

As $\unicode[STIX]{x1D6E5}=2\unicode[STIX]{x1D70B}^{2}\sum _{k\geqslant 2}(k^{2}-1)c_{k}^{2}$ (cf. [Reference Cufí and Reventós2]),

$$\begin{eqnarray}K_{m}\leqslant \frac{\unicode[STIX]{x1D70B}^{m-1}}{4}\bigg(1-\bigg(\frac{3}{4}\bigg)^{m}\bigg)\unicode[STIX]{x1D6E5},\end{eqnarray}$$

which gives the desired lower bound.

From formula (26) applied to a circle, it follows that $M_{m}\geqslant \unicode[STIX]{x1D70B}^{m}/2$ and so

$$\begin{eqnarray}\displaystyle \int _{P\notin K}(\unicode[STIX]{x1D714}^{m}-\sin ^{m}\unicode[STIX]{x1D714})\,dP & {\geqslant} & \displaystyle -\unicode[STIX]{x1D70B}^{m}F+M_{m}\frac{L^{2}}{2\unicode[STIX]{x1D70B}}-\frac{\unicode[STIX]{x1D70B}^{m-1}}{4}\bigg(1-\bigg(\frac{3}{4}\bigg)^{m}\bigg)\unicode[STIX]{x1D6E5}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \frac{\unicode[STIX]{x1D70B}^{m-1}}{4}\bigg(\frac{3}{4}\bigg)^{m}\unicode[STIX]{x1D6E5}\geqslant 0.\nonumber\end{eqnarray}$$

Finally, if some $c_{k}\not =0$ , we have strict inequality in (29), and hence the first inequality becomes an equality only for circles.◻