Help with a troublesome double integral

Solution 1:

The integral to evaluate is

$$I(s):=-2i\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos{\left(t\log{(1-ix)}\right)}-\cos{\left(t\log{(1+ix)}\right)}}{t\left(e^{2\pi x}-1\right)\left(e^{2\pi t/s}-1\right)}\mathrm{d}x\mathrm{d}t,$$

where $s\in\mathbb{C}$ is a complex parameter. Right away we notice that the integration with respect to the variable $x$ is much more formidable than that with respect to $t$, so the first thing we do is interchange the order of integration:

$$\begin{align}I(s)&=-2i\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos{\left(t\log{(1-ix)}\right)}-\cos{\left(t\log{(1+ix)}\right)}}{t\left(e^{2\pi x}-1\right)\left(e^{2\pi t/s}-1\right)}\mathrm{d}t\mathrm{d}x\\ &=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{0}^{\infty}\frac{\cos{\left(t\log{(1-ix)}\right)}-\cos{\left(t\log{(1+ix)}\right)}}{t\left(e^{2\pi t/s}-1\right)}\mathrm{d}t.\end{align}$$

The integration w.r.t. $t$ is still too complicated to evaluate, but notice that the term in the numerator of the integrand involving a difference of cosine functions is highly suggestive of the Fundamental Theorem of Calculus. Indeed, using the simple identity $\int_{a}^{b}\sin{(t\,\omega)}\,\mathrm{d}\omega=\frac{\cos{(t\,a)}-\cos{(t\,b)}}{t}$ we can write,

$$\int_{\log{(1-ix)}}^{\log{(1+ix)}}\sin{(t\,\omega)}\,\mathrm{d}\omega=\frac{\cos{(t\,\log{(1-ix)})}-\cos{(t\,\log{(1+ix)})}}{t}.$$

Substituting this expression back into the integral over $t$ not only absorbs the problematic factor of $t$ in the denominator of the integrand, it also gives us the option of performing another change-of-order-of-integration magic trick.

$$\begin{align}I(s)&=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{0}^{\infty}\frac{\mathrm{d}t}{\left(e^{2\pi t/s}-1\right)}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\sin{(t\,\omega)}\,\mathrm{d}\omega\\ &=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{0}^{\infty}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\frac{\sin{(t\,\omega)}}{\left(e^{2\pi t/s}-1\right)}\,\mathrm{d}\omega\mathrm{d}t\\ &=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{\left(e^{2\pi t/s}-1\right)}\,\mathrm{d}t\mathrm{d}\omega\end{align}$$

So we turn our attention to solving the integral $\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{e^{2\pi t/s}-1}\mathrm{d}t$. This integral (see notes at bottom) is,

$$\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{e^{2\pi t/s}-1}\mathrm{d}t=\frac{1}{2\omega}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right),$$

and the integral becomes,

$$\begin{align}I(s)&=-2i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)}\int_{\log{(1-ix)}}^{\log{(1+ix)}}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{2\omega}\\ &=-i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)} \int_{\log{(1-ix)}}^{\log{(1+ix)}} \left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{\omega}\\ &=-i\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(e^{2\pi x}-1\right)} G(x,s).\end{align}$$

See appendix 2 for details on the function $G(x,s)$. It is seen to have the form $G(x,s)=f(ix)-f(-ix)$, and so we can apply the Abel-Plana formula to the final integral $I(s)$:

$$\begin{align} I(s)&=-i\int_{0}^{\infty}\frac{f(ix)-f(-ix)}{\left(e^{2\pi x}-1\right)}\mathrm{d}x\\ &=\int_{0}^{\infty}f(x)\,\mathrm{d}x+\frac12f(0)-\sum_{n=0}^{\infty}f(n) \end{align}$$

Appendix 1:

$\tau=\frac{2\pi t}{s}$, $t=\frac{s}{2\pi}\tau$, $\alpha:=\frac{s\omega}{2\pi}$

$$\begin{align}\int_{0}^{\infty}\frac{\sin{(t\,\omega)}}{e^{2\pi t/s}-1}\mathrm{d}t&=\frac{s}{2\pi}\int_{0}^{\infty}\frac{\sin{(\omega\frac{s}{2\pi}\tau)}}{e^{\tau}-1}\mathrm{d}\tau\\ &=\frac{s}{2\pi}\int_{0}^{\infty}\frac{\sin{(\alpha\,\tau)}}{e^{\tau}-1}\mathrm{d}\tau\\ &=\frac{s}{2\pi}\int_{0}^{\infty}\frac{\sin{(\alpha\,\tau)}\,e^{-\tau}}{1-e^{-\tau}}\mathrm{d}\tau\\ &=\frac{s}{2\pi}\int_{0}^{\infty}\sin{(\alpha\,\tau)}\,e^{-\tau}\sum_{n=0}^{\infty}e^{-n\tau}\mathrm{d}\tau\\ &=\frac{s}{2\pi}\sum_{n=0}^{\infty}\int_{0}^{\infty}\sin{(\alpha\,\tau)}\,e^{-(n+1)\tau}\mathrm{d}\tau\\ &=\frac{s}{2\pi}\sum_{n=0}^{\infty}\frac{\alpha}{\alpha^2+(n+1)^2}\\ &=\frac{s}{2\pi}\frac{\pi\alpha\coth{(\pi\alpha)}-1}{2\alpha}\\ &=\frac{1}{2\omega}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\end{align}.$$

Appendix 2:

$$\int\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{\omega}=\log{\left(\sinh{\left(\frac{s\omega}{2}\right)}\right)}-\log{\omega}+\text{constant}$$

$$\begin{align} G(x,s):&=\int_{\log{(1-ix)}}^{\log{(1+ix)}}\left(\frac{s\omega}{2}\coth{\left(\frac{s\omega}{2}\right)}-1\right)\frac{\mathrm{d}\omega}{\omega}\\ &=\left(\log{\left(\sinh{\left(\frac{s\log{(1+ix)}}{2}\right)}\right)}-\log{\log{(1+ix)}}\right)-\left(\log{\left(\sinh{\left(\frac{s\log{(1-ix)}}{2}\right)}\right)}-\log{\log{(1-ix)}}\right)\\ &=f(ix)-f(-ix), \end{align}$$

where $f(z):=\left(\log{\left(\sinh{\left(\frac{s\log{(1+z)}}{2}\right)}\right)}-\log{\log{(1+z)}}\right)$.

In response to OP's edit #2:

For $\Re{(\gamma)}>|\Re{(\beta)}|$,

$$\int_{0}^{\infty}\frac{\cos{\left(\alpha\,t\right)}\sinh{\left(\beta\,t\right)}}{e^{\gamma\,t}-1}\,\mathrm{d}t = \frac{\beta}{2\left(\alpha^2+\beta^2\right)}-\frac{\pi}{2\gamma}\cdot\frac{\sin{\left(\frac{2\pi\beta}{\gamma}\right)}}{\cosh{\left(\frac{2\pi\alpha}{\gamma}\right)}-\cos{\left(\frac{2\pi\beta}{\gamma}\right)}}.$$

The above integral is formula $4.132.4$ of Gradstheyn's Table of integrals.