Proof Substituting (3.3) to (3.4), applying [Reference Olver, Lozier, Boisvert and Clarke8, 15.8.1] and making the change of variable $x=(1+t)^{-1}$ , we obtain

(3.6) $$ \begin{align} \frac{\hat{g}_{2}(u,v,\rho,k;w)}{2(2\pi)^{2u}\cos(\pi u)}&= \int_0^{\infty}t^{w-v+\rho/2-1/2-k}(1+t)^{v-w+\rho/2-1/2} \nonumber\\& \quad \times {}_{2}F_{1} \left( \begin{matrix} k-u+v,k+u+v \\ 2k \end{matrix}; -t^{-1} \right) dt. \end{align} $$

Using the Mellin integral representation [Reference Olver, Lozier, Boisvert and Clarke8, 15.6.6] of the hypergeometric function in (3.6), changing the orders of integration and evaluating the integral over t with the help of [Reference Olver, Lozier, Boisvert and Clarke8, 5.12.3], we find

(3.7) $$ \begin{align} \frac{\hat{g}_{2}(u,v,\rho,k;w)}{2(2\pi)^{2u}\cos(\pi u)} & = \frac{\Gamma(k-u- v)}{\Gamma(k+u+v)} \frac{1}{2\pi i} \int_{(\Delta)}\frac{\Gamma(k-u+v+s)\Gamma(k+u+v+s)}{\Gamma(2k+s)} \nonumber\\ &\quad\times \frac{\Gamma(1/2+\rho/2-v+w-k-s)}{\Gamma(1/2-\rho/2-v+w)} \Gamma\left(-s\right)\Gamma\left(k-\rho+s\right) ds, \end{align} $$

where

$$\begin{align*}\max(\Re{(\rho)}-k,|\Re{u}|-\Re{(v)}-k)<\Delta<\min(1/2+\Re{(\rho/2-v+w)}-k,0).\end{align*}$$

Finally, making the change of variable $s=z-k$ , we obtain (3.5).

Proof Moving the line of integration in (3.5) to the right on $\Re {z}=\sigma <k$ , we cross poles at the points

(3.9) $$ \begin{align} z_p(j)=1/2+\rho/2-v+ir+j \quad\text{for}\quad j=0,1,2,\ldots. \end{align} $$

The residues are estimated in the following way:

(3.10) $$ \begin{align} \frac{\Gamma(k-u- v)}{\Gamma(k+u+v)} \frac{\Gamma(k-1/2-\rho/2+v-ir-j)}{\Gamma(k+1/2+\rho/2-v+ir+j)} &\frac{\Gamma\left(1/2-\rho/2-v+ir+j\right)}{\Gamma(1/2-\rho/2-v+ir)}\nonumber\\\times\ \Gamma\left(1/2+\rho/2+ir+ j-u\right)&\Gamma\left(1/2+\rho/2+ir+j+u\right)\ll \nonumber\\ \ll \frac{k^{4\delta}(1+|r|)^{3j+\delta}e^{-\pi|r|}}{(k+|r|)^{1+2j-3\delta}} &\ll \frac{(k(|r|+1))^{\epsilon}}{k(1+|r|)^A}. \end{align} $$

Therefore,

(3.11) $$ \begin{align} \hat{g}_{2}&(u,v,\rho,k;ir)\ll\frac{(k(|r|+1))^{\epsilon}}{k(1+|r|)^A}+ \bigg|\frac{k^{4\delta}}{2\pi i} \int_{(\sigma)}\frac{\Gamma(k-z)}{\Gamma(k+z)} \nonumber\\ &\times \frac{\Gamma(1/2+\rho/2-v+ir-z)}{\Gamma(1/2-\rho/2-v+ir)} \Gamma\left(z-u+v\right)\Gamma\left(z+u+v\right)\Gamma\left(z-\rho\right) dz\bigg|. \end{align} $$

Let $z=\sigma +iy$ and $y_0=C\log ^2(k(1+|r|))$ , where the constant C is chosen in such way that $|r-y_0|>r/2$ . To estimate the integral (3.11), we apply (3.2) and split it in two parts, the first one over $|y|>y_0$ and the second over $|y|\le y_0$ . The first integral is smaller than

(3.12) $$ \begin{align} \int_{|y|>y_0}\frac{|y|^{3\sigma+3\delta-3/2}e^{-\pi(3|y|+|y-r|-|r|)/2}} {(|y-r|+1)^{\sigma-3\delta/2}(|r|+1)^{-3\delta/2}(k+|y|)^{2\sigma}}dy\ll \frac{1}{(k(1+|r|))^A}. \end{align} $$

The second integral is smaller than

(3.13) $$ \begin{align} &\int_{|y|\le y_0} \frac{|y|^{3\sigma+3\delta-3/2}(|y-r|+1)^{-\sigma+3\delta/2}} {(|r|+1)^{-3\delta/2}(k+|y|)^{2\sigma}}dy\ll \nonumber\\ &\quad\ll \int_{|y|\le y_0} \frac{(k(1+|r|))^{\epsilon}|y|^{3\sigma-3/2}dy} {(k+|y|)^{2\sigma}(|y-r|+1)^{\sigma}}\ll \frac{(k(1+|r|))^{\epsilon}}{(k^2(1+|r|))^{\sigma}}. \end{align} $$

Substituting (3.12) and (3.13) to (3.11), we prove (3.8).

