1 Introduction
Let
$\mathbb {S}^{2n-1}$
denote the unit sphere in
$\mathbb {C}^n$
, where
$n\geq 2$
. The hypersurface
$\mathbb {S}^{2n-1}$
is an embedded CR manifold, and the tangential Cauchy–Riemann operators
$\overline {\partial }_b$
and
$\overline {\partial }_b^*$
are defined on the corresponding Hilbert spaces. Furthermore, the Kohn Laplacian on
$L^2(\mathbb {S}^{2n-1})$
$$ \begin{align*}\square_b=\overline{\partial}_b^*\overline{\partial}_b\end{align*} $$
is a linear self-adjoint densely defined closed operator. We refer to [Reference BoggessBog91, Reference Chen and ShawCS01] for the detailed definitions.
Our inspiration is the celebrated Weyl’s law for Riemannian manifolds. In that setting, the eigenvalue counting function of the Laplace–Beltrami operator on a manifold M has leading coefficient proportional to the volume of M with a constant depending only on the dimension of M. There is also a corresponding result for the extension of this operator (also called the Hodge Laplacian, or the Laplace–de Rham operator) to differential forms.
Motivated by the spectral theory for the Laplacian on Riemannian manifolds, one can investigate the spectrum of
$\square _b$
on CR manifolds and its relation to the complex geometry of the underlying manifold. For example, in [Reference FuFu05, Reference FuFu08], Fu studied the spectrum of the
$\overline {\partial }$
-Neumann Laplacian
$\square $
on smooth pseudoconvex domains
$\Omega $
and related the distribution of the eigenvalues of
$\square $
to the D’Angelo type of
$b\Omega $
. In [Reference FollandFol72], Folland computed the spectrum of
$\square _b$
on
$\mathbb {S}^{2n-1}$
on all differential form levels. Recently, in [Reference Ahn, Bansil, Brown, Cardin and ZeytuncuABB+19, Reference Abbas, Brown, Ramasami and ZeytuncuABRZ19], the authors used the spectrum of the Kohn Laplacian to prove the nonembeddability of the Rossi sphere.
For
$\lambda>0$
, let
$N(\lambda )$
denote the number of positive eigenvalues (counting multiplicity) of
$\square _b$
on
$L^2(\mathbb {S}^{2n-1})$
that are less than or equal to
$\lambda $
. It was noted in [Reference Ahn, Bansil, Brown, Cardin and ZeytuncuABB+19] that
$N(\lambda )$
grows on the order of
$\lambda ^n$
and later, in [Reference Bansil and ZeytuncuBZ20] (see the erratum), the leading coefficient in the asymptotic expansion was calculated as an infinite sum. In particular, [Reference Bansil and ZeytuncuBZ20] obtained
$$ \begin{align*} \lim_{\lambda \to \infty} \frac{N(\lambda)}{\lambda^n} = \frac{1}{2^n n!}\sum_{q=1}^\infty \frac{1}{q^n} \bigg[\binom{n+q-2}{q}+{q-1 \choose n -2}\bigg] \end{align*} $$
by using careful counting arguments.
1.1 Main result
In this paper, we obtain the same leading coefficient by another argument, namely by Karamata’s Tauberian Theorem. We highlight that this technique is different than the one in [Reference Bansil and ZeytuncuBZ20]. Furthermore, we formulate the leading coefficient as a product of the volume of
$\mathbb {S}^{2n-1}$
and an integral that only depends on n. This representation resonates better with Weyl’s law and is more amenable to generalization to other CR manifolds.
Theorem 1.1 Let
$N(\lambda )$
be the eigenvalue counting function for
$\square _b$
on
$L^2(\mathbb {S}^{2n-1})$
as above. Then,
$$ \begin{align*} \lim_{\lambda \to \infty} \frac{N(\lambda)}{\lambda^n} &= \frac{1}{2^n n!}\sum_{q=1}^\infty \frac{1}{q^n} \bigg[\binom{n+q-2}{q}+{q-1 \choose n -2}\bigg]\\ &=\mathrm{vol}(\mathbb{S}^{2n-1})\frac{n-1}{n(2\pi)^n\Gamma(n+1)}\int_{-\infty}^{\infty}\left(\frac{t}{\sinh t}\right)^ne^{-(n-2)t}dt. \end{align*} $$
The integral representation is strikingly similar to (and indeed inspired by) a similar formula for the leading coefficient of the eigenvalue counting function for
$\square _b$
acting on
$(0,q)$
-forms (where
$q\geq 1$
) in [Reference Stanton and TartakoffST84, Reference StantonSta84]. However, we note that the expression in Theorem 1.1 is not a special case of the expression in [Reference Stanton and TartakoffST84]. Stanton and Tartakoff’s formula does not cover the case of functions; the improper integral in their representation diverges for
$q=0$
. Furthermore, in the case of functions, the kernel of
$\square _b$
is infinite-dimensional (unlike
$(0,q)$
-forms where
$q\geq 1$
). Therefore, a careful regularization idea and some nontrivial adjustments are necessary to formulate the correct expression for the case
$q=0$
.
We anticipate that, for a general pseudoconvex CR manifold M of hypersurface type embedded in
$\mathbb {C}^n$
, the leading coefficient of the Kohn Laplacian on functions should be given by
$$ \begin{align} \lim_{\lambda \to \infty} \frac{N(\lambda)}{\lambda^n} = \text{vol}(M) \frac{n-1}{n(2\pi)^n\Gamma(n+1)}\int_{-\infty}^{\infty}\left(\frac{t}{\sinh t}\right)^ne^{-(n-2)t}dt. \end{align} $$
This conjecture is modeled after Theorem 1.1 and is further supported by a connection with Stanton and Tartakoff’s formula. In the last section, we recover formula (1.1) by extending Stanton and Tartakoff’s formula to a meromorphic function and evaluating the nonsingular part at
$q=0$
. Although in Section 3 we present only the computation for the case of spheres, the same computation holds for general hypersurfaces in
$\mathbb {C}^n$
.
1.2 Ingredients
Before we present a proof of Theorem 1.1 in the next section, we state some known facts about the eigenvalues of
$\square _b$
on
$\mathbb {S}^{2n-1}$
and Karamata’s Tauberian Theorem.
The Kohn Laplacian acts on the space of
$L^2$
differential forms on the sphere. In [Reference FollandFol72], Folland uses unitary representations to explicitly compute the eigenvalues and corresponding eigenspaces for the Kohn Laplacian on
$(0,j)$
-forms. In particular, he shows that eigenforms
$$ \begin{align*}\bar{z_1}^{q-1} z_n^p \sum_{i=1}^{j+1} d\bar{z_1} \wedge \cdots \wedge \widehat{d\bar{z_i}} \wedge \cdots \wedge d\bar{z}_{j+1}\end{align*} $$
have corresponding eigenvalues
$2 (q+j) (p+n-j-1)$
, where the hat over a form indicates its exclusion from the wedge product.
We are interested in the case where
$j=0$
, corresponding to functions on the sphere. Folland (see also [Reference Ahn, Bansil, Brown, Cardin and ZeytuncuABB+19, Reference Abbas, Brown, Ramasami and ZeytuncuABRZ19]) explicitly shows that, in this case,
$$ \begin{align*}L^2(\mathbb{S}^{2n-1}) = \bigoplus_{k=0}^\infty \mathcal{H}_{k}(\mathbb{S}^{2n-1}) = \bigoplus_{p,q=0}^\infty \mathcal{H}_{p,q}(\mathbb{S}^{2n-1}),\end{align*} $$
where
$\mathcal {H}_{p,q}(\mathbb {S}^{2n-1})$
is the space of spherical harmonics of bidegree
$p,q$
. Furthermore, each space
$\mathcal {H}_{p,q}(\mathbb {S}^{2n-1})$
is an eigenspace of
$\square _b$
with dimension
$$ \begin{align*}\mathrm{dim}(\mathcal{H}_{p,q}(\mathbb{S}^{2n-1}))=\binom{n+p-1}{p} \binom{n+q-1}{q}-\binom{n+p-2}{p-1}\binom{n+q-2}{q-1},\end{align*} $$
as computed by an inclusion–exclusion argument (see [Reference KlimaKli04]). These spaces have corresponding eigenvalues
$2q (p+n-1)$
.
Karamata’s Tauberian Theorem has been used in [Reference KacKac66, Reference Stanton and TartakoffST84] to understand the distribution of eigenvalues. We follow the statement in [Reference Arendt, Nittka, Peter, Steiner, Arendt and SchleichANPS09, Theorem 1.1, p. 57], where the reader can find further references.
Theorem 1.2 (Karamata)
Let
$\{\lambda _j\}_{j\in \mathbb {N}}$
be a sequence of positive real numbers such that
$\sum _{j\in \mathbb {N}}e^{-\lambda _j t}$
converges for every
$t>0$
. Then, for
$n>0$
and
$a\in \mathbb {R}$
, the following are equivalent.
