Monday Math 77

Find $\int_0^{\infty}\frac{x^{n-1}}{\sinh{x}}\,dx$ .

Using $\sinh{x}=\frac{e^x-e^{-x}}{2}=\frac{e^{2x}-1}{2e^x}$ , we see
$\int_0^{\infty}\frac{x^{n-1}}{\sinh{x}}\,dx=2\int_0^{\infty}\frac{x^{n-1}e^x}{e^{2x}-1}\,dx$
Next, we try to eliminate the e^x term in the numerator. By noting that $e^{2x}-1=(e^x-1)(e^x+1)$ , we do so as follows:
$\begin{eqnarray}\int_0^{\infty}\frac{x^{n-1}e^x}{e^{2x}-1}\,dx&=&\int_0^{\infty}\frac{x^{n-1}(e^x+1-1)}{e^{2x}-1}\,dx\\&=&\int_0^{\infty}\frac{x^{n-1}(e^x+1)}{e^{2x}-1}-\frac{x^{n-1}}{e^{2x}-1}\,dx\\&=&\int_0^{\infty}\frac{x^{n-1}}{e^x-1}\,dx-\int_0^{\infty}\frac{x^{n-1}}{e^{2x}-1}\,dx\end{eqnarray}$ .
For the second integral, we make the u-substitution u=2x to get
$\begin{eqnarray}\int_0^{\infty}\frac{x^{n-1}}{e^{2x}-1}\,dx&=&\int_0^{\infty}\frac{\left(\frac{u}{2}\right)^{n-1}}{e^u-1}\frac{du}{2}\\&=&\frac{1}{2^n}\int_0^{\infty}\frac{u^{n-1}}{e^u-1}\,du\end{eqnarray}$
and renaming the variable back to x, we get
$\begin{eqnarray}\int_0^{\infty}\frac{x^{n-1}e^x}{e^{2x}-1}\,dx&=&\int_0^{\infty}\frac{x^{n-1}}{e^x-1}\,dx-\frac{1}{2^n}\int_0^{\infty}\frac{x^{n-1}}{e^x-1}\,dx\\&=&\left(1-\frac{1}{2^n}\right)\int_0^{\infty}\frac{x^{n-1}}{e^x-1}\,dx\end{eqnarray}$ .

Now, we showed here that $\int_0^{\infty}\frac{x^{n-1}}{e^x-1}\,dx=\operatorname{\Gamma}(n)\zeta(n)$ . Thus, our integral is
$\begin{eqnarray}\int_0^{\infty}\frac{x^{n-1}}{\sinh{x}}\,dx&=&2\left(1-\frac{1}{2^n}\right)\operatorname{\Gamma}(n)\zeta(n)\\&=&\frac{2^n-1}{2^{n-1}}\operatorname{\Gamma}(n)\zeta(n)\end{eqnarray}$ .

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Tags: Gamma Function, Hyperbolic Sine, Integral, Math, Monday Math, Riemann Zeta Function

This entry was posted on June 22, 2009 at 8:11 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Monday Math 77”

Monday Math 78 « Twisted One 151’s Weblog Says:
June 29, 2009 at 1:50 am | Reply
[...] parts, with and ; then and Now, using and , integration by parts tells us: . Now, we proved here that , so the above is just this for the case n=2, so since and (see here), then our integral is [...]

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Monday Math 77

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