1. Hard Integral

Hi, I have another integral problem that is difficult for me. I was wondering if anyone here can help. Thanks.

Question: Evaluate $\displaystyle \displaystyle\int_0^{\infty}\frac{\sin xt}{e^t-1}dt$

2. Is the sin function dependant on t & x or is the x a typo?

3. Use the infinite geometric series $\displaystyle \sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$ for $\displaystyle |x|\le1$ to get that $\displaystyle \frac{1}{e^t-1}=\frac{e^{-t}}{1-e^{-t}}=\sum_{n=1}^{\infty}e^{-nt}$ for t>0.

Thus $\displaystyle \int_0^{\infty}\frac{\sin{xt}}{e^t-1}\,dt=\int_0^{\infty}\sum_{n=1}^{\infty}e^{-nt}\sin{xt}\,dt$
$\displaystyle \;\;=\sum_{n=1}^{\infty}\int_0^{\infty}e^{-nt}\sin{xt}\,dt$
And you can find $\displaystyle \int{e^{-nt}}\sin{xt}\,dt$ either via integration by parts or via a table of integrals. Integrating each term thus gives you an infinite series for your answer.

--Kevin C.

4. Hi TwistedOne151, thank you for replying to my question. However, I was wondering if one can find a closed form expression for this integral i.e. not equate the integral with an infinite series.

5. Besides complex variables , i have a nice solution for this question .

I'd like to make use of these two lemmas :

$\displaystyle \zeta(s) \Gamma(s) = \int_0^{\infty} \frac{ x^{s-1}}{ e^x - 1 } dx$

and

$\displaystyle \sum_{n=1}^{\infty} \zeta(2n) x^{2n} = \frac{ 1 - \pi x \cot{ \pi x} }{2}$

$\displaystyle \int_{0}^{\infty} \frac{\sin{xt} }{e^t-1}dt$

$\displaystyle = \int_{0}^{\infty} \frac{dt}{e^t -1 } \sum_{n=0}^{\infty} \frac{ (-1)^n (xt)^{2n+1} }{ (2n+1)!}$

$\displaystyle = \int_0^{\infty} \frac{dt}{e^t -1 } [ \frac{xt}{1!} - \frac{(xt)^3}{3!} + .... ]$

$\displaystyle = \frac{x}{1!} \zeta(2) \Gamma(2) - \frac{x^3}{3!} \zeta(4) \Gamma(4) + .......$

$\displaystyle = \sum_{n=1}^{\infty} \frac{ (-1)^{n+1} x^{2n-1} \zeta(2n) \Gamma(2n) }{(2n-1)! }$

Since $\displaystyle \Gamma(2n) = (2n-1)!$
the series becomes :

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1} x^{2n-1} \zeta(2n)$

$\displaystyle = \frac{-1}{x} \sum_{n=1}^{\infty} (ix)^{2n} \zeta(2n) ~ , i^2 = -1$

Use the lemma given , replace $\displaystyle x~$ by $\displaystyle ix$

$\displaystyle = \frac{-1}{x}( \frac{ 1 - \pi {ix} \cot{\pi ix}}{2})$

We have $\displaystyle \cot{\pi ix} = \frac{ \coth{\pi x }}{i}$

The integral $\displaystyle = \frac{ \pi x \coth{\pi x} - 1}{2x}$

Although you may think it is a stupid method , i think it is very interesting indeed !

6. Hi simplependulum, I like your solution. I know the first lemma that you used, but can you please provide me with a reference for your second lemma.
Do you have a reference for a proof/derivation of $\displaystyle \sum_{n=1}^{\infty} \zeta(2n) x^{2n} = \frac{ 1 - \pi x \cot{ \pi x} }{2}$

7. Originally Posted by nonsingular
Hi simplependulum, I like your solution. I know the first lemma that you used, but can you please provide me with a reference for your second lemma.
Do you have a reference for a proof/derivation of $\displaystyle \sum_{n=1}^{\infty} \zeta(2n) x^{2n} = \frac{ 1 - \pi x \cot{ \pi x} }{2}$

Of course , consider this identity :

$\displaystyle \sin{\pi x } = \pi x \prod_{k=1}^{\infty}( 1 - \frac{x^2}{k^2} )$

Then take logarithmic derivative to obtain this equation ,

$\displaystyle \pi \cot{\pi x} = \frac{1}{x} - \sum_{k=1}^{\infty} \frac{2x}{k^2 - x^2}$

$\displaystyle 1 - \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2} = \pi x \cot{\pi x}$

And consider

$\displaystyle \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2}$

$\displaystyle = \sum_{k=1}^{\infty} 2 ( \frac{x^2}{k^2} )( \frac{1}{ 1- \frac{x^2}{k^2}} )$

$\displaystyle = 2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} ( \frac{x^2}{k^2} )^n$

$\displaystyle = 2 \sum_{n=1}^{\infty} x^{2n} \zeta(2n)$

Finally put this back to the equation in line 3 , the required lemma is then obtained .