-
(1)
$\lim _{t\rightarrow {0}^+}t^n\sum _{j\in \mathbb {N}}e^{-\lambda _jt}=a$
; -
(2)
$\lim _{\lambda \rightarrow {\infty }}\frac {N(\lambda )}{\lambda ^n}=\frac {a}{\Gamma (n+1)}$
,
where
$N(\lambda )=\#\{\lambda _j:\lambda _j\leq \lambda \}$
is the counting function.
The next section is dedicated to the proof of Theorem 1.1. First, we prove the expression for the leading coefficient as a series (recovering the result in [Reference Bansil and ZeytuncuBZ20]). Next, we express the leading coefficient as the volume of
$\mathbb {S}^{2n-1}$
times an integral.
2 Tonelli’s tactic to tie together two Tauberian terms
Instead of considering
$N(\lambda )$
directly, we define the function
$G(t)=\sum _{j\in \mathbb {N}} e^{-\lambda _j t}$
and consider
$$ \begin{align*}\lim_{t\to 0^+} t^n\sum_{j\in \mathbb{N}} e^{-\lambda_j t},\end{align*} $$
where the sequence
$\{\lambda _j\}_{j\in \mathbb {N}}$
is the sequence of all positive eigenvalues (with multiplicity) of
$\square _b$
on the sphere. Once we compute this limit, we can invoke Karamata’s Theorem. Using the preliminaries from Section 1.2, we have
$$ \begin{align*} G(t)&=\sum_{j\in \mathbb{N}} e^{-\lambda_j t}= \sum_{q=1}^\infty \sum_{p=0}^\infty \dim(\mathcal{H}_{p,q}) e^{-2q(p+n-1) t}\\ &=\sum_{q=1}^\infty \sum_{p=0}^\infty \left(\binom{n+p-1}{p}\binom{n+q-1}{q}-\binom{n+p-2}{p-1}\binom{n+q-2}{q-1}\right) (e^{-2 t q})^{p+n-1} . \end{align*} $$
Applying the standard recursive formula for binomial coefficients to the first product of binomial coefficients gives usFootnote 1
$$ \begin{align*} \binom{n+p-1}{p}\binom{n+q-1}{q} &= \binom{n+p-1}{p}\left[\binom{n+q-2}{q} + \binom{n+q-2}{q-1}\right] \\ &= \binom{n+p-1}{p} \binom{n+q-2}{q} + \binom{n+p-1}{p} \binom{n+q-2}{q-1} \\ &= \binom{n+p-1}{p} \binom{n+q-2}{q}\\ &\quad + \left[\binom{n + p - 2}{p} + \binom{n + p - 2}{p - 1} \right] \binom{n+q-2}{q-1} \\ &= \binom{n+p-1}{p} \binom{n+q-2}{q} + \binom{n + p - 2}{p}\binom{n+q-2}{q-1}\\ & \quad + \binom{n + p - 2}{p - 1} \binom{n+q-2}{q-1}. \end{align*} $$
This allows us to rewrite
$G(t)$
as a sum of two positive pieces by noticing that
$$ \begin{align*}G(t) = \sum_{q=1}^\infty \sum_{p=0}^\infty \left(\binom{n+p-1}{p} \binom{n+q-2}{q} + \binom{n + p - 2}{p}\binom{n+q-2}{q-1}\right) e^{-2 t q(p+n-1)}.\end{align*} $$
We label the parts as
$$ \begin{align*} G_1(t) = \sum_{q=1}^\infty \sum_{p=0}^\infty \binom{n+p-1}{p} \binom{n+q-2}{q} e^{-2 t q(p+n-1)}\end{align*} $$
and
$$ \begin{align*}G_2(t) = \sum_{q=1}^\infty \sum_{p=0}^\infty \binom{n + p - 2}{p}\binom{n+q-2}{q-1} e^{-2 t q(p+n-1)},\end{align*} $$
so that
$G(t)= G_1(t)+ G_2(t)$
.
Our goal is to calculate
$\lim _{t \to 0^+} t^n G(t)$
, which we do by calculating
$ \lim _{t \to 0^+} t^n G_1(t)$
and
$\lim _{t \to 0^+} t^n G_2(t)$
separately.
Theorem 1.1 claims that
$\lim _{\lambda \to \infty } \frac {N(\lambda )}{\lambda ^n}$
can be written either as an infinite series or as an improper integral. The key to obtaining these two distinct forms is Tonelli’s Theorem (see [Reference RudinRud87, Theorem 8.8]). Because the summands in
$G_1(t)$
and
$G_2(t)$
are positive, we can exchange the order of the infinite sums in each. When the outer sum is over q, we can show that
$\lim _{t\to 0^+} G_1(t)$
is an infinite series and
$\lim _{t\to 0^+} G_2(t)$
is an improper integral; exchanging the order of summation allows us to express
$\lim _{t\to 0^+} G_1(t)$
as an integral and
$\lim _{t\to 0^+} G_2(t)$
as an infinite series.
2.1 Serious series for spherical spectra
In this part, we prove the infinite series formula for
$\lim _{\lambda \to \infty } \frac {N(\lambda )}{\lambda ^n}$
. The computations for
$G_1(t)$
and
$G_2(t)$
are quite similar. In each instance, we will apply the Dominated Convergence Theorem by leveraging Lemma 2.2.
In each computation, we eventually exchange the limit as
$t\to 0^+$
with a sum, and in each case, we require the following limit calculation, which follows from L’Hopital’s rule.
Lemma 2.1 Let
$\alpha> 0$
. Then,
$$ \begin{align*}\lim_{t\to 0} \frac{t^n}{(1-e^{-\alpha t})^n}=\alpha^{-n}.\end{align*} $$
In order to exchange limits with sums, we will apply the Dominated Convergence Theorem. The next technical lemma will come in handy for showing the conditions of the Dominated Convergence Theorem are satisfied in each case.
Lemma 2.2 For
$n\in \mathbb {N}$
, there exists
$M>0$
such that
$$ \begin{align*}f(x)=\frac{x^n e^{2x(n-1)}}{(e^{2x}-1)^n}<M,\end{align*} $$
for all
$x\in (0,\infty )$
.
Proof The function f is continuous on
$(0,\infty )$
. If we show that
$\lim _{x\to \infty } f(x)<\infty $
and
$\lim _{x\to 0^+} f(x)<\infty $
, it will follow that there exists some M such that
$f(x)<M$
on
$(0,\infty )$
. Indeed,
$\lim _{x\to \infty } f(x) = 0$
. Furthermore, L’Hopital’s rule gives
$\lim _{x\to 0^+}\frac {x}{e^{2x}-1}=\frac {1}{2}$
and so
$\lim _{x\to 0^+}\frac {x^n e^{2x(n-1)}}{(e^{2x}-1)^n} =\frac {1}{2^n}$
. ▪
Now, we are ready to compute
$\lim _{t\to 0^+}t^n G_1(t)$
.
Proposition 2.3
$$ \begin{align*}\lim_{t\to 0^+} t^n G_1(t) = \frac{1}{2^n}\sum_{q=1}^\infty \binom{n+q-2}{q}\frac{1}{q^n}.\end{align*} $$
Proof We start by coming up with an expression for
$G_1(t)$
containing only a single sum. For
$|z| < 1$
, there is the Taylor series expansion
$$ \begin{align*}\frac{1}{(1-z)^n} = \sum_{p=0}^\infty \binom{n+p-1}{p}z^p = \sum_{p=1}^\infty \binom{n+p-2}{p-1}z^{p-1}.\end{align*} $$
Let
$z = e^{-2tq}$
; then,
$$ \begin{align*} G_1(t) &= \sum_{q=1}^\infty \sum_{p=0}^\infty \binom{n+p-1}{p} \binom{n+q-2}{q} e^{-2 t q(p+n-1)}\\ &= \sum_{q=1}^\infty \left(\binom{n+q-2}{q}\sum_{p=0}^\infty \binom{n+p-1}{p} (e^{-2 t q})^{p+n-1}\right) \\ &=\sum_{q=1}^\infty \binom{n+q-2}{q}\sum_{p=0}^\infty \binom{n+p-1}{p} z^{p+n-1}\\ &=\sum_{q=1}^\infty \binom{n+q-2}{q}z^{n-1}\sum_{p=0}^\infty \binom{n+p-1}{p} z^{p} \\ &=\sum_{q=1}^\infty \binom{n+q-2}{q}\frac{z^{n-1}}{(1-z)^n}\\ &=\sum_{q=1}^\infty \binom{n+q-2}{q}\frac{(e^{-2 t q})^{n-1}}{(1-e^{-2 t q})^n}\\ &=\sum_{q=1}^\infty \binom{n+q-2}{q}\frac{e^{2 t q}}{(e^{2 t q}-1)^n}. \end{align*} $$
For positive t,
$\frac {t^n(e^{2 t q})}{(e^{2 t q}-1)^n}\leq \frac {1}{q^n}\frac {(qt)^n(e^{2 t q (n-1)})}{(e^{2 t q}-1)^n} =\frac {f(qt)}{q^n}\leq \frac {M}{q^n}$
, where f is as in Lemma 2.2. Hence,
$t^n \binom {n+q-2}{q} \frac {(e^{2 t q})}{(e^{2 t q}-1)^n}\leq M \binom {n+q-2}{q} \frac {1}{q^n}$
, for all t, and because
$\sum _{q=1}^\infty M \binom {n+q-2}{q} \frac {1}{q^n}$
converges, we are free to apply the Dominated Convergence Theorem to
$\lim _{t\to 0^+}t^n G_1(t)$
.
