Integral #4 (and hopefully a bit more challenging)

May 2009
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362
Challenge Problem:

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

Moderator editor: Approved Challenge question.
 

simplependulum

MHF Hall of Honor
Jan 2009
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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


\(\displaystyle \Gamma(a+1)\zeta(a+1) \left( 2 - \frac{1}{2^a} \right) \) ?
 

Drexel28

MHF Hall of Honor
Nov 2009
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Challenge Problem:

\(\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx, \ a>0 \)
\(\displaystyle \int_0^{\infty}\frac{x^a}{\sinh(x)}dx=\int_0^{\infty}\frac{x^a}{\frac{e^x-e^{-x}}{2}}=2\int_0^{\infty}\frac{e^{-x}x^a}{1-\left(e^{-x}\right)^2}dx\). 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 \(\displaystyle 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{a+1}}\int_0^{\infty}z^ae^{-z}dz=2\sum_{n=0}^{\infty}\frac{\Gamma(a+1)}{(2n+1)^{a+1}}=\Gamma(a+1)\zeta(a+1)\left(2-2^{-a}\right)\)
 

Drexel28

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\(\displaystyle \int_0^{\infty}\frac{x^a}{\sinh(x)}dx=\int_0^{\infty}\frac{x^a}{\frac{e^x-e^{-x}}{2}}=2\int_0^{\infty}\frac{e^{-x}x^a}{1-\left(e^{-x}\right)^2}dx\). 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, \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n+1)^a}=\zeta(a+1)\left(1-2^{-(a+1)}\right)=\frac{1}{2}\zeta(a+1)\left(2-2^{-a}\right)\)
 
May 2009
959
362
You guys are too good. I give up. (Tongueout)


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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Drexel28

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You guys are too good. I give up. (Tongueout)


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 \(\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n,\text{ }n\in\mathbb{N}\)
 

Bruno J.

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Jun 2009
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Canada
Here's one that I have admittedly never tried but looks difficult. How about \(\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n,\text{ }n\in\mathbb{N}\)
\(\displaystyle dn\), right? (Talking)
 

Drexel28

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Nov 2009
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\(\displaystyle dn\), right? (Talking)
Haha, \(\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n{\color{red}dx},\text{ }n\in\mathbb{N}\)
 

NonCommAlg

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May 2008
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Haha, \(\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n{\color{red}dx},\text{ }n\in\mathbb{N}\)
can' be done because i couldn't do it! (Rofl) 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!
 

simplependulum

MHF Hall of Honor
Jan 2009
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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 .