$\displaystyle \sum_{n=1}^{\infty} x^{2n} \zeta(2n) = \frac{1 - \pi x \cot{\pi x} }{2}$

8. What about extending the problem to the complex t-plane, and using contour integration with Cauchy's Residue Theorem?

9. OK, you asked for it.

Let's integrate the function $\displaystyle f(z)=\frac{\textrm e^{\textrm ixz}}{\textrm e^z-1}$ around the rectangular contour $\displaystyle C$ with vertices at $\displaystyle 0$, $\displaystyle R$, $\displaystyle R+2\pi\textrm i$ and $\displaystyle 2\pi\textrm i$ in the complex plane , indented at $\displaystyle 0$ and at $\displaystyle 2\pi\textrm i$ with quarter circles of radius $\displaystyle r<\min(\pi,R)$. $\displaystyle f(z)$ is analytic inside and on $\displaystyle C$ and so $\displaystyle \oint_C f(z)\textrm dz=0$. This leads to

$\displaystyle \int_r^R\frac{\textrm e^{\textrm ixt}}{\textrm e^t-1}\textrm dt+\int_0^{2\pi}\frac{\textrm e^{\textrm ix(R+\textrm it)}}{\textrm e^{R+\textrm it}-1}\textrm i\,\textrm dt+\int_R^r\frac{\textrm e^{\textrm ix(t+2\pi\textrm i)}}{\textrm e^{t+2\pi\textrm i}-1}\textrm dt+\gamma(x,r,2\pi\textrm i)+\int_{2\pi-r}^r\frac{\textrm e^{\textrm ix(\textrm it)}}{\textrm e^{\textrm it}-1}\textrm i\,\textrm dt$$\displaystyle {}+\gamma(x,r,0)=0 where \displaystyle \gamma(x,r,z) represents the integral around the indentation at \displaystyle z. Simplify, simplify... (*) \displaystyle (1-\textrm e^{-2\pi x})\int_r^R\frac{\textrm e^{\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\int_r^{2\pi-r}\frac{\textrm e^{-xt}}{\textrm e^{\textrm it}-1}\textrm dt=I_R(x)-\gamma(x,r,0)-\gamma(x,r,2\pi\textrm i) where \displaystyle I_R(x)=-\int_0^{2\pi}\frac{\textrm e^{-xt+\textrm ixR}}{\textrm e^{R+\textrm it}-1}\textrm dt. It's easy to show that \displaystyle I_R(x)\to 0 as \displaystyle R\to\infty irrespective of the value of \displaystyle x, and by the indentation theorem, \displaystyle \lim_{r\to 0}\gamma(x,r,0)=-\frac\pi2\textrm i\,\textrm{Res}_{z=0}f(z)=-\frac\pi2\textrm i and \displaystyle \lim_{r\to 0}\gamma(x,r,2\pi\textrm i)=-\frac\pi2\textrm i\,\textrm{Res}_{z=2\pi\textrm i}f(z)=-\frac\pi2\textrm i\textrm e^{-2\pi x}. Obviously we have a problem letting \displaystyle r tend to zero in the first two terms of (*). The key observation here is the identity \displaystyle \frac1{\textrm e^{\textrm it}-1}+\frac1{\textrm e^{-\textrm it}-1}=-1. So we change \displaystyle x to \displaystyle -x in (*) to get \displaystyle (1-\textrm e^{2\pi x})\int_r^R\frac{\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\int_r^{2\pi-r}\frac{\textrm e^{xt}}{\textrm e^{\textrm it}-1}\textrm dt=I_R(-x)-\gamma(-x,r,0)-\gamma(-x,r,2\pi\textrm i). Now changing \displaystyle t to \displaystyle 2\pi-t in the second integral yields \displaystyle (1-\textrm e^{2\pi x})\int_r^R\frac{\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\,\textrm e^{2\pi x}\int_r^{2\pi-r}\frac{\textrm e^{-xt}}{\textrm e^{-\textrm it}-1}\textrm dt=I_R(-x)-\gamma(-x,r,0)-\gamma(-x,r,2\pi\textrm i), and dividing by \displaystyle \textrm e^{2\pi x} gives (**) \displaystyle (\textrm e^{-2\pi x}-1)\int_r^R\frac{\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\int_r^{2\pi-r}\frac{\textrm e^{-xt}}{\textrm e^{-\textrm it}-1}\textrm dt$$\displaystyle {}=\textrm e^{-2\pi x}I_R(-x)-\textrm e^{-2\pi x}\gamma(-x,r,0)-\textrm e^{-2\pi x}\gamma(-x,r,2\pi\textrm i)$.

Are you still watching this? Good. Now add (*) and (**) to see what happens. Don't forget the identity!

$\displaystyle (1-\textrm e^{-2\pi x})\int_r^R\frac{\textrm e^{\textrm ixt}-\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt+\textrm i\int_r^{2\pi-r}\textrm e^{-xt}\textrm dt$$\displaystyle {}=I_R(x)-\gamma(x,r,0)-\gamma(x,r,2\pi\textrm i)+\textrm e^{-2\pi x}I_R(-x)-\textrm e^{-2\pi x}\gamma(-x,r,0)-\textrm e^{-2\pi x}\gamma(-x,r,2\pi\textrm i) Before rigour mortis sets in we should let \displaystyle r\to 0 and \displaystyle R\to\infty to arrive at these equations: \displaystyle (1-\textrm e^{-2\pi x})\int_0^\infty\frac{2\textrm i\sin xt}{\textrm e^t-1}\textrm dt+\textrm i\int_0^{2\pi}\textrm e^{-xt}\textrm dt$$\displaystyle {}=0+\frac\pi2\textrm i+\frac\pi2\textrm i\textrm e^{-2\pi x}+0+\textrm e^{-2\pi x}\frac\pi2\textrm i+\textrm e^{-2\pi x}\frac\pi2\textrm i\textrm e^{2\pi x}=\pi\textrm i(1+\textrm e^{-2\pi x})$

The second term on the left is $\displaystyle \frac{\textrm i(1-\textrm e^{-2\pi x})}x$, so finally we have full view of the answer:

$\displaystyle \int_0^\infty\frac{\sin xt}{\textrm e^t-1}\textrm dt=\frac{\pi(1+\textrm e^{-2\pi x})}{2(1-\textrm e^{-2\pi x})}-\frac1{2x}=\frac\pi2\coth \pi x-\frac1{2x}$.

Isn't Cauchy wonderful?

You can see how much easier the series method is. I'm off for a nice cup of tea.