This technical justification allows us to exchange the order of the limit and summation in
$\lim _{t\to 0^+}t^n G_1(t)$
and conclude that
$$ \begin{align*} \lim_{t\to 0^+}\sum_{q=1}^\infty \binom{n+q-2}{q}\frac{t^n(e^{-2 t q})^{n-1}}{(1-e^{-2 t q})^n} &= \sum_{q=1}^\infty \lim_{t\to 0^+} \binom{n+q-2}{q}\frac{t^n(e^{-2 t q})^{n-1}}{(1-e^{-2 t q})^n}\\ &=\sum_{q=1}^\infty \binom{n+q-2}{q}\frac{1}{(2q)^n}, \end{align*} $$
where the final equality follows from Lemma 2.1. ▪
Next, we move to the second piece and compute
$\lim _{t\to 0^+} t^n G_2(t)$
.
Proposition 2.4
$$ \begin{align*}\lim_{t\to 0^+} t^n G_2(t) = \frac{1}{2^n}\sum_{q=1}^\infty \binom{q-1}{n-2}\frac{1}{q^n}.\end{align*} $$
Proof We start out by manipulating the form of
$G_2(t)$
as we did with
$G_1(t)$
, but this time, we apply Tonelli’s Theorem to switch the order of summation. In our calculation, we will make use of the substitutions
$z=e^{-2t(p+n-1)}$
and
$w=p+n-1$
, and we will apply the power series expansion of
$\frac {1}{(1-z)^{n-1}}$
.
$$ \begin{align*} G_2(t)&=\sum_{q=1}^\infty \sum_{p=0}^\infty \binom{n+p-2}{p}\binom{n+q-2}{q-1}e^{-2tq(p+n-1)} \\&=\sum_{p=0}^\infty \binom{n+p-2}{p} \sum_{q=1}^\infty \binom{n+q-2}{q-1}e^{-2tq(p+n-1)}\\ &=\sum_{p=0}^\infty \binom{n+p-2}{p} z \sum_{q=1}^\infty \binom{n+q-2}{q-1}z^{q-1}=\sum_{p=0}^\infty \binom{n+p-2}{p} \frac{z}{(1-z)^n}\\ &=\sum_{p=0}^\infty \binom{n+p-2}{n-2} \frac{e^{-2t(p+n-1)}}{(1-e^{-2t(p+n-1)})^n}=\sum_{w=n-1}^\infty \binom{w-1}{n-2}\frac{e^{-2tw}}{(1-e^{-2tw})^n}\\ &=\sum_{w=1}^\infty \binom{w-1}{n-2}\frac{e^{-2tw}}{(\frac{e^{2tw}-1}{e^{2tw}})^n}=\sum_{w=1}^\infty \binom{w-1}{n-2} \frac{e^{2tw(n-1)}}{(e^{2tw}-1)^n}. \end{align*} $$
By Lemma 2.2,
$t^n \binom {w-1}{n-2} \frac {e^{2tw(n-1)}}{(e^{2tw}-1)^n}=\binom {w-1}{n-2}\frac {f(tw)}{w^n} \leq M\binom {w-1}{n-2}\frac {1}{w^n}.$
As
$\sum_{w=0}^\infty M \binom {w-1}{n-2}\frac {1}{w^n}.$
converges, we apply the Dominated Convergence Theorem to
$\lim _{t\to 0^+} t^nG_2(t)$
in order to exchange the limit and the sum. Thus,
$$ \begin{align*}\lim_{t\to 0^+} t^n G_2(t) = \sum_{w=1}^\infty \lim_{t\to 0^+}\binom{w-1}{n-2}\frac{t^n e^{-2tw}}{(\frac{e^{2tw}-1}{e^{2tw}})^n}=\frac{1}{2^n}\sum_{w=1}^\infty \binom{w-1}{n-2}\frac{1}{w^n}. \end{align*} $$
▪
Applying the Tauberian Theorem 1.2 in combination with Propositions 2.3 and 2.4 proves that
$$ \begin{align*}\lim_{\lambda\to \infty} \frac{N(\lambda)}{\lambda^n} = \frac{1}{2^n n!}\sum_{q=1}^\infty \frac{1}{q^n}\left[\binom{n+q-2}{q}+\binom{q-1}{n-2} \right].\end{align*} $$
This concludes the proof of the first identity in Theorem 1.1.
2.2 Is it not interesting: intense integral is identical in immensity
In this subsection, we will prove the second part of Theorem 1.1. We start off by reanalyzing
$G_1(t)$
:
$$ \begin{align*} G_1(t) &= \sum_{q=1}^\infty \sum_{p=0}^\infty \binom{n+p-1}{p} \binom{n+q-2}{q} e^{-2 t q(p+n-1)} \\ &= \sum_{p=0}^\infty \sum_{q=1}^\infty \binom{n+p-1}{p} \binom{n+q-2}{q} e^{-2 t q(p+n-1)} \\ &= \sum_{p=0}^{\infty}\binom{n+p-1}{p}\sum_{q=1}^{\infty}\binom{n+q-2}{q} (e^{-2 t (p + n - 1)})^{q} \\ &= \sum_{p=0}^{\infty}\binom{n+p-1}{p}\left( \frac{1}{(1 - e^{-2t(p + n - 1)})^{n-1}} - 1 \right) \\ &= \sum_{w=n-1}^{\infty}\binom{w}{n-1}\left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right), \end{align*} $$
where the switching of the order of summation is justified by Tonelli’s Theorem and we have taken
$w = n + p - 1$
. To calculate the limit as
$t\to 0^+$
, we need the following lemma.
Lemma 2.5 Fix an integer
$r \geq 1$
and consider the function
$$ \begin{align*}f_{r}(x) = x^r \left( \frac{1}{(1 - e^{-2x})^r} - 1 \right)\end{align*} $$
defined for
$x> 0$
. Then,
-
(1)
$f_{r}(x)> 0$
. -
(2)
$f_{r}^{\prime }(x) < 0$
for sufficiently large x. -
(3)
$f_{r}$
is bounded and
$\int _{0}^\infty f_{r}(x) dx < \infty .$
Proof We can tell
$f_{r}(x)>0$
because
$e^{-2x}<1$
, so
$0<1-e^{-2x}<1$
, and hence
$\frac {1}{(1-e^{-2x})^r}>1$
, proving (1).
To establish (2), a derivative calculation shows that
$$ \begin{align*} f_r^{\prime}(x) &= rx^{r-1}\left(\frac{1}{(1 - e^{-2x})^r} - 1\right) - x^r\frac{2re^{-2x}}{(1 - e^{-2x})^{r+1}} \\ &= \frac{r x^{r-1}}{(1 - e^{-2x})^{r+1}}\left[1 - e^{-2x} - (1 - e^{-2x})^{r+1} - 2xe^{-2x} \right] \\ &= \frac{r x^{r-1}}{(1 - e^{-2x})^{r+1}e^{2x}}\left[e^{2x}- 1 - (1 - e^{-2x})^{r+1}e^{2x} - 2x \right]. \end{align*} $$
Define
$l(x) = e^{2x} - 1 - (1 - e^{-2x})^{r+1}e^{2x} - 2x.$
To prove (2), it suffices to show that
$l(x) < 0$
for sufficiently large x. For this, let
$y = e^{-2x}$
and note that
$$ \begin{align*}\lim_{x \to \infty} e^{2x}\kern-0.5pt -\kern-0.5pt 1\kern-0.5pt -\kern-0.5pt (1\kern-0.5pt -\kern-0.5pt e^{-2x})^{r\kern-0.5pt + \kern-0.5pt 1}e^{2x} = \lim_{y \to 0^+} \frac{1\kern-0.5pt -\kern-0.5pt y\kern-0.5pt -\kern-0.5pt (1\kern-0.5pt -\kern-0.5pt y)^{r+1}}{y} = \lim_{y \to 0^+} -1 + (r+1)(1-y)^{r} = r.\end{align*} $$
Looking at the definition of l, the first three terms converge to r and the last goes to
$-\infty $
, so
$l(x) < 0$
for sufficiently large x.
