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Thread: Integral #4 (and hopefully a bit more challenging)

  1. #1
    Super Member Random Variable's Avatar
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    Integral #4 (and hopefully a bit more challenging)

    Challenge Problem:

    $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx, \ a>0 $

    Moderator editor: Approved Challenge question.
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    Challenge Problem:

    $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx, \ a>0 $

    Moderator editor: Approved Challenge question.

    I am sitting at the library and the time is near to be up , is the answer


    $\displaystyle \Gamma(a+1)\zeta(a+1) \left( 2 - \frac{1}{2^a} \right) $ ?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Challenge Problem:

    $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx, \ a>0 $
    $\displaystyle \int_0^{\infty}\frac{x^a}{\sinh(x)}dx=\int_0^{\inf ty}\frac{x^a}{\frac{e^x-e^{-x}}{2}}=2\int_0^{\infty}\frac{e^{-x}x^a}{1-\left(e^{-x}\right)^2}dx$. But, this is equal to $\displaystyle 2\int_0^{\infty}e^{-x}x^a\sum_{n=0}^{\infty}e^{-2nx}=2\sum_{n=0}^{\infty}\int_0^\infty x^ae^{-(2n+1)x}$. But, making the sub $\displaystyle (2n+1)x=z$ gives $\displaystyle 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{a+1}}\int_0^{ \infty}z^ae^{-z}dz=2\sum_{n=0}^{\infty}\frac{\Gamma(a+1)}{(2n+1) ^{a+1}}=\Gamma(a+1)\zeta(a+1)\left(2-2^{-a}\right)$
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    $\displaystyle \int_0^{\infty}\frac{x^a}{\sinh(x)}dx=\int_0^{\inf ty}\frac{x^a}{\frac{e^x-e^{-x}}{2}}=2\int_0^{\infty}\frac{e^{-x}x^a}{1-\left(e^{-x}\right)^2}dx$. But, this is equal to $\displaystyle 2\int_0^{\infty}e^{-x}x^a\sum_{n=0}^{\infty}e^{-2nx}=2\sum_{n=0}^{\infty}\int_0^\infty x^ae^{-(2n+1)x}$. But, making the sub $\displaystyle (2n+1)x=z$ gives $\displaystyle 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{a+1}}\int_0^{ \infty}z^ae^{-z}dz=2\sum_{n=0}^{\infty}\frac{\Gamma(a+1)}{(2n+1) ^{a+1}}=\Gamma(a+1)\zeta(a+1)\left(2-2^{-a}\right)$
    I guess I should probably justify that last sum.

    $\displaystyle \zeta(a+1)=\sum_{n=1}^{\infty}\frac{1}{n^a}=\sum_{ n=1}^{\infty}\frac{1}{(2n)^a}+\sum_{n=1}^{\infty}\ frac{1}{(2n+1)^a}=\frac{1}{2^a}\zeta(a)+\sum_{n=1} ^{\infty}\frac{1}{(2n+1)^a}$.

    Thus, $\displaystyle \zeta(a)-\frac{1}{2^a}\zeta(a)=\left(1-2^{-a}\right)\zeta(a)=\sum_{n=1}^{\infty}\frac{1}{(2n+ 1)^a}$.

    So, $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n+1)^a}=\zeta(a+1)\l eft(1-2^{-(a+1)}\right)=\frac{1}{2}\zeta(a+1)\left(2-2^{-a}\right)$
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  5. #5
    Super Member Random Variable's Avatar
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    You guys are too good. I give up.


