Hard Integral

nonsingular · #1 $Old$ July 1st, 2009, 02:17 PM

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

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

pickslides · #2 $Old$ July 1st, 2009, 03:51 PM

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

TwistedOne151 · #3 $Old$ July 1st, 2009, 04:31 PM

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

Thus $\int_0^{\infty}\frac{\sin{xt}}{e^t-1}\,dt=\int_0^{\infty}\sum_{n=1}^{\infty}e^{-nt}\sin{xt}\,dt$
$\;\;=\sum_{n=1}^{\infty}\int_0^{\infty}e^{-nt}\sin{xt}\,dt$
And you can find $\int{e^{-nt}}\sin{xt}\,dt$ 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.

nonsingular · #4 $Old$ July 1st, 2009, 05:32 PM

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.

simplependulum · #5 $Old$ July 2nd, 2009, 12:13 AM

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

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

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

and

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

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

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

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

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

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

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

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

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

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

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

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

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

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

nonsingular · #6 $Old$ July 2nd, 2009, 07:08 PM

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

simplependulum · #7 $Old$ July 2nd, 2009, 10:15 PM

Quote:

Originally Posted by nonsingular $View Post$

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

Of course , consider this identity :

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

Then take logarithmic derivative to obtain this equation ,

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

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

And consider

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

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

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

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

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

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

ldawg5962 · #8 $Old$ July 3rd, 2009, 10:30 AM

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

halbard · #9 $Old$ July 4th, 2009, 07:03 AM

OK, you asked for it.

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

$\int_r^R\frac{\textrm e^{\textrm ixt}}{\textrm e^t-1}\textrm dt+\int_0^{2\pi}\frac{\textrm e^{\textrm ix(R+\textrm it)}}{\textrm e^{R+\textrm it}-1}\textrm i\,\textrm dt+\int_R^r\frac{\textrm e^{\textrm ix(t+2\pi\textrm i)}}{\textrm e^{t+2\pi\textrm i}-1}\textrm dt+\gamma(x,r,2\pi\textrm i)+\int_{2\pi-r}^r\frac{\textrm e^{\textrm ix(\textrm it)}}{\textrm e^{\textrm it}-1}\textrm i\,\textrm dt$ ${}+\gamma(x,r,0)=0$

where $\gamma(x,r,z)$ represents the integral around the indentation at $z$ . Simplify, simplify...

(*) $(1-\textrm e^{-2\pi x})\int_r^R\frac{\textrm e^{\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\int_r^{2\pi-r}\frac{\textrm e^{-xt}}{\textrm e^{\textrm it}-1}\textrm dt=I_R(x)-\gamma(x,r,0)-\gamma(x,r,2\pi\textrm i)$ where $I_R(x)=-\int_0^{2\pi}\frac{\textrm e^{-xt+\textrm ixR}}{\textrm e^{R+\textrm it}-1}\textrm dt$ .

It's easy to show that $I_R(x)\to 0$ as $R\to\infty$ irrespective of the value of $x$ , and by the indentation theorem,

$\lim_{r\to 0}\gamma(x,r,0)=-\frac\pi2\textrm i\,\textrm{Res}_{z=0}f(z)=-\frac\pi2\textrm i$ and $\lim_{r\to 0}\gamma(x,r,2\pi\textrm i)=-\frac\pi2\textrm i\,\textrm{Res}_{z=2\pi\textrm i}f(z)=-\frac\pi2\textrm i\textrm e^{-2\pi x}$ .

Obviously we have a problem letting $r$ tend to zero in the first two terms of (*). The key observation here is the identity

$\frac1{\textrm e^{\textrm it}-1}+\frac1{\textrm e^{-\textrm it}-1}=-1$ . So we change $x$ to $-x$ in (*) to get

$(1-\textrm e^{2\pi x})\int_r^R\frac{\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\int_r^{2\pi-r}\frac{\textrm e^{xt}}{\textrm e^{\textrm it}-1}\textrm dt=I_R(-x)-\gamma(-x,r,0)-\gamma(-x,r,2\pi\textrm i)$ .

Now changing $t$ to $2\pi-t$ in the second integral yields

$(1-\textrm e^{2\pi x})\int_r^R\frac{\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\,\textrm e^{2\pi x}\int_r^{2\pi-r}\frac{\textrm e^{-xt}}{\textrm e^{-\textrm it}-1}\textrm dt=I_R(-x)-\gamma(-x,r,0)-\gamma(-x,r,2\pi\textrm i)$ , and dividing by $\textrm e^{2\pi x}$ gives

(**) $(\textrm e^{-2\pi x}-1)\int_r^R\frac{\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt-\textrm i\int_r^{2\pi-r}\frac{\textrm e^{-xt}}{\textrm e^{-\textrm it}-1}\textrm dt$ ${}=\textrm e^{-2\pi x}I_R(-x)-\textrm e^{-2\pi x}\gamma(-x,r,0)-\textrm e^{-2\pi x}\gamma(-x,r,2\pi\textrm i)$ .

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

$(1-\textrm e^{-2\pi x})\int_r^R\frac{\textrm e^{\textrm ixt}-\textrm e^{-\textrm ixt}}{\textrm e^t-1}\textrm dt+\textrm i\int_r^{2\pi-r}\textrm e^{-xt}\textrm dt$ ${}=I_R(x)-\gamma(x,r,0)-\gamma(x,r,2\pi\textrm i)+\textrm e^{-2\pi x}I_R(-x)-\textrm e^{-2\pi x}\gamma(-x,r,0)-\textrm e^{-2\pi x}\gamma(-x,r,2\pi\textrm i)$

Before rigour mortis sets in we should let $r\to 0$ and $R\to\infty$ to arrive at these equations:

$(1-\textrm e^{-2\pi x})\int_0^\infty\frac{2\textrm i\sin xt}{\textrm e^t-1}\textrm dt+\textrm i\int_0^{2\pi}\textrm e^{-xt}\textrm dt$ ${}=0+\frac\pi2\textrm i+\frac\pi2\textrm i\textrm e^{-2\pi x}+0+\textrm e^{-2\pi x}\frac\pi2\textrm i+\textrm e^{-2\pi x}\frac\pi2\textrm i\textrm e^{2\pi x}=\pi\textrm i(1+\textrm e^{-2\pi x})$

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

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

Isn't Cauchy wonderful?

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