To prove (3), we will show that
$f_r(x)$
is bounded by a constant near 0 and bounded by an exponentially decaying function for large x. The function
$f_{r}(x)$
is not defined at
$x=0$
, but Lemma 2.1 implies that
$\lim _{x\to 0} f_{r}(x)<\infty $
, so the discontinuity at 0 is removable. This implies that
$f_r(x)$
is bounded on
$[0,\frac {\log (2)}{2}]$
. To bound
$f_r(x)$
on
$[\frac {\log 2}{2},\infty )$
, we note
$$ \begin{align*}\frac{1}{(1-e^{-2x})^r}-1=\frac{1}{((e^{2x}-1)/e^{2x})^r}-1= \frac{(e^{2x})^r-(e^{2x}-1)^r}{(e^{2x}-1)^r}. \end{align*} $$
Using the formula for a difference of rth powers, we obtain the following expression for the numerator of the above fraction:
$$ \begin{align*} (e^{2x})^r-(e^{2x}-1)^r&=(e^{2x}-(e^{2x}-1)) \sum_{i=0}^{r-1} (e^{2x})^i (e^{2x}-1 )^{r-1-i}\\ &\leq \sum_{i=0}^{r-1} (e^{2x})^i (e^{2x})^{r-1-i}=re^{2x(r-1)}. \end{align*} $$
Hence,
$$ \begin{align*}\frac{1}{(1-e^{-2x})^r}-1 \leq \frac{re^{2x(r-1)}}{(e^{2x}-1)^r}.\end{align*} $$
Note
$e^{2x}-1 \geq \frac {e^{2x}}{2}$
for
$x \geq \frac {\log (2)}{2}$
, so
$$ \begin{align*}\frac{1}{(1-e^{-2x})^r}-1 \leq \frac{re^{2x(r-1)}}{(e^{2x}/2)^r}=\frac{r2^r}{e^{2x}}.\end{align*} $$
Applying this to
$f_r(x)$
gives that when
$x \geq \frac {\log (2)}{2},$
$$ \begin{align*}f_r(x)\leq \frac{r2^rx^r}{e^{2x}}.\end{align*} $$
Thus,
$f_r(x)$
is dominated by an exponentially decaying function for sufficiently large x, so
$\int _0^\infty f_r(x)\,dx<\infty .$
▪
It will be helpful to make the following definition.
Definition 2.6 For real
$\alpha \neq 0$
, define the scaled ceiling function
$\lceil \cdot \rceil _{\alpha }: \mathbb {R} \to \mathbb {R}$
by
$$ \begin{align*}\lceil x \rceil_{\alpha} = \alpha\lceil x/\alpha \rceil.\end{align*} $$
For example,
$\lceil 7 \rceil _{3} = 3 \lceil \frac {7}{3} \rceil = 3 \cdot 3 = 9$
and
$\lceil 6 \rceil _{2} = 2 \lceil \frac {6}{2} \rceil = 2 \cdot 3 = 6$
. Then, we have the following proposition.
Proposition 2.7
$\lceil x \rceil _{\alpha }$
is x rounded up to the nearest integral multiple of
$\alpha $
, i.e.,
$$ \begin{align*}\lceil x \rceil_{\alpha} = \min\{n\alpha: n \in \mathbb{Z}, n\alpha \geq x\}.\end{align*} $$
Proof
$$ \begin{align*}\lceil x \rceil_{\alpha} = \alpha \min\{n \in \mathbb{Z}: n \geq x/\alpha\} = \alpha \min\{n \in \mathbb{Z}: \alpha n \geq x\}, \end{align*} $$
and the result follows. ▪
The scaled ceiling function has the following properties:
-
(1) Fix
$x \in \mathbb {R}$
,
$\alpha> 0$
. Then,
$0 \leq \lceil x \rceil _{\alpha } - x < \alpha $
. -
(2) Fix
$x \in \mathbb {R}$
. Then,
$\lim _{\alpha \to 0^+} \lceil x \rceil _{\alpha } = x$
. -
(3) Let
$f: [a,b] \to \mathbb {R}$
be monotonically decreasing. Fix
$0 < \alpha \leq b - a$
. Then, for all
$x \in [a,b-\alpha ]$
,
$f(\lceil x \rceil _{\alpha }) \leq f(x)$
.
The next few propositions will allow us to compute
$\lim _{t\to 0^+} t^n G_1(t)$
.
Proposition 2.8
$$ \begin{align*} \lim_{t \to 0^+} t^n \sum_{w=n-1}^{\infty}w^{n-1} \left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right) = \int_{0}^\infty x^{n-1} \left( \frac{1}{(1 - e^{-2x})^{n-1}} - 1 \right)dx. \end{align*} $$
Proof Manipulating the sum, we have
$$ \begin{align*} t^n \sum_{w=n-1}^{\infty} w^{n-1}& \left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right) = \sum_{w=n-1}^{\infty}t^{n} w^{n-1} \left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right) \\ &= \sum_{w=n-1}^{\infty} \int_{w-1}^w t^{n} \lceil w' \rceil ^{n-1} \left( \frac{1}{(1 - e^{-2t\lceil w' \rceil})^{n-1}} - 1 \right) dw' \\ &= \int_{n-2}^\infty t^{n} \lceil w' \rceil ^{n-1} \left( \frac{1}{(1 - e^{-2t\lceil w' \rceil})^{n-1}} - 1 \right) dw' \\ &= \int_{t(n-2)}^\infty (t\lceil x/t \rceil) ^{n-1} \left( \frac{1}{(1 - e^{-2t\lceil x/t \rceil})^{n-1}} - 1 \right) dx \\ &= \int_{t(n-2)}^\infty \lceil x \rceil_{t} ^{n-1} \left( \frac{1}{(1 - e^{-2 \lceil x \rceil_{t}})^{n-1}} - 1 \right) dx\\ &=\int_{t(n-2)}^\infty f_{n-1}(\lceil x \rceil_{t}) dx. \end{align*} $$
Fix C such that
$f_{n-1}^{\prime }(x) < 0$
for
$x \geq C$
and fix M such that
$f_{n-1}(x)<M$
. Then,
$$ \begin{align*}f_{n-1}(\lceil x \rceil_{t}) \leq M \mathbb{1}_{\{x < C\}} + f_{n-1}(x), \end{align*} $$
for all
$x> 0$
. Because the integral of the right-hand side is finite, we may apply dominated convergence to see that
$$ \begin{align*}\lim_{t \to 0^+} \sum_{w=n-1}^{\infty}t^n w^{n-1} \left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right)dx = \int_{0}^\infty x^{n-1} \left( \frac{1}{(1 - e^{-2x})^{n-1}} - 1 \right)dx, \end{align*} $$
which completes the proof. ▪
The following proposition will help us compute the rest of the limits we need before we can tackle
$\lim _{t\to 0^+}t^nG(t)$
in its entirety.
Proposition 2.9 Suppose that
$a_t(w)$
is a positive function of real t and integer
$w\geq 1$
such that
$\lim _{t\to 0^+}a_t(w)=0$
for each w and
$\lim _{t\to 0^+}\sum _{w=1}^\infty a_t(w) =M<\infty $
. Then,
$$ \begin{align*}\lim_{t\to 0^+} \sum_{w=1}^\infty \frac{a_t(w)}{w} = 0.\end{align*} $$
Proof Let
$\epsilon>0$
be arbitrary and let
$k>2\frac {M}{\epsilon }-1$
be an integer. Because
$\lim _{t\to 0^+} a_t(w)=0$
for each w, there exists some T such that for all
$0<t<T$
and all
$w\leq k$
,
$a_t(w)\leq \frac {\epsilon }{2k}$
. Thus, for
$t<T$
, we have
$$ \begin{align*} \sum_{w=1}^\infty \frac{a_t(w)}{w}&=\sum_{w=1}^k \frac{a_t(w)}{w}+\sum_{w=k+1}^\infty \frac{a_t(w)}{w}\\ &\leq\sum_{w=1}^k a_t(w) +\frac{1}{k+1}\sum_{w=k+1}^\infty a_t(w)\leq \sum_{w=1}^k a_t(w) +\frac{1}{k+1}\sum_{w=1}^\infty a_t(w)\\ &\leq \sum_{w=1}^k \frac{\epsilon}{2k} +\frac{M}{k+1}\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \end{align*} $$
Therefore,
$\lim _{t\to 0^+} \sum _{w=1}^\infty \frac {a_t(w)}{w} = 0.$
▪
Proposition 2.10 Fix
$0 \leq k < n -1$
. Then,
$$ \begin{align*} \lim_{t \to 0^+} t^n \sum_{w=n-1}^{\infty}w^k \left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right) = 0.\end{align*} $$
Proof The observation that
$0\leq \sum _{w=n-1}^{\infty }t^n w^k ( \frac {1}{(1 - e^{-2tw})^{n-1}} - 1 )\leq \sum _{w=n-1}^{\infty }\frac {1}{w} t^n w^{n-1} ( \frac {1}{(1 - e^{-2tw})^{n-1}} - 1 )$
allows us to apply Propositions 2.8 and 2.9 and arrive at the desired conclusion.▪
With Propositions 2.8 and 2.10 in hand, we can move forward. We break
$\binom {w}{n-1}$
into a polynomial in w as follows:
$$ \begin{align*}\binom{w}{n-1} = \frac{w!}{(n-1)! (w-n+1)!} = \frac{w(w-1)\cdots(w-n+2)}{(n-1)!} = \frac{w^{n-1}}{(n-1)!} + \sum_{k=0}^{n-2} a_k w^k.\end{align*} $$
We write
$$ \begin{align} \lim_{t \to 0^+} t^n G_1(t) &= \lim_{t \to 0^+} t^n \sum_{w=n-1}^{\infty} \binom{w}{n-1}\left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right) \nonumber\\ &= \lim_{t \to 0^+} \bigg[ \frac{t^n}{(n-1)!}\sum_{w=n-1}^{\infty}w^{n-1}\left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right) \end{align} $$
$$ \begin{align}\phantom{\lim_{t \to 0^+} t^n G_1} &\quad + t^n \sum_{k=0}^{n-2}a_k \sum_{w=n-1}^{\infty}w^{k}\left( \frac{1}{(1 - e^{-2tw})^{n-1}} - 1 \right)\bigg] \nonumber\\ &= \frac{1}{(n-1)!}\int_{0}^\infty x^{n-1} \left( \frac{1}{(1 - e^{-2x})^{n-1}} - 1 \right)dx. \end{align} $$
The final equality was obtained by applying Propositions 2.8 and 2.10.