    Here's what I did:


    $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx $


    $\displaystyle = 2 \int^{\infty}_{0} \frac{x^a}{e^{x}-e^{-x}} \ dx = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{e^{2x}-1} \ dx $


    $\displaystyle = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{(e^{x}+1)(e^{x}-1)} \ dx = 2 \int^{\infty}_{0} \frac{(e^{x}+1-1)x^a}{(e^{x}+1)(e^{x}-1)} \ du $

    $\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - 2 \int^{\infty}_{0} \frac{x^{a}}{e^{2x}-1} \ dx $

    $\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - \int^{\infty}_{0} \frac{(\frac{u}{2})^{a}}{e^{u}-1} \ dx $


    $\displaystyle = \Big(2- \frac{1}{2^{a}} \Big) \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx $


    since $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx = \Gamma(a+1) \zeta(a+1)$ (an integral which has been done on here before)


    $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx = \Big(2 - \frac{1}{2^{a}} \Big) \Gamma(a+1) \zeta(a+1) $
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    You guys are too good. I give up.


    Here's what I did:


    $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx $


    $\displaystyle = 2 \int^{\infty}_{0} \frac{x^a}{e^{x}-e^{-x}} \ dx = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{e^{2x}-1} \ dx $


    $\displaystyle = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{(e^{x}+1)(e^{x}-1)} \ dx = 2 \int^{\infty}_{0} \frac{(e^{x}+1-1)x^a}{(e^{x}+1)(e^{x}-1)} \ du $

    $\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - 2 \int^{\infty}_{0} \frac{x^{a}}{e^{2x}-1} \ dx $

    $\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - \int^{\infty}_{0} \frac{(\frac{u}{2})^{a}}{e^{u}-1} \ dx $


    $\displaystyle = \Big(2- \frac{1}{2^{a}} \Big) \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx $


    since $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx = \Gamma(a+1) \zeta(a+1)$ (an integral which has been done on here before)


    $\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx = \Big(2 - \frac{1}{2^{a}} \Big) \Gamma(a+1) \zeta(a+1) $
    Here's one that I have admittedly never tried but looks difficult. How about $\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\righ t)^n,\text{ }n\in\mathbb{N}$
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Here's one that I have admittedly never tried but looks difficult. How about $\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\righ t)^n,\text{ }n\in\mathbb{N}$
    $\displaystyle dn$, right?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    $\displaystyle dn$, right?
    Haha, $\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\righ t)^n{\color{red}dx},\text{ }n\in\mathbb{N}$
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    Haha, $\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\righ t)^n{\color{red}dx},\text{ }n\in\mathbb{N}$
    can' be done because i couldn't do it! even for n = 1, the value of the integral is in terms of Catalan's constant. so ... i'm just going to leave this integral in peace!
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  10. #10
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    I think we can obtain a good looking reduction formula separately for even and odd numbers $\displaystyle n$ .The key is that can we obtain it for $\displaystyle n=1,2,3 $ . For $\displaystyle n =2 $ the integral is $\displaystyle \pi \ln{2} $ but for $\displaystyle n= 1,3$ the result looks quite bad .... I am looking for a reduction formula for the integral which only consists of two integrals and it only reduces the power of the csc function .
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  11. #11
    Super Member Deadstar's Avatar
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    I got bored so did a power series expansion which is reasonable for $\displaystyle 0 \leq x \leq \pi/2$...

    $\displaystyle \frac{x}{\sin(x)} \approx 1 + \frac{1}{6}x^2 + \frac{7}{360}x^4 + \frac{31}{15120}x^6 + \frac{127}{604800}x^8 + ...$

    Integrating this with $\displaystyle n=2$ for example gives you $\displaystyle 2.176549896$ while $\displaystyle \pi \ln(2) \approx 2.177586091$.

    These were the first 10 values for n=1..10. Guessing n=1 must be wrong then...
    1.831577148
    2.176549896
    2.638905767
    3.266244113
    4.127190286
    5.321167770
    6.992867753
    9.353729126
    12.71390133
    17.52991142
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  12. #12
    Super Member Deadstar's Avatar
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    And a wee bit more playing around gives a rough term approximation for my series (the x/sin(x) one) as...

    $\displaystyle a_n \approx 2\bigg{(} \frac{1}{\pi}\bigg{)}^{2n}$...
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