Proof Using (3.5) with $\Delta =1-\epsilon _0$ ( $\epsilon _0$ is chosen in such way that all poles of $\Gamma (1-v\pm u-z)$ are located to the right of the line $\Re {z}=\Delta $ ), writing $z=\Delta +iy$ and applying (3.2), we obtain

(3.15) $$ \begin{align} \hat{g}_{2}(u,v,\rho,k;1/2-\rho/2\pm u)\ll k^{4\delta} \int_{-\infty}^{\infty}\frac{(1+|y|)^{2\Delta+3\delta-1}e^{-2\pi|y|}} {(k+|y|)^{2\Delta}}dy\ll \frac{k^{\epsilon}}{k^2}. \end{align} $$

Note that estimating the integral representation [Reference Frolenkov6, (4.15)] of $\hat {g}_{2}(\ldots )$ by absolute value with the use of the Stirling bound (3.2) on Gamma factors, we are able to obtain only a weaker estimate $k^{-1+\epsilon }$ . This estimate is sufficient for our further computations.

Proof It follows from [Reference Gradshteyn, Ryzhik, Jeffrey and Zwillinger7, 6.574.1] that for $-\Re {(k+u)}<\Re {(v)}<1/2,$ we have

(3.20) $$ \begin{align} F(v,u;y)=(1-y)^v2^{-2v}\frac{\Gamma(k-u-v)\Gamma(1+2u)}{\Gamma(k+u+v)} \int_{0}^{\infty}J_{2u}(t\sqrt{y})J_{2k-1}(t)t^{2v}dt. \end{align} $$

Substituting (3.20) to (3.17), we obtain, for $-\Re {(k-|u|)}<\Re {(v)}<1/2,$

(3.21) $$ \begin{align} \tilde{\phi}_k(1-y;u,v)&=(-1)^k\frac{(2\pi)^{2u}\pi(1-y)^{v}2^{-2v}} {\sin{(\pi v)}\sin{(2\pi u)}}\frac{\Gamma(k-v-u)}{\Gamma(k+v+u)} \nonumber\\& \quad \times \sum_{\pm}\pm\sin{\pi(v\pm u)} \int_{0}^{\infty}J_{\pm2u}(t\sqrt{y})J_{2k-1}(t)t^{2v}dt. \end{align} $$

Applying [Reference Olver, Lozier, Boisvert and Clarke8, 1.14.36] together with [Reference Erdelyi, Magnus, Oberhettinger and Tricomi3, p. 326 f.(1)], we prove that

(3.22) $$ \begin{align} \int_{0}^{\infty}J_{2u}(t\sqrt{y})J_{2k-1}(t)t^{2v}dt= \frac{1}{2\pi i}\int_{(c)} \frac{\Gamma(k+v-s/2)\Gamma(u+s/2)}{\Gamma(k-v+s/2)\Gamma(1+u-s/2)} \frac{2^{2v-1}ds}{y^{s/2}}. \end{align} $$

Next, we change the variable s to $1+2v-s$ , and then apply [Reference Olver, Lozier, Boisvert and Clarke8, 5.5.3] to transform $\Gamma (1/2+u-v+s/2)$ . As a result,

(3.23) $$ \begin{align} \int_{0}^{\infty}&J_{2u}(t\sqrt{y})J_{2k-1}(t)t^{2v}dt= \frac{1}{2\pi i}\int_{(c)} \frac{\Gamma(k-1/2+s/2)}{\Gamma(k+1/2-s/2)} \nonumber\\ &\times \Gamma(1/2+u+v-s/2)\Gamma(1/2-u+v-s/2)\cos\pi(u-v+s/2) \frac{2^{2v-1}ds}{\pi y^{1/2+v-s/2}}. \end{align} $$

Substituting (3.23) to (3.21), we have

(3.24) $$ \begin{align} \tilde{\phi}_k(1-y;u,v)&= (-1)^k \frac{(2\pi)^{2u}(1-y)^{v}}{2\sin{(\pi v)}y^{1/2+v}} \frac{\Gamma(k-v-u)}{\Gamma(k+v+u)}\nonumber\\ & \quad \times \frac{1}{2\pi i}\int_{(c)} \frac{\Gamma(k-1/2+s/2)}{\Gamma(k+1/2-s/2)}\Gamma(1/2+u+v-s/2) \nonumber\\ &\quad\times \Gamma(1/2-u+v-s/2)\cos\pi(2v-s/2) y^{s/2}ds. \end{align} $$

According to [Reference Olver, Lozier, Boisvert and Clarke8, 5.12.1] for $\Re {(-\rho /2-1/2-v)}<\Re {w}<\Re {(\rho /2+s/2-v)},$ we have

(3.25) $$ \begin{align} &\int_0^1(y(1-y))^{\rho/2-1/2} (1-y) {}^{w+v}y^{s/2-1/2-v-w}dy = \frac{\Gamma(\rho/2+s/2-v-w)\Gamma(\rho/2+1/2+v+w)}{\Gamma(\rho+s/2+1/2)}. \end{align} $$

Using (3.16), (3.18), (3.24) and evaluating the resulting integral over y using (3.25), we prove (3.19).