Now, we move on to
$G_2(t)$
. We have
$$ \begin{align*} G_2(t) &= \sum_{q=1}^\infty \sum_{p=0}^\infty \binom{n + p - 2}{p}\binom{n+q-2}{q-1} e^{-2 t q(p+n-1)} \\ &= \sum_{q=1}^\infty\binom{n+q-2}{q-1} e^{-2tq(n-1)}\sum_{p=0}^\infty \binom{n + p - 2}{p} (e^{-2 t q})^p\end{align*} $$
$$\begin{align*} &= \sum_{q=1}^\infty\binom{n+q-2}{q-1} \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} \\ &= \sum_{q=1}^\infty\binom{n+q-2}{n-1} \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}}. \end{align*} $$
The next lemma plays a role in our analysis of
$G_2(t)$
, which is analogous to the role of Lemma 2.5 in our analysis of
$G_1(t)$
.
Lemma 2.11 Fix integers
$r \geq 0$
and
$s \geq 1$
, and consider the function
$$ \begin{align*}g_{r,s}(x) = x^r \frac{e^{-2xs}}{(1-e^{-2x})^{s}}\\[-18pt] \end{align*} $$
defined for
$x> 0$
. Then:
-
(1)
$g_{r,s}(x)> 0.$
-
(2)
$g^{\prime }_{r,r} (x)< 0.$
-
(3) If
$r \geq s$
,
$g_{r,s}(x)$
is bounded and
$\int _{0}^\infty g_{r,s}(x) dx < \infty $
. -
(4) If
$r \geq s + 1$
,
$\lim _{x \to 0} g_{r,s}(x) = 0$
.
Proof Suppose
$r\geq 0$
and
$s\geq 1$
are integers and that
$x>0$
. As
$x^r>0$
,
$e^{-2xs}>0$
, and
$1 - e^{-2x}> 0$
, we have
$g_{r,s}(x)> 0$
, which shows (1). To prove (2), notice that we may write
$$ \begin{align*}g_{r,r}(x)=x^r \frac{e^{-2xr}}{(1-e^{-2x})^{r}}=\frac{x^r}{(e^{2x}-1)^r}.\end{align*} $$
Because the function
$x/(e^{2x}-1)$
is decreasing, it follows that
$g^{\prime }_{r,r}(x)<0$
.
For part (3), fix
$M \in \mathbb {R}$
such that for all
$x \geq M$
,
$e^{2x} - 1 \geq \frac {e^{2x}}{2}$
and
$(e^{2x})^{1/2} \geq x^r$
. Then, for
$x \geq M$
,
$$ \begin{align*}g_{r,s}(x) = \frac{x^r}{(e^{2x} - 1)^s} \leq \frac{x^r}{\left( \frac{e^{2x}}{2} \right)^s} = 2^s \frac{x^r}{(e^{2x})^s} \leq 2^s \frac{(e^{2x})^{1/2}}{(e^{2x})^s} = \frac{2^s}{(e^{2x})^{s-1/2}}.\end{align*} $$
As
$s \geq 1$
, the right-hand side has finite integral over
$[M,\infty )$
and hence
$\int _M^\infty g_{r,s}(x)\ dx < \infty $
. The integral of
$g_{r,s}(x)$
over
$[0,M]$
is also finite, because we can extend
$g_{r,s}(x)$
to a continuous, bounded function on this compact interval. Adding these two parts together shows that
$$ \begin{align*}\int_0^\infty g_{r,s}(x)\ dx < \infty.\end{align*} $$
It remains to show
$\lim _{x\rightarrow 0}g_{r,s}(x)=0$
whenever
$r\geq s+1$
. First, notice that Lemma 2.1 can be used to show that
$$ \begin{align*}\lim_{x\rightarrow 0}\frac{x^r}{(1-e^{-2x})^s}=\lim_{x\rightarrow 0}x^{r-s}\lim_{x\rightarrow 0}\frac{x^s}{(1-e^{-2x})^s}=0\cdot2^{-n}=0.\end{align*} $$
Therefore, when
$r\geq s+1$
, we have
$$ \begin{align*}\lim_{x\rightarrow 0}x^r\frac{e^{-2sx}}{(1-e^{-2x})^s}=\lim_{x\rightarrow 0}e^{-2xs}\lim_{x\rightarrow 0}\frac{x^r}{(1-e^{-2x})^s}=1\cdot0=0\end{align*} $$
as we needed to show, this proves (4). ▪
Now, we move on to propositions which, similarly to our strategy for analyzing
$G_1(t)$
, allow us to break up the binomial coefficient in
$G_2(t)$
into a polynomial and analyze the resulting sums separately.
Proposition 2.12 Fix
$0 \leq k < n -1$
. Then,
$$ \begin{align*} \lim_{t \to 0^+} t^n \sum_{q=1}^{\infty}q^k \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} = 0.\end{align*} $$
Proof We recognize that this expression can be rewritten in terms of
$g_{n,n-1}(tq)$
:
$$ \begin{align*} t^n\sum_{q=1}^{\infty}q^k \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^n} = \sum_{q=1}^{\infty}\frac{1}{q^{n-k}}(tq)^n \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} = \sum_{q=1}^\infty \frac{1}{q^{n-k}} g_{n,n-1}(tq).\end{align*} $$
Because
$g_{n,n-1}$
is bounded and
$n-k \geq 2$
, the summand is dominated by some constant times
$\frac {1}{q^{2}}$
. Therefore, we can apply dominated convergence to conclude that the limit is 0, because the summand converges pointwise to 0 by the previous lemma. ▪
Proposition 2.13
$$ \begin{align*}\lim_{t \to 0^+} t^n \sum_{q=1}^{\infty}q^{n-1} \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} = \int_{0}^\infty x^{n-1} \frac{e^{-2x(n-1)}}{(1 - e^{-2x})^{n-1}} dx.\end{align*} $$
Proof Write
$$ \begin{align*} t^n \sum_{q=1}^{\infty}q^{n-1} \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} &= \sum_{q=1}^{\infty}\int_{q-1}^q t^n \lceil q' \rceil^{n-1} \frac{e^{-2t\lceil q' \rceil(n-1)}}{(1 - e^{-2t \lceil q' \rceil})^{n-1}} dq' \\ &= \int_{0}^\infty t^n \lceil q' \rceil^{n-1} \frac{e^{-2t\lceil q' \rceil(n-1)}}{(1 - e^{-2t \lceil q' \rceil})^{n-1}} dq' \\ &= \int_{0}^\infty (t \lceil x/t \rceil)^{n-1} \frac{e^{-2t\lceil x/t \rceil(n-1)}}{(1 - e^{-2t \lceil x/t \rceil})^{n-1}} dx \\ &= \int_{0}^\infty \lceil x \rceil_{t}^{n-1} \frac{e^{-2\lceil x \rceil_{t}(n-1)}}{(1 - e^{-2\lceil x \rceil_{t}})^{n-1}} dx. \end{align*} $$
The integrand is exactly
$g_{n-1,n-1}(\lceil x \rceil _{t})$
, and is thus dominated by
$g_{n-1,n-1}$
, because
$g_{n-1,n-1}$
is a decreasing function. Therefore, we can apply dominated convergence to find that
$$ \begin{align*}\lim_{t \to 0^+} t^n \sum_{q=1}^{\infty}q^{n-1} \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} = \int_{0}^\infty x^{n-1} \frac{e^{-2x(n-1)}}{(1 - e^{-2x})^{n-1}} dx.\end{align*} $$
▪
We can expand
$\binom {n+q-2}{n-1}$
as
$$ \begin{align*}\binom{n+q-2}{n-1} = \frac{q^{n-1}}{(n-1)!} + \sum_{k=0}^{n-2} b_k q^k.\end{align*} $$
Therefore,
$$ \begin{align*} \lim_{t \to 0^+} &t^n G_2(t) = \lim_{t \to 0^+} t^n \sum_{q=1}^\infty \binom{n+q-2}{n-1} \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} \\ &= \lim_{t \to 0^+} \left[ \frac{1}{(n-1)!} t^n \sum_{q=1}^\infty q^{n-1} \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}}+ t^n \sum_{k=0}^{n-2} b_k \sum_{q=1}^\infty q^k \frac{e^{-2tq(n-1)}}{(1 - e^{-2tq})^{n-1}} \right] \\ &= \frac{1}{(n-1)!} \int_{0}^\infty x^{n-1} \frac{e^{-2x(n-1)}}{(1 - e^{-2x})^{n-1}} dx \\ &= \frac{1}{(n-1)!} \int_{0}^\infty x^{n-1} \frac{1}{(e^{2x} - 1)^{n-1}} dx. \end{align*} $$
Next, we put the two parts of G back together in the limit. This gives us
$$ \begin{align*} \lim_{t \to 0^+} &t^n G(t) = \lim_{t \to 0^+} t^n G_1(t) + \lim_{t \to 0^+} t^n G_2(t) \\ &= \frac{1}{(n-1)!} \left[\int_{0}^\infty x^{n-1} \left( \frac{1}{(1 - e^{-2x})^{n-1}} - 1 \right)dx + \int_{0}^\infty x^{n-1} \frac{1}{(e^{2x} - 1)^{n-1}} dx \right] \\ &= \frac{1}{(n-1)!}\int_{0}^\infty x^{n-1} \left( \frac{1}{(1 - e^{-2x})^{n-1}} - 1 + \frac{1}{(e^{2x} - 1)^{n-1}} \right) dx. \end{align*} $$
We now have an expression for
$\lim _{t\to 0^+} G(t)$
in terms of an integral, so we could have chosen to stop here. Instead, we will press on and apply a few more tricks in order to arrive at the expression in Theorem 1.1. One reason we prefer the expression in Theorem 1.1 is its similarity to a closely related result in [Reference Stanton and TartakoffST84].
