MHB Evaluation of an Integral (using contours or otherwise)

[roninpro]

Hello everyone - this is my first post here since the move. I had to deal with a few engineering students yesterday; they asked me how to exactly evaluate the integral

[math]\int_{0}^\infty \frac{x^3}{e^x-1} \text{d}x[/math]

It looked like this integral would best be handled by treating it as complex-valued and setting up a contour so that the complex parts dropped out. However, the students did not know any complex analysis and were looking for a more elementary approach. It wasn't obvious how to do it using real integrals (and I tried to use tricks as in the evaluation of [math]\int_{-\infty}^\infty e^x \text{d}x[/math]). I also discussed this with another graduate student, and we were actually even unable to select a decent contour to take the integral. (Though we did put it into Mathematica which computed [math]\frac{\pi^4}{15}[/math].) We're basically baffled. I was hoping to find some advice about this integral here. Any help would be appreciated!

 

[Opalg]

Hello everyone - this is my first post here since the move. I had to deal with a few engineering students yesterday; they asked me how to exactly evaluate the integral

[math]\int_{0}^\infty \frac{x^3}{e^x-1} \text{d}x[/math]

It looked like this integral would best be handled by treating it as complex-valued and setting up a contour so that the complex parts dropped out. However, the students did not know any complex analysis and were looking for a more elementary approach. It wasn't obvious how to do it using real integrals (and I tried to use tricks as in the evaluation of [math]\int_{-\infty}^\infty e^x \text{d}x[/math]). I also discussed this with another graduate student, and we were actually even unable to select a decent contour to take the integral. (Though we did put it into Mathematica which computed [math]\frac{\pi^4}{15}[/math].) We're basically baffled. I was hoping to find some advice about this integral here. Any help would be appreciated!

Write it as $\displaystyle\int_{0}^\infty \frac{x^3}{e^x-1} \,dx = \int_{0}^\infty \frac{x^3}{e^x(1-e^{-x})}\,dx = \int_{0}^\infty x^3e^{-x}\sum_{n=0}^\infty e^{-nx}\,dx$ (binomial series). Then (if you can justify integrating term by term) this is equal to $\displaystyle\sum_{n=1}^\infty \int_{0}^\infty x^3e^{-nx}\,dx.$

Integrate by parts several times to find that $\displaystyle \int_0^\infty x^3e^{-nx}\,dx = \frac 6{n^4}.$ The result then follows from the fact that $\displaystyle\sum_{n=1}^\infty\frac1{n^4} = \frac{\pi^4}{90}.$

 

[roninpro]

We actually originally tried to interpret the term [math]\frac{1}{e^x-1}[/math] as a geometric series, but convergence was an issue. Your rearrangement fixes that issue. Thanks very much, this works for me!

 

[chisigma]

With systematic application of integration by part You arrive to the general formula...

$\displaystyle \int t^{m}\ \ln ^{n} t\ dt = n!\ t^{m+1}\ \sum_{k=0}^{n}\frac{(-1)^{k}\ \ln^{n-k} t}{(m+1)^{k}\ (n-k)!} +c$ (1)

... and from (1) with simple steps...

$\displaystyle \zeta(n)= \sum_{k=1}