Proof The proof is similar to the one of [Reference Frolenkov6, Lemma 4.8]. We move the line of integration in (3.19) to the left, crossing the pole at $s=\pm 2v+2w-\rho $ . Evaluating the residues and letting $w=1/2-\rho /2\pm v,$ we obtain (3.26). To prove (3.29), we estimate Gamma factors in (3.27) and (3.28) by means of the Stirling formula (3.2). Taking into account that $|v|+|u|+|\rho |<\epsilon $ and choosing $\Delta _1=\Delta _2=0,$ we have

$$ \begin{align*} J_{1}(u,\pm v,\rho,k)\ll k^{\epsilon} \int_{-\infty}^{\infty}\frac{(k+|y|)^{\Delta_1-1}}{(1+|y|)^{\Delta_1+1}}dy\ll \frac{k^{\epsilon}}{k}, \end{align*} $$

$$ \begin{align*} J_{2}(u,\pm v,\rho,k)\ll k^{\epsilon} \int_{-\infty}^{\infty}\frac{(k+|y|)^{\Delta_2-1}}{(1+|y|)^{\Delta_2+1\pm2\Re{v}}}dy\ll \frac{k^{\epsilon}}{k}, \end{align*} $$

thus proving (3.29).

Proof Moving the line of integration in [Reference Frolenkov6, (4.28)] to the left and evaluating residues at the points

$$ \begin{align*} s_1(j)=1-2k-2j,\quad s_2(j)=-\rho-2u-2w-2j\quad j=0,1,2\ldots, \end{align*} $$

we obtain (see also [Reference Frolenkov6, Lemma 4.11])

(3.31) $$ \begin{align} &\hat{g}_{1}(u,v,\rho,k;w)=-2(-1)^k(2\pi)^{2u} \Gamma\left(\frac{1+\rho}{2}-u-w\right) \frac{\sin(\pi\rho)\cos(\pi u)} {\cos\pi(\rho/2+u+w)}\nonumber\\& \quad\times {}_{3}\mathrm{I}_{2} \left( \begin{matrix} k-u+v,k-u-v, k-\rho \\ 2k, k+\frac{1-\rho}{2}-u-w \end{matrix}; 1 \right) + 2(-1)^k(2\pi)^{2u} \Gamma\left(\frac{1+\rho}{2}-u-w\right) \nonumber\\& \quad \times \frac{\sin\pi(\rho/2+w)\cos\pi\left(\rho/2-u-w\right)} {\cos\pi\left(\rho/2+u+w\right)} {}_{3}\mathrm{I}_{2} \left( \begin{matrix} \frac{1+\rho}{2}+v+w, \frac{1+\rho}{2}-v+w, \frac{1-\rho}{2}+u+w, \\ k+\frac{1+\rho}{2}+u+w, \frac{3+\rho}{2}+u+w-k \end{matrix}; 1 \right).\nonumber\\ \end{align} $$

Rewriting (3.31) in terms of $\alpha _1,\alpha _2,\alpha _3$ using (2.1) and applying (1.10), [Reference Frolenkov6, (4.16)], we prove (3.30).

Proof Applying [Reference Prudnikov, Brychkov and Marichev9, (7.4.4.2)] with

$$ \begin{align*} a=\frac{1+\alpha_1-\alpha_2+\alpha_3}{2}-w,\, b=\frac{1+\alpha_1+\alpha_2-\alpha_3}{2}-w,\, c=\frac{1-\alpha_1+\alpha_2+\alpha_3}{2}-w, \end{align*} $$

$$ \begin{align*} d=\frac{3+\alpha_1+\alpha_2+\alpha_3}{2}-w-k,\, e=k+\frac{1+\alpha_1+\alpha_2+\alpha_3}{2}-w,\, s=\frac{1+\alpha_1+\alpha_2+\alpha_3}{2}+w, \end{align*} $$

we deduce

(3.35) $$ \begin{align} \mathcal{H}_1(\bar{\alpha},k;w)& = \frac{\Gamma\left(\frac{1+\alpha_1+\alpha_2-\alpha_3}{2}-w\right) \Gamma\left(\frac{1-\alpha_1+\alpha_2+\alpha_3}{2}-w\right)}{ \Gamma\left(1+\alpha_1+\alpha_2\right)\Gamma\left(1+\alpha_2+\alpha_3\right)} \nonumber\\& \quad \times \Gamma\left(\frac{1+\alpha_1+\alpha_2+\alpha_3}{2}+w\right) {}_{3}F_{2} \left( \begin{matrix} 1+\alpha_2-k,k+\alpha_2, \frac{1+\alpha_1+\alpha_2+\alpha_3}{2}+w \\ 1+\alpha_1+\alpha_2,1+\alpha_2+\alpha_3 \end{matrix}; 1 \right). \end{align} $$

This formula can be viewed as an analog of [Reference Frolenkov6, (4.41)]. For the hypergeometric function on the right-hand side of (3.35), we use the integral representation [Reference Olver, Lozier, Boisvert and Clarke8, 16.5.2] with

$$ \begin{align*} a_0=\frac{1+\alpha_1+\alpha_2+\alpha_3}{2}+w,\quad b_0=1+\alpha_1+\alpha_2, \end{align*} $$

thus proving (3.34).

