# Help with a troublesome double integral

I'm having difficulty with a double integral

$$-2i\int_{0}^{\infty}\int_{0}^{\infty}\frac{dxdt}{t(e^{2\pi x}-1)(e^{2\pi t/s}-1)}\left[\cos(t\log(1-ix))-\cos(t\log(1+ix))\right]$$

where $s\in\text{D}\subset \mathbb{C}$, but i am not sure of the domain of convergence !

EDIT

It can be shown that the above integral is equivalent to : $$-4\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sinh\left(t\tan^{-1}(x)\right )\sin\left(t\log\left(\sqrt{1+x^{2}} \right ) \right )}{t(e^{2\pi x}-1)(e^{2\pi t/s}-1)}dxdt$$

EDIT2

Doing the integral with respect to $t$ first appears to be a promising approach. Particularly, we wish to compute the integral of the form: $$\int_{0}^{\infty}\frac{\sinh\left(at\right )\sin\left(bt \right )}{t(e^{2\pi tz}-1)}dt\;\;\;\;\;\left(a,b\in \mathbb{R}^{ \geq 0}\;\;,\;\;z\in\text{D}\subset \mathbb{C}\right)$$

• i tried doing the integral WRT $t$ first,and expanding $\frac{1}{e^{2 \pi t /s}-1}$. not sure how to evaluate :$$\int_{0}^{\infty}\left[\cos(t\log(1-ix))-\cos(t\log(1+ix))\right]\frac{e^{-2 \pi n t /s}}{t}dt$$ – Mohammad Al Jamal Jun 5 '14 at 19:36
• @DavidH there's a problem with TEX, i can't view your formula ! – Mohammad Al Jamal Aug 31 '14 at 11:49
• @MohammadAlJamal Ack, sorry about that, and sorry for taking so long to get back to you. I can't figure out why the TEX won't work inside the comment box, but it works fine below so I've included the integral in an edit. – David H Sep 8 '14 at 6:29

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.

• that's beautiful ... for the last step, i don't know about you, but i would use the Abel-Plana formula ! i would love to see how you will proceed . – Mohammad Al Jamal Jun 12 '14 at 20:06
• @MohammadAlJamal I think the Abel-Plana formula is the perfect to wrap up this problem also. Right now I'm looking for possible intermediate steps that will give me a cleaner integral to apply Abel-Plana to. Right now I'm toying with differentiating with respect to the paramater $s$. – David H Jun 12 '14 at 21:13
• the Abel-Plana rep. is very attractive to be abandoned, but i haven't found a way to use it so that we may have an explicit solution ! have you @DavidH ? – Mohammad Al Jamal Jun 15 '14 at 2:03
• @MohammadAlJamal Unfortunately I haven't either. I got stuck and had to set the problem down for a while. But this problem has got under my skin so I'm gonna give it another shot after dinner. – David H Jun 15 '14 at 22:55
• an idea has just crossed my mind. using the Weierstrass product representation: $$\frac{\sinh \left(s\log\sqrt{1+x} \right )}{\log (1+x) }=\frac{s}{2}\prod_{n=1}^{\infty}\left(1+\frac{\left(s\log\sqrt{1+x} \right )^{2}}{\pi ^{2}n^{2}} \right )$$ we have: $$I(s)=-i\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{\log \left(1+\frac{\left(s\log\sqrt{1+ix} \right )^{2}}{\pi ^{2}n^{2}} \right )-\log \left(1+\frac{\left(s\log\sqrt{1-ix} \right )^{2}}{\pi ^{2}n^{2}} \right )}{e^{2\pi x}-1}dx$$ and we can apply the Abel-Plana formula to the individual terms (if possible !) – Mohammad Al Jamal Jun 16 '14 at 20:17