To continue on our path of manipulating the combined integral, we apply integration by parts with
$$ \begin{align*} u &= \frac{1}{(1 - e^{-2x})^{n-1}} - 1 + \frac{1}{(e^{2x} - 1)^{n-1}},\\ dv&=x^{n-1}dx,\\ du &= -(n-1)\left(\frac{2e^{-2x}}{(1-e^{-2x})^n} + \frac{2e^{2x}}{(e^{2x} - 1)^n}\right) dx\\& = -2(n-1)\frac{e^{(n-2)x} + e^{-(n-2)x}}{(e^x - e^{-x})^n}dx = -\frac{n-1}{2^{n-2}} \frac{\cosh((n-2)x)}{\sinh(x)^n} dx, \end{align*} $$
and
$v=n^{-1}x^n$
. This gives us
$$ \begin{align*} \lim_{t \to 0^+} t^n G(t) &= \frac{1}{n!} x^n \left(\frac{1}{(1 - e^{-2x})^{n-1}} - 1 + \frac{1}{(e^{2x} - 1)^{n-1}} \right)\bigg|_{0}^\infty \\&\quad + \frac{n-1}{2^{n-2}n!}\int_0^\infty x^n \frac{\cosh((n-2)x)}{\sinh(x)^n} dx \\ &= \frac{n-1}{2^{n-2}n!}\int_0^\infty \frac{x^n \cosh((n-2)x)}{\sinh(x)^n} dx. \end{align*} $$
The boundary term clearly vanishes at
$\infty $
, because
$x^n \left ( \frac {1}{(1 - e^{-2x})^{n-1}} - 1 \right )$
and
$\frac {x^n}{(e^{2x} - 1)^{n-1}}$
each vanish at
$\infty $
. To see that it vanishes at 0, apply L’Hopital’s rule to the quotients
$\frac {x}{1-e^{-2x}}$
and
$\frac {x}{e^{2x}-1}$
. Because the integrand is even, we further have
$$ \begin{align*} \lim_{t \to 0^+} t^n G(t) &= \frac{n-1}{2^{n-2}n!}\int_0^\infty \frac{x^n \cosh((n-2)x)}{\sinh(x)^n} dx \\ &= \frac{n-1}{2^{n-1}n!}\int_{-\infty}^\infty \frac{x^n \cosh((n-2)x)}{\sinh(x)^n} dx \\ &= \frac{n-1}{2^{n-2}n!} \left( \int_{-\infty}^\infty \frac{x^n}{\sinh(x)^n} e^{x(n-2)} dx + \int_{-\infty}^\infty \frac{x^n}{\sinh(x)^n} e^{-x(n-2)} dx \right). \end{align*} $$
As
$\frac {x}{\sinh (x)}$
is an even function, the value of the first integral is unchanged if we change the
$e^{x(n-2)}$
in the integrand to
$e^{-x(n-2)}$
. Thus,
$$ \begin{align*} \lim_{t \to 0^+} t^n G(t) &= \frac{n-1}{2^{n-2}n!} \left( \int_{-\infty}^\infty \frac{x^n}{\sinh(x)^n} e^{-x(n-2)} dx + \int_{-\infty}^\infty \frac{x^n}{\sinh(x)^n} e^{-x(n-2)} dx \right) \\ &=\frac{n-1}{2^{n-1}n!} \int_{-\infty}^\infty \left( \frac{x}{\sinh(x)} \right)^n e^{-x(n-2)} dx\\ &= \frac{2\pi^n}{(n-1)!}\cdot \frac{n-1}{ (2\pi)^n n} \int_{-\infty}^\infty \left( \frac{x}{\sinh(x)} \right)^n e^{-x(n-2)} dx \\ &= \mathrm{vol}(\mathbb{S}^{2n-1}) \frac{n-1}{ (2\pi)^n n}\int_{-\infty}^\infty \left( \frac{x}{\sinh(x)} \right)^n e^{-x(n-2)} dx. \end{align*} $$
Therefore, by the Tauberian Theorem due to Karamata, we obtain the following limit:
$$ \begin{align*}\lim_{\lambda \to \infty} \frac{N(\lambda)}{\lambda^n} = \mathrm{vol}(\mathbb{S}^{2n-1}) \frac{(n-1)}{n (2\pi)^n \Gamma(n+1) } \int_{-\infty}^\infty \left( \frac{x}{\sinh(x)} \right)^n e^{-x(n-2)} dx,\end{align*} $$
and complete the proof of Theorem 1.1.
3 The formula for functions versus the formula for forms
In [Reference Stanton and TartakoffST84], Stanton and Tartakoff prove the following formula, reminiscent of Weyl’s law, for the Kohn Laplacian on CR manifolds of hypersurface type.
Theorem 3.1 [Reference Stanton and TartakoffST84, Theorem 6.1]
Let M be a CR submanifold of
$\mathbb {C}^n$
,
$n \geq 3$
. Let
$N(\lambda )$
be the eigenvalue counting function of
$\square _b$
on M acting on
$(p,q)$
-forms,
$0 \leq p < n$
,
$0 < q < n-1$
. Then, we have the asymptotic equivalence
$$ \begin{align*}\lim_{\lambda \to \infty} \frac{N(\lambda)}{\lambda^n} = c_n \mathrm{vol}(M), \end{align*} $$
where
$$ \begin{align*}c_n = \binom{n-1}{p} \binom{n-1}{q} \frac{1}{(2\pi)^{n} \Gamma(n+1)} \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{n-1} e^{-(n-1-2q)\tau} \,d\tau, \end{align*} $$
and the Kohn Laplacian and the volume of M are defined with respect to a Levi metric.
For spheres embedded in
$\mathbb {C}^n$
, the induced metric is a Levi metric (see Definition 1.5 and the following remark in [Reference Stanton and TartakoffST84]). However, as stated, this only applies to
$(p,q)$
-forms with
$q\geq 1$
. We analyze how this expression relates to Theorem 1.1. Toward this end, we define the function
$$ \begin{align*}f(q) = \binom{n-1}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1})}{(2\pi)^{n} \Gamma(n+1)} \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{n-1} e^{-(n-1-2q)\tau} \,d\tau,\end{align*} $$
which, for
$q=1,\dots ,n-2$
, is the leading coefficient on
$\lambda ^{n}$
for the asymptotic growth of
$N(\lambda )$
, the eigenvalue counting function of
$\square _b$
on M acting on
$(0,q)$
-forms.
The following statement shows that this function is closely related to our formula.