Proof We apply [Reference Farid Khwaja and Olde Daalhuis4, Theorem 3.2] with

$$ \begin{align*} z=-\cos(2y),\,\lambda=k-\frac{1}{2},\, a=\frac{1}{2}+\alpha_2,\,c=1+\alpha_2+\alpha_3. \end{align*} $$

In this case, [Reference Farid Khwaja and Olde Daalhuis4, (3.9)] becomes

$$ \begin{align*} \xi=\log\left(-z-i\sqrt{1-z^2} \right)=-2iy, \end{align*} $$

and [Reference Farid Khwaja and Olde Daalhuis4, (3.10)] proves (3.37). Finally, the K-Bessel functions of purely imaginary argument in [Reference Farid Khwaja and Olde Daalhuis4, (3.8)] can be transformed into combination of Y and J-Bessel functions by means of [Reference Olver, Lozier, Boisvert and Clarke8, 10.27.8 and 10.4.3]

$$ \begin{align*} K_{\nu}(2iy\lambda)=\frac{-\pi i}{2}e^{-\nu\pi i/2}H_{\nu}^{(2)}(2y\lambda)=\frac{-\pi i}{2}e^{-\nu\pi i/2} \left( J_{\nu}(2y\lambda)-iY_{\nu}(2y\lambda)\right), \end{align*} $$

$$ \begin{align*} K_{\nu}(-2iy\lambda)=\frac{\pi i}{2}e^{\nu\pi i/2}H_{\nu}^{(1)}(2y\lambda)=\frac{\pi i}{2}e^{\nu\pi i/2}\left( J_{\nu}(2y\lambda)+iY_{\nu}(2y\lambda) \right). \end{align*} $$

After some straightforward computations, we obtain (3.36).

Proof This is [Reference Farid Khwaja and Olde Daalhuis5, (48)]. The estimate (3.41) of the error term can be obtained in the same way as in [Reference Farid Khwaja and Olde Daalhuis4, Theorem 3.2].

Proof This is [Reference Farid Khwaja and Olde Daalhuis5, (74)]. The estimate (3.44) of the error term can be obtained in the same way as in [Reference Farid Khwaja and Olde Daalhuis4, Theorem 3.2].

Proof Using [Reference Temme10, (10.1.9) and (10.3.46)], we first show that

(3.51) $$ \begin{align} {}_{1}F_{1} \left( \begin{matrix} \alpha-a \\ c \end{matrix}; az \right) = \frac{\Gamma(1-\alpha+a)\Gamma(c)}{\Gamma(c+a-\alpha)}\frac{e^{az/2}}{2\pi i}\int_{L}e^{(c+a-\alpha)s-az/s}f(-az,-s)\frac{ds}{s^c}, \end{align} $$

where L is the Hankel contour (see [Reference Temme10, Figure 2.2]), and (see [Reference Temme10, (10.3.28)])

(3.52) $$ \begin{align} f(z,s)=e^{zg_0(s)}\left(\frac{s}{1-e^{-s}}\right)^{c},\quad g_0(s)=\frac{1}{s}-\frac{1}{e^s-1}-\frac{1}{2}. \end{align} $$

Note that

(3.53) $$ \begin{align} e^{cs}f(-z,-s)=f(az,s), \end{align} $$

and therefore,

(3.54) $$ \begin{align} e^{az/2}e^{(c+a)s-az/s}f(-az,-s)=e^{as-az/(e^s-1)}\left(\frac{s}{1-e^{-s}}\right)^{c}. \end{align} $$

Substituting (3.54) to (3.51), we obtain an analog of [Reference Temme10, (27.4.65) and (27.4.66)]:

(3.55) $$ \begin{align} {}_{1}F_{1} \left( \begin{matrix} \alpha-a \\ c \end{matrix}; az \right) = \frac{\Gamma(1-\alpha+a)\Gamma(c)}{\Gamma(c+a-\alpha)}\frac{1}{2\pi i}\int_{L}e^{a\psi(w)}\tilde{g}(w)\frac{dw}{w^c}, \end{align} $$

where

(3.56) $$ \begin{align} \psi(w)=w-\frac{z}{e^w-1},\quad \tilde{g}(w)=e^{-\alpha w}\left(\frac{w}{1-e^{-w}}\right)^{c}. \end{align} $$

The only difference with [Reference Temme10, (27.4.66)] is the presence of the multiple $e^{-\alpha w}$ in (3.56). Proceeding as in [Reference Temme10], we obtain [Reference Temme10, (27.4.69)] with $p(t)$ being replaced by

