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Math Help - Hard Integral

  1. #1
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    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\int_0^{\infty}\frac{\sin xt}{e^t-1}dt
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  2. #2
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    Is the sin function dependant on t & x or is the x a typo?
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  3. #3
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    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.
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  4. #4
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    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.
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  5. #5
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    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 !
    Last edited by simplependulum; July 1st 2009 at 11:42 PM.
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  6. #6
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    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}
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  7. #7
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    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}
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  8. #8
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    What about extending the problem to the complex t-plane, and using contour integration with Cauchy's Residue Theorem?
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  9. #9
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    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.
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