Theorem 3.2 The definition of f given above is convergent for complex q satisfying
$0 < \Re (q) < n-1$
. Furthermore, f is holomorphic on this strip, and has an analytic continuation to a meromorphic function on the strip
$-1 < \Re (q) < n-1$
whose only pole is at
$q=0$
. Finally, the Laurent expansion of f about 0 is
$$ \begin{align*}f(q) = a_n/q^n + \mathrm{vol}(\mathbb{S}^{2n-1}) \frac{(n-1)}{n (2\pi)^n \Gamma(n+1) } \int_{-\infty}^\infty \left( \frac{x}{\sinh(x)} \right)^n e^{-x(n-2)} dx,\end{align*} $$
where
$a_n \neq 0$
.
In other words, the constant term in the Laurent expansion of f about 0 is the expression from Theorem 1.1.
Proof For notational convenience, let
$m = n - 1$
. Then, we may write f as
$$ \begin{align*}f(q) = \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1 })}{(2\pi)^{n} \Gamma(n+1)} \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{m} e^{-(m-2q)\tau} \,d\tau. \end{align*} $$
We first prove that the integrand is integrable (
$L^1$
) whenever
$0 < \Re (q) < m$
. For this, choose
$C> 0$
and
$\alpha> 0$
, so that whenever
$|x| \geq C$
,
$|\sinh (x)| \geq \alpha e^{|x|}$
. Therefore, for
$|x| \geq C$
, we have
$$ \begin{align*}\frac{x}{\sinh(x)} = \left|\frac{x}{\sinh(x)}\right| \leq \frac{|x|}{\alpha e^{|x|}}.\end{align*} $$
Because
$x/\sinh (x)$
has a removable singularity at 0, it is continuous and thus bounded on
$[-C,C].$
Hence,
$x^m/\sinh (x)^m$
is bounded on
$[-C,C]$
as well, so we can choose D, so that
$x^m/\sinh (x)^m \leq D$
whenever
$|x| \leq C$
. Thus, we have the bound
$$ \begin{align*}\frac{x^m}{\sinh(x)^m} \leq \frac{|x|^m}{\alpha^m e^{m |x|}} \mathbb{1}_{|x| \geq C} + D \mathbb{1}_{|x| \leq C}, \end{align*} $$
for all
$x \in \mathbb {R}$
.
Therefore, we may write
$$ \begin{align*}\int_{-\infty}^\infty& \left| \left(\frac{\tau}{\sinh \tau}\right)^{m} e^{-(m-2q)\tau} \right| \,d\tau = \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{m} e^{-(m-2\Re(q))\tau} \,d\tau \\ &\leq \alpha^{-m}\int_{-\infty}^C |\tau|^m e^{-m|\tau| - (m-2\Re(q))\tau} \, d\tau + \alpha^{-m} \int_C^{\infty} \tau^m e^{-m\tau - (m-2\Re(q))\tau} \, d\tau + \int_{-C}^{C} D \, d\tau \\ &= \alpha^{-m}\int_{C}^\infty \tau^m e^{-m\tau + (m-2\Re(q))\tau} \, d\tau + \alpha^{-m} \int_C^{\infty} \tau^m e^{-m\tau - (m-2\Re(q))\tau} \, d\tau + 2CD \\ &= \alpha^{-m}\int_C^\infty \tau^m e^{-2 \tau \Re(q)} \, d\tau + \alpha^{-m} \int_C^{\infty} \tau^m e^{-2\tau(m-\Re(q))} \, d\tau + 2CD. \end{align*} $$
A repeated integration by parts shows that these integrals are finite if
$0 < \Re (q) < m.$
To define the binomial coefficient, use the gamma function, i.e.,
$$ \begin{align*}\binom{m}{q} = \frac{m!}{\Gamma(q+1)\Gamma(m-q+1)}, \end{align*} $$
which is defined because
$\Gamma $
has no zeros. Thus, f is well defined on its domain of definition.
Now, rewrite f as follows:
$$ \begin{align*} f(q) &= \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1 })}{(2\pi)^{n} \Gamma(n+1)} \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{m} e^{-(m-2q)\tau} \,d\tau \\ &= \frac{1}{2} \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1 })}{(2\pi)^{n} \Gamma(n+1)} \Bigg( \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{m} e^{-(m-2q)\tau} \,d\tau\\ &\quad + \int_{-\infty}^\infty\left(\frac{\tau}{\sinh \tau}\right)^{m} e^{-(m-2q)\tau} \,d\tau \Bigg). \end{align*} $$
As
$\frac {x}{\sinh (x)}$
is an even function, the value of the second integral is unchanged if we change the
$e^{-(m-2q)\tau }$
in the integrand to
$e^{(m-2q)\tau }$
. Therefore,
$$ \begin{align*} f(q) &= \frac{1}{2}\binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1 })}{(2\pi)^{n} \Gamma(n+1)} \Bigg( \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{m} e^{-(m-2q)\tau} \,d\tau\\&\quad + \int_{-\infty}^{\infty} \left(\frac{\tau}{\sinh \tau}\right)^{m} e^{(m-2q)\tau} \,d\tau \Bigg) \\ &= \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1 })}{(2\pi)^{n} \Gamma(n+1)} \int_{-\infty}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{m} \cosh((m-2q)\tau) \,d\tau \\ &= 2 \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1 })}{(2\pi)^{n} \Gamma(n+1)} \int_{0}^\infty \left(\frac{\tau}{\sinh \tau}\right)^{m} \cosh((m-2q)\tau) \,d\tau, \end{align*} $$
where the last step follows because the integrand is even. Now, define the function g as follows:
$$ \begin{align*} g(q) = 2 \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1})}{(2\pi)^{n} \Gamma(n+1)} \int_{0}^{\infty} \tau^m \left(\frac{\cosh((m-2q)\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2q\tau} \right) \,d\tau. \end{align*} $$
We claim that g is holomorphic on the strip
$-1 < \Re (q) < m.$
Assuming this for now, consider
$$ \begin{align*} f(q) - g(q) = 2^m \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1})}{(2\pi)^{n} \Gamma(n+1)} \int_{0}^{\infty} \tau^m e^{-2q\tau} \,d\tau \end{align*} $$
defined for
$0 < \Re (q) < m$
. This is easy to evaluate explicitly with the substitution
$u = 2q\tau$
:
$$ \begin{align*} \int_{0}^{\infty} \tau^m e^{-2q\tau} \,d\tau = \frac{1}{(2q)^m}\int_{0}^{\infty} u^m e^{-u}\,du = \frac{\Gamma(m+1)}{(2q)^m}, \end{align*} $$
so
$$ \begin{align*} f(q) - g(q) = \binom{m}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1})}{(2\pi)^{n}(m+1)} \frac{1}{q^{m+1}} = \binom{n-1}{q} \frac{\mathrm{vol}(\mathbb{S}^{2n-1})}{n(2\pi)^{n}} \frac{1}{q^{n}}.\end{align*} $$
This is meromorphic in q on the whole complex plane. Thus, the function
$g + (f - g)$
is meromorphic in q for
$-1 < \Re (q) < n-1$
, and equal to f if
$0 < \Re (q) < n-1$
. Because f converges on its domain of definition, f is holomorphic in q, and
$g + (f -g)$
is the desired continuation. To complete the proof, we need to show that
$g(0)$
is our formula. We have
$$ \begin{align*} g(0) &= 2 \frac{\mathrm{vol}(\mathbb{S}^{2n-1})}{(2\pi)^{n} \Gamma(n+1)} \int_{0}^{\infty} \tau^m \left(\frac{\cosh(m\tau)}{(\sinh \tau)^m} - 2^{m-1} \right) \,d\tau \\ &= 2 \frac{2\pi^n/\Gamma(n)}{(2\pi)^{n} \Gamma(n+1)} \int_{0}^{\infty} \tau^m \left(\frac{e^{m\tau} + e^{-m\tau}}{2^{m-1}(e^\tau - e^{-\tau})^m} - 2^{m-1} \right) \,d\tau \\ &= \frac{1}{\Gamma(n)\Gamma(n+1)} \int_{0}^{\infty} \tau^m \left(\frac{e^{m\tau} + e^{-m\tau}}{(e^\tau - e^{-\tau})^m} - 1 \right) \,d\tau \\ &= \frac{1}{(n-1)! \,\Gamma(n+1)} \int_{0}^{\infty} \tau^m \left(\frac{e^{m\tau}}{(e^\tau - e^{-\tau})^m} + \frac{e^{-m\tau}}{(e^\tau - e^{-\tau})^m} - 1 \right) \,d\tau \\ &= \frac{1}{(n-1)! \, \Gamma(n+1)} \int_{0}^{\infty} \tau^m \left(\frac{1}{(1 - e^{-2\tau})^m} + \frac{1}{(e^{2\tau} - 1)^m} - 1 \right) \,d\tau \\ &= \frac{1}{\Gamma(n+1)} \lim_{t \to 0} t^n G(t) \\ &= \lim_{\lambda \to \infty} \frac{N(\lambda)}{\lambda^n}, \end{align*} $$
where we have used the expression appearing in the discussion after Proposition 2.13, and Karamata’s Theorem. Thus, we have our theorem, modulo showing that g is holomorphic for
$-1 < \Re (q) < m$
. We move on to this now. It clearly suffices to show that the expression
$$ \begin{align*} h(q) = \int_{0}^{\infty} \tau^m \left(\frac{\cosh((m-2q)\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2q\tau} \right) \,d\tau \end{align*} $$
is holomorphic for
$-1 < \Re (q) < m$
.