(3.57) $$ \begin{align} \tilde{p}(t)=\left(\frac{w}{t}\right)^{-c}\tilde{g}(w)\frac{dw}{dt}= e^{-\alpha w}\left(\frac{1-e^{-w}}{t}\right)^{-c+2}\frac{t^2+\gamma^2}{(1-e^{-w})^2+ze^{-w}}, \end{align} $$

and [Reference Temme10, (27.4.71)] with $A_k,B_k$ being replaced by

(3.58) $$ \begin{align} \tilde{p}_j(t)=-t^c\frac{d}{dt}(t^{1-c}\tilde{q}_{j-1}(t))=\tilde{A}_j(a,z)+\tilde{B}_j(a,z)t+(t+\gamma^2/t)\tilde{q}_{j}(t), \end{align} $$

(3.59) $$ \begin{align} \tilde{A}_j(a,z)=\frac{\tilde{p}_j(i\gamma)+\tilde{p}_j(i\gamma)}{2},\quad \tilde{B}_j(a,z)=\frac{\tilde{p}_j(i\gamma)-\tilde{p}_j(i\gamma)}{2i\gamma}, \end{align} $$

where $\tilde {p}_0(t)=\tilde {p}(t).$ The closed formulas for the first coefficients are (see [Reference Temme10, (27.4.74)]):

(3.60) $$ \begin{align} \tilde{A}_0(a,z)=\left(\frac{\gamma}{2\sin\theta}\right)^{c}\sqrt{\frac{2\tan\theta}{\gamma}}\cos(c-2\alpha)\theta, \end{align} $$

(3.61) $$ \begin{align} \tilde{B}_0(a,z)=\left(\frac{\gamma}{2\sin\theta}\right)^{c}\sqrt{\frac{2\tan\theta}{\gamma}}\frac{\sin(c-2\alpha)\theta}{\gamma}. \end{align} $$

Finally, the estimate (3.50) of the error term can be obtained in the same way as in [Reference Farid Khwaja and Olde Daalhuis4, Theorem 3.2].

Proof Using [Reference Temme10, (10.1.11) and (10.3.61)], we prove an analog of [Reference Temme10, (27.4.80)]:

(3.65) $$ \begin{align} \frac{U(\alpha-a,c,az)}{\Gamma(c+a-\alpha)}&= \frac{e^{\mp\pi i(a-\alpha)}}{\Gamma(c)} {}_{1}F_{1} \left( \begin{matrix} \alpha-a \\ c \end{matrix}; az \right) \nonumber\\ &\quad - \frac{e^{az}e^{\pm\pi ic}}{\Gamma(\alpha-a)}U(c+a-\alpha,c,aze^{\pm\pi i}). \end{align} $$

For the $_{1}F_{1}$ function in (3.65), we can apply (3.48). Therefore, we are left to consider $U(c+a-\alpha ,c,az)$ (at the end, we will replace $az$ by $aze^{\pm \pi i}$ ). Applying [Reference Temme10, (10.3.27)] and (3.53), we prove an analog of [Reference Temme10, (10.3.62)]:

(3.66) $$ \begin{align} U(c+a-\alpha,c,az)=\frac{e^{az/2}}{\Gamma(c-\alpha+a)}\int_0^{\infty}e^{-as-az/s}e^{\alpha s}f(-az,-s)\frac{ds}{s^c}, \end{align} $$

where the function $f(\cdot ,\cdot )$ is defined by (3.52). Simplifying the function under the integral in (3.66), we have the following analog of [Reference Temme10, (27.4.81)]:

(3.67) $$ \begin{align} U(c+a-\alpha,c,az)=\frac{1}{\Gamma(c-\alpha+a)}\int_0^{\infty}e^{-a\varphi(w)} \tilde{g}(-w)\frac{dw}{w^c}, \end{align} $$

where

(3.68) $$ \begin{align} \varphi(w)=w+\frac{z}{e^w-1},\quad \tilde{g}(-w)=e^{\alpha w}\left(\frac{e^{w}-1}{w}\right)^{-c}. \end{align} $$

The only difference with [Reference Temme10, (27.4.81)] is the presence of the factor $e^{\alpha w}$ . Arguing in the same way as in [Reference Temme10, Section 27.4.1], making the change of variable [Reference Temme10, (27.4.32)], we obtain an analog of [Reference Temme10, (27.4.38) and (27.4.82)], namely,

(3.69) $$ \begin{align} U(c+a-\alpha,c,az)=\frac{e^{az/2}}{\Gamma(c-\alpha+a)}\int_0^{\infty}e^{-a(t+\beta^2/t)}\tilde{f}(-t)\frac{dt}{t^c}, \end{align} $$

where

(3.70) $$ \begin{align} \beta=\frac{w_0+\sinh w_0}{2},\quad w_0=2\operatorname{\mathrm{arcsinh}}\frac{\sqrt{z}}{2} \end{align} $$

and

(3.71) $$ \begin{align} \tilde{f}(-t)=e^{\alpha w(t)}\left(\frac{e^{w(t)}-1}{t}\right)^{-c+2}\frac{t^2-\beta^2}{(e^{w(t)}-1)^2-ze^{w(t)}} \end{align} $$

is an analog of [Reference Temme10, (27.4.41)]. Note that $\tilde {f}(t)$ coincides with $\tilde {p}(t)$ defined in (3.57) after changing z to $-z$ . Therefore, we show the following analog of [Reference Temme10, (27.4.84)]:

