12
$\begingroup$

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: $$\int_{0}^{\infty}\frac{\sinh\left(at\right )\sin\left(bt \right )}{t(e^{2\pi tz}-1)}dt\;\;\;\;\;\left(a,b\in \mathbb{R}^{ \geq 0}\;\;,\;\;z\in\text{D}\subset \mathbb{C}\right)$$

$\endgroup$
  • $\begingroup$ i tried doing the integral WRT $t$ first,and expanding $\frac{1}{e^{2 \pi t /s}-1}$. not sure how to evaluate :$$\int_{0}^{\infty}\left[\cos(t\log(1-ix))-\cos(t\log(1+ix))\right]\frac{e^{-2 \pi n t /s}}{t}dt$$ $\endgroup$ – Mohammad Al Jamal Jun 5 '14 at 19:36
  • $\begingroup$ @DavidH there's a problem with TEX, i can't view your formula ! $\endgroup$ – Mohammad Al Jamal Aug 31 '14 at 11:49
  • $\begingroup$ @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. $\endgroup$ – David H Sep 8 '14 at 6:29
9
+50
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 . $\endgroup$ – Mohammad Al Jamal Jun 12 '14 at 20:06
  • 1
    $\begingroup$ @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 $s$. $\endgroup$ – David H Jun 12 '14 at 21:13
  • $\begingroup$ 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 ? $\endgroup$ – Mohammad Al Jamal Jun 15 '14 at 2:03
  • $\begingroup$ @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. $\endgroup$ – David H Jun 15 '14 at 22:55
  • $\begingroup$ 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 !) $\endgroup$ – Mohammad Al Jamal Jun 16 '14 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.