How to compute the integral below: $$ \int _0 ^ \infty \frac {dx} {x^3 (e^\frac{\pi}{x} -1)}$$ I have trouble in dealing with the item of $e^ \frac{\pi} {x} -1$ without the method of Taylor expansion.
Any idea will be helpful.
Evaluate integral $ \int _0 ^ \infty \frac {dx} {x^3 (e^\frac{\pi}{x} -1)}$ without Taylor expansion
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Did you try $u=\pi/x$? – Matthew H. May 14 '21 at 01:40
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I do try that, but the item of $e^x-1$ is also the trouble. – yufeng lu May 14 '21 at 01:44
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@yufenglu are you allowed to use techniques from multivariable calculus (multiple integrals and a change of variables, specifically) in your solution? – Alann Rosas May 14 '21 at 02:19
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Maybe not I think. – yufeng lu May 14 '21 at 03:55
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Substitute $t=\frac {\pi}x$ $$ \int_0^{\infty}\frac{dx}{x^3(e^{\pi/x}-1)}=\frac 1{\pi^2}\int_0^{\infty}\frac{t}{e^{t}-1}\,dt. $$ Use integral representation of riemann zeta function $$ \zeta(s)=\frac 1{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}}{e^t-1}\,dt. $$ Put $s=2$, we have \begin{align} \frac 1{\pi^2}\int_0^{\infty}\frac{t}{e^{t}-1}\,dt&=\frac 1{\pi^2}\zeta(2)\\ &=\frac 16 \end{align}
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Reexpress the integral with the substitution $y=e^{-\frac\pi x}$
$$ \int _0 ^ \infty \frac {dx} {x^3 (e^\frac{\pi}{x} -1)} =\frac1{\pi^2}\int_0^1 \frac{\ln y }{y-1}dy$$ Then, evaluate $\int_0^1 \frac{\ln y }{y-1}dy = \frac{\pi^2}6$ without Tayler expansion.
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