(3.72) $$ \begin{align} U(\alpha-a,c,aze^{\pm\pi i})& = 2\gamma^{1-c}e^{\pm \pi i(1-c)/2}\frac{e^{-az/2}}{\Gamma(c-\alpha+a)} \nonumber\\ &\quad \times\left( K_{1-c}(\pm 2i\gamma a) \sum_{j=0}^{\infty}\frac{\tilde{A}_{j}(a,z)}{a^{j}}\mp i\gamma K_{c-2}(\pm2i\gamma a) \sum_{j=0}^{\infty}\frac{\tilde{B}_{j}(a,z)}{a^{j}} \right). \end{align} $$

Using the relation [Reference Olver, Lozier, Boisvert and Clarke8, (10.27.3)], i.e., $K_{-\nu }(z)=K_{\nu }(z)$ , and [Reference Temme10, (10.3.64)], we infer

(3.73) $$ \begin{align} e^{\pm \pi i(c+1)/2}K_{c-1}(\pm 2i\gamma a)=\frac{\pi}{2}\left( Y_{c-1}(2\gamma a)\pm iJ_{c-1}(2\gamma a)\right) \end{align} $$

(3.74) $$ \begin{align} e^{\pm \pi i(c+1)/2}K_{c-2}(\pm 2i\gamma a)=\frac{\pi}{2}e^{\pm \pi i/2}\left( Y_{c-2}(2\gamma a)\pm iJ_{c-2}(2\gamma a)\right). \end{align} $$

Truncating the series (3.72) at the point $j=n$ with the error (3.64) (see the proof of [Reference Farid Khwaja and Olde Daalhuis4, Theorem 3.2]), we obtain

(3.75) $$ \begin{align} &U(\alpha-a,c,aze^{\pm\pi i}) = \pi\gamma^{1-c}e^{\mp\pi ic}\frac{e^{-az/2}}{\Gamma(c-\alpha+a)}\nonumber\\& \quad \times \left( O(\Phi^C_n(a,z)) + \left( Y_{c-1}(2\gamma a)\pm iJ_{c-1}(2\gamma a)\right) \sum_{j=0}^{n-1}\frac{\tilde{A}_{j}(a,z)}{a^{j}}\right.\nonumber\\& \left.\qquad\quad +\ \gamma \left( Y_{c-2}(2\gamma a)\pm iJ_{c-2}(2\gamma a)\right) \sum_{j=0}^{n-1}\frac{\tilde{B}_{j}(a,z)}{a^{j}} \right). \end{align} $$

Substituting (3.48) and (3.75) into (3.65), we have

(3.76) $$ \begin{align} &\frac{U(\alpha-a,c,az)}{\Gamma(c+a-\alpha)} = \frac{\Gamma(1-\alpha+a)}{\Gamma(c+a-\alpha)}e^{az/2}\gamma^{1-c} \nonumber\\&\quad\times \left(O(\Phi^C_n(a,z))+ C_{c-1}(2\gamma a) \sum_{j=0}^{n-1}\frac{\tilde{A}_{j}(a,z)}{a^{j}}+ \gamma C_{c-2}(2\gamma a) \sum_{j=0}^{n-1}\frac{\tilde{B}_{j}(a,z)}{a^{j}} \right), \end{align} $$

where

(3.77) $$ \begin{align} C_{\nu}(2\gamma a)=e^{\mp\pi i(a-\alpha)}J_{\nu}(2\gamma a)-\sin\pi(\alpha-a)\left( Y_{\nu}(2\gamma a)\pm iJ_{\nu}(2\gamma a)\right). \end{align} $$

Opening the brackets in (3.77), we complete the proof of (3.63).

Proof of Proposition 3.7

First, we decompose the integral in (3.34) smoothly (say by inserting functions $\chi _0(x),\chi _{1/2}(x),\chi _1(x)$ ) into three ranges:

(3.78) $$ \begin{align} 0<x<\delta_0,\quad \delta_0<x<1-\delta_1,\quad 1-\delta_1<x<1, \end{align} $$

where $\delta _0,\delta _1$ are some small constants.

Consider first the part over $0<x<\delta _0$ . In this case, we apply Lemma 3.10 followed by Lemma 3.12. Taking n in (3.39) and (3.48) sufficiently large, we obtain that the contribution of the error terms is negligible, and thus, it is enough to investigate only the contribution of the main term (all other terms will have the same structure and will be smaller) which is bounded by the sum of the integrals of the following type:

(3.79) $$ \begin{align} &e^{-\pi|\Im{w}|/2}(1+|w|)^{\epsilon}k^{\epsilon} \sum_{j=0}^2\int_0^{1}\chi_0(x) (1-x)^{\frac{-1+\alpha_1+\alpha_2-\alpha_3}{2}-w} \nonumber\\ &\quad\times x^{\frac{-1+\alpha_1+\alpha_2+\alpha_3}{2}+w}A_j(x)\gamma(x)^{-\nu_j} J_{\nu_j}(2\gamma(x) k)dx, \end{align} $$

where $\nu =-j+\alpha _2+\alpha _3$ for $j=0,1,2$ and $A_j(x)$ are products of coefficients in (3.39) and (3.48). Note that

(3.80) $$ \begin{align} \gamma(x)=\frac{\arccos(1-2x)}{2}=\sqrt{x}+O(x^{3/2}), \end{align} $$

since we first perform transformations (3.40) and then (3.49) with $z=2\zeta _0$ . Now, we decompose the integral (3.79) smoothly at the point $x=k^{-2+\epsilon }$ . For $0<x<k^{-2+\epsilon }$ using [Reference Olver, Lozier, Boisvert and Clarke8, (10.7.3)] and (3.80), we show that $ \gamma (x)^{-\nu _j}J_{\nu _j}(2\gamma (x) k)\ll k^{\nu _j}$ , and therefore,

(3.81) $$ \begin{align} \int_0^{k^{-2+\epsilon}} (1-x)^{\frac{-1+\alpha_1+\alpha_2-\alpha_3}{2}-w} x^{\frac{-1+\alpha_1+\alpha_2+\alpha_3}{2}+w}A_j(x)\gamma(x)^{-\nu_j} J_{\nu_j}(2\gamma(x) k)dx\ll k^{-1+\epsilon}. \end{align} $$

In the case of $k^{-2+\epsilon }<x<\delta _0$ , we apply the asymptotic formula [Reference Gradshteyn, Ryzhik, Jeffrey and Zwillinger7, 8.451.1] for the Bessel function. The error term is again negligible and we are left to estimate the integral of the following type (here, we set $w=ir$ ):

(3.82) $$ \begin{align} \int_{k^{-2+\epsilon}}^{1}\chi_0(x)W(x) x^{\frac{-1+\alpha_1+\alpha_2+\alpha_3}{2}}\gamma(x)^{\alpha_2+\alpha_3} e^{ig_{\pm}(x)}\frac{dx}{\sqrt{\gamma(x) k}}, \end{align} $$

where $W(x)=(1-x)^{\frac {-1+\alpha _1+\alpha _2-\alpha _3}{2}}A_j(x)$ and the phase function is equal to

(3.83) $$ \begin{align} g_{\pm}(x)=\pm2\gamma(x) k+r\log x-r\log(1-x). \end{align} $$

Estimating it by absolute value using (3.79), we prove (3.32). For $|r|\ll \log ^2k$ , we have $g^{\prime }_{\pm }(x)\gg k/\sqrt {x}$ , and therefore, using the first derivative test, we estimate (3.82) as

(3.84) $$ \begin{align} \max_{k^{-2+\epsilon}<x<\delta_0} \frac{x^{\frac{\alpha_1+\alpha_2+\alpha_3}{2}}\gamma(x)^{\alpha_2+\alpha_3} }{k\sqrt{\gamma(x) k}}\ll\frac{k^{\epsilon}}{k}. \end{align} $$

As a result, in view of (3.79), we prove (3.33).

Consider the third part of the integral (3.34) over $1-\delta _1<x<1$ . In this case, we apply Lemma 3.11 followed by Lemma 3.13. Again the contribution of the error terms is negligible, and thus, it is enough to investigate only the contribution of the main term (all other terms will have the same structure and will be smaller), which is the sum of the integrals of the following type:

(3.85) $$ \begin{align} & e^{-\pi|\Im{w}|/2}(1+|w|)^{\epsilon}k^{\epsilon} \sum_{j=0}^2\int_0^{1}\chi_1(x) (1-x)^{\frac{-1+\alpha_1+\alpha_2-\alpha_3}{2}-w} \nonumber\\ & \qquad\qquad\qquad\qquad\qquad \times x^{\frac{-1+\alpha_1+\alpha_2+\alpha_3}{2}+w}D_j(x)\gamma(x)^{-\nu_j} C_{\nu_j}(2\gamma_1(x) k)dx, \end{align} $$

where $\nu _j=-j+\alpha _2-\alpha _3$ for $j=0,1,2$ , $D_j(x)$ are products of coefficients in (3.42) and (3.62), and by (3.63), we have

(3.86) $$ \begin{align} C_{\nu_j}(2\gamma_1(x) k)= (-1)^{k-1}\cos(\pi\alpha_2)J_{\nu_j}(2\gamma_1(x) k)+(-1)^k\sin(\pi\alpha_2)Y_{\nu_j}(2\gamma_1(x) k). \end{align} $$

Note that now since we first perform transformations (3.43) and then (3.49) with $z=2\zeta _1$ , we have

(3.87) $$ \begin{align} \gamma_1(x)=\frac{\arccos(2x-1)}{2}=\sqrt{1-x}+O((1-x)^{3/2}). \end{align} $$