As a stepping stone toward proving that
$h(q)$
is holomorphic, we will first prove that, for
$\beta>0$
,
$$ \begin{align*} \phi(\beta) = \int_0^{\infty} e^{-2\beta \tau} \left(\frac{\tau}{1 - e^{-2\tau}}\right)^m d\tau\end{align*} $$
is convergent and continuous. By differentiating with respect to
$\beta $
, we see that the integrand is motonically increasing, for each
$\tau $
, as
$\beta \to 0.$
Fix some
$\beta _0> 0$
. Now, for
$\tau $
near 0, the integrand is bounded because it has a limit at 0. For large
$\tau $
, it is bounded by a constant times
$e^{-2 \beta \tau } \tau ^m$
, which is integrable on
$[0,\infty ).$
To see continuity at
$\beta _0$
, fix some
$\beta _1$
with
$0 < \beta _1 < \beta _0$
, and note that the integrand at
$\beta _1$
dominates the integrand at
$\beta $
for all
$\beta> \beta _1$
. Continuity of
$\phi $
at
$\beta _0$
follows then by dominated convergence. Thus, we have shown that
$\phi $
is convergent and continuous for
$\beta>0$
.
Next, we compute
$$ \begin{align*} \int_{0}^{\infty} \left| \tau^m \left(\frac{\cosh((m-2q)\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2q\tau} \right) \right| \,d\tau\end{align*} $$
$$ \begin{align*}= 2^{m-1}\int_{0}^{\infty} \tau^m \left|\frac{e^{(m-2q)\tau} + e^{-(m-2q)\tau}}{(e^\tau - e^{-\tau})^m} - e^{-2q\tau} \right| \,d\tau \end{align*} $$
$$ \begin{align*} \leq 2^{m-1}\int_{0}^{\infty} \tau^m \left|\frac{ e^{-(m-2q)\tau}}{(e^\tau - e^{-\tau})^m} \right| \,d\tau + 2^{m-1}\int_{0}^{\infty} \tau^m \left|\frac{e^{(m-2q)\tau} }{(e^\tau - e^{-\tau})^m} - e^{-2q\tau} \right| \,d\tau. \end{align*} $$
To analyze the first term, we have
$$ \begin{align*} \int_{0}^{\infty} \tau^m \left|\frac{ e^{-(m-2q)\tau}}{(e^\tau - e^{-\tau})^m} \right| \,d\tau &= \int_{0}^{\infty} \tau^m \frac{ e^{-(m-2\Re(q))\tau}}{(e^\tau - e^{-\tau})^m} \,d\tau\\& = \int_{0}^{\infty} \tau^m \frac{ e^{-(2m-2\Re(q))\tau}}{(1 - e^{-2\tau})^m} \,d\tau = \phi(m - \Re(q)). \end{align*} $$
For the second term, we have
$$ \begin{align*}\int_{0}^{\infty} \tau^m \left|\frac{e^{(m-2q)\tau} }{(e^\tau - e^{-\tau})^m} - e^{-2q\tau} \right| \,d\tau &= \int_{0}^{\infty} \tau^m \left|e^{-2q\tau} \left(\frac{e^{m\tau} }{(e^\tau - e^{-\tau})^m} - 1\right) \right| \,d\tau\\ &= \int_{0}^{\infty} \tau^m e^{-2\Re(q)\tau} \left| \frac{1}{(1 - e^{-2\tau})^m} - 1 \right| \,d\tau. \end{align*} $$
We bound the latter expression in the integrand by
$$ \begin{align*}\left| \frac{1}{(1 - e^{-2\tau})^m} - 1 \right| &= \left| \frac{1 - (1- e^{-2\tau})^m}{(1 - e^{-2\tau})^m} \right|\\& = \left| \frac{1 - \sum_{k=0}^m (-1)^k \binom{m}{k} e^{-2k\tau}}{(1 - e^{-2\tau})^m} \right| = \left| \frac{- \sum_{k=1}^m (-1)^k \binom{m}{k} e^{-2k\tau}}{(1 - e^{-2\tau})^m} \right|\\ &\leq \frac{ \sum_{k=1}^m \binom{m}{k} e^{-2k\tau}}{(1 - e^{-2\tau})^m} = e^{-2\tau} \frac{ \sum_{k=1}^m \binom{m}{k} e^{-2(k-1)\tau}}{(1 - e^{-2\tau})^m}\\ &\leq e^{-2\tau} \frac{ \sum_{k=1}^m \binom{m}{k}}{(1 - e^{-2\tau})^m} \leq e^{-2\tau}\frac{ 2^m}{(1 - e^{-2\tau})^m}.\end{align*} $$
Thus, for the second term, we have
$$ \begin{align*}\int_{0}^{\infty} \tau^m \left|\frac{e^{(m-2q)\tau} }{(e^\tau - e^{-\tau})^m} - e^{-2q\tau} \right| \,d\tau &\leq \int_{0}^{\infty} \tau^m e^{-2\Re(q)\tau} e^{-2\tau}\frac{ 2^m}{(1 - e^{-2\tau})^m}\\& = 2^m \phi(\Re(q) + 1).\end{align*} $$
In total, we have shown that
$$ \begin{align*}\int_{0}^{\infty} \left| \tau^m \left(\frac{\cosh((m-2q)\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2q\tau} \right) \right| \,d\tau &\leq 2^{m-1} \phi(m - \Re(q))\\ &\quad + 2^{2m-1} \phi(\Re(q) + 1). \end{align*} $$
Using this bound, we now show that h is holomorphic via Morera’s Theorem. Fix some triangle
$\Delta \subset \{q \in \mathbb {C}: -1 < \Re (q) < m\}.$
Parameterize
$\Delta $
by arc length with the piecewise differentiable curve
$\gamma (t)$
,
$a \leq t \leq b$
(so
$|\gamma '(t)| = 1 )$
for all t). Then,
$$ \begin{align*}\int_{\Delta} h(q) \,dq = \int_a^b \int_{0}^\infty h(\gamma(t)) \gamma'(t) \,d\tau \,dt.\end{align*} $$
The estimate above,
$$ \begin{align*}\int_{0}^{\infty}& \left| \tau^m \left(\frac{\cosh((m-2 \gamma(t) )\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2 \gamma(t) \tau} \right) \gamma'(t) \right| \,d\tau\\ &\leq 2^{m-1} \phi(m - \Re(\gamma(t))) + 2^{2m-1} \phi(\Re(\gamma(t)) + 1),\end{align*} $$
is uniformly bounded in t by the compactness of
$\Delta $
and the continuity of the upper bound in t, so
$$ \begin{align*} \int_a^b \int_{0}^\infty \left| \tau^m \left(\frac{\cosh((m-2 \gamma(t) )\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2 \gamma(t) \tau} \right) \gamma'(t) \right| \,d\tau \,dt < \infty. \end{align*} $$
Therefore, we may apply Fubini’s Theorem to see that
$$ \begin{align*} \int_a^b \int_{0}^\infty \tau^m \left(\frac{\cosh((m-2 \gamma(t) )\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2 \gamma(t) \tau} \right) \gamma'(t) \,d\tau \,dt \end{align*} $$
$$ \begin{align*} = \int_{0}^{\infty} \int_{a}^b \tau^m \left(\frac{\cosh((m-2 \gamma(t) )\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2 \gamma(t) \tau} \right) \gamma'(t) \,dt \,d\tau = \int_{0}^{\infty} 0 \, d\tau = 0\end{align*} $$
by Cauchy’s Theorem. This shows (by Morera’s Theorem) that
$$ \begin{align*} h(q) = \int_{0}^{\infty} \tau^m \left(\frac{\cosh((m-2q)\tau))}{(\sinh \tau)^m} - 2^{m-1} e^{-2q\tau} \right) \,d\tau \end{align*} $$
is a holomorphic function of q for
$-1 < \Re (q) < m$
, and thus
$g(q)$
is holomorphic for
$-1 < \Re (q) < m$
, completing the proof. ▪
Acknowledgment
We would like to thank the referees for constructive feedback. We would also like to thank Mohit Bansil and Michael Dabkowski for careful comments on an earlier version of this paper. This research was conducted at the NSF REU Site (DMS-1950102 and DMS-1659203) in Mathematical Analysis and Applications at the University of Michigan–Dearborn. We would also like to thank the National Science Foundation, National Security Agency, and University of Michigan–Dearborn for their support.