Now, we decompose the integral in (3.85) smoothly $x=1-k^{-2+\epsilon }$ . For $1-k^{-2+\epsilon }<x<1$ using [Reference Olver, Lozier, Boisvert and Clarke8, (10.7.3)–(10.7.5)] and (3.87), we have

(3.88) $$ \begin{align} \gamma_1(x)^{-\nu_j}C_{\nu_j}(2\gamma_1(x) k)\ll \gamma_1(x)^{-\nu_j}(\gamma_1(x) k)^{-|\nu_j|}\ll (k(1-x))^{\epsilon}, \end{align} $$

and therefore,

(3.89) $$ \begin{align} \int_{1-k^{-2+\epsilon}}^1 (1-x)^{\frac{-1+\alpha_1+\alpha_2-\alpha_3}{2}-w} x^{\frac{-1+\alpha_1+\alpha_2+\alpha_3}{2}+w}D_j(x)\gamma_1(x)^{-\nu_j} C_{\nu_j}(2\gamma_1(x) k)dx\ll k^{-1+\epsilon}. \end{align} $$

To handle the case of $1-\delta _1<x<1-k^{-2+\epsilon }$ , we apply an asymptotic formulas [Reference Gradshteyn, Ryzhik, Jeffrey and Zwillinger7, 8.451.1 and 8.451.2] for the Bessel functions. Using them, one can write $C_{\nu _j}(2\gamma _1(x) k)$ as

(3.90) $$ \begin{align} C_{\nu_j}(y)=\sum_{\pm}\frac{e^{\pm iy}}{\sqrt{y}}V_{\pm}(y), \end{align} $$

where $V_{\pm }(y)$ smoothed non-oscillating functions uniformly bounded by a constant. So we are left to estimate integral of the following type (here, we set $w=ir$ ):

(3.91) $$ \begin{align} \int_0^{1-k^{-2+\epsilon}}\chi_1(x)W(x) (1-x)^{\frac{-1+\alpha_1+\alpha_2-\alpha_3}{2}} \gamma(x)^{-\alpha_2+\alpha_3} e^{ig_{\pm}(x)}\frac{dx}{\sqrt{\gamma_1(x) k}}, \end{align} $$

where $W(x)=x^{\frac {-1+\alpha _1+\alpha _2+\alpha _3}{2}}V_{\pm }(2\gamma (x) k)D_j(x)$ and the phase function

(3.92) $$ \begin{align} g_{\pm}(x)=\pm2\gamma_1(x) k+r\log x-r\log(1-x). \end{align} $$

Estimating the integral by absolute value and in view of (3.85), we obtain the estimate (3.32). For $|r|\ll \log ^2k,$ we have $g^{\prime }_{\pm }(x)\gg k/\sqrt {1-x}$ and thus using the first derivative test, we estimate (3.91) as

(3.93) $$ \begin{align} \max_{1-\delta_1<x<1-k^{-2+\epsilon}} \frac{(1-x)^{\frac{\alpha_1+\alpha_2-\alpha_3}{2}}\gamma_1(x)^{-\alpha_2+\alpha_3} }{k\sqrt{\gamma_1(x) k}}\ll\frac{k^{\epsilon}}{k}. \end{align} $$

Finally, in view of (3.85), we prove the estimate (3.33).

Consider the second part of the integral (3.34) over $\delta _0<x<1-\delta _1$ . In this case, we make the change of variable $x=\cos ^2y$ and apply Lemma 3.9. Again the contribution of the error terms is negligible and thus it is enough to investigate only the contribution of the main term (all other terms will have the same structure and will be smaller). Note that now y is bounded away from 0 and $\pi /2$ . So the arguments of Y and J-Bessel functions in (3.36) are comparable with k and we can apply for their combination an asymptotic formula similar to (3.90). Finally, we obtain the integrals

(3.94) $$ \begin{align} \frac{e^{-\pi|\Im{w}|/2}(1+|w|)^{\epsilon}}{k^{1/2+\alpha_2+\alpha_3}}\int_0^{1}\chi_{1/2}(\cos^2y)W_{\pm}(y) e^{if_{\pm}(y)}dy, \end{align} $$

where $W_{\pm }(y)$ are smoothed non-oscillating functions uniformly bounded by a constant and

(3.95) $$ \begin{align} f_{\pm}(y)=\pm(2k-1)y-2r\log(\tan y). \end{align} $$

Estimating the integral (3.94) by absolute value, we obtain the estimate (3.32). For $|r|\ll \log ^2k$ , we have $f^{\prime }_{\pm }(y)\gg k$ , and thus using the first derivative test, we estimate (3.94) as

(3.96) $$ \begin{align} \frac{e^{-\pi|\Im{w}|/2}(1+|w|)^{\epsilon}}{k^{1/2+\alpha_2+\alpha_3}}\max_{0<y<\pi/2} \frac{\chi_{1/2}(\cos^2y)W_{\pm}(y)}{k}\ll\frac{e^{-\pi|\Im{w}|/2}(1+|w|)^{\epsilon}k^{\epsilon}}{k}, \end{align} $$

completing the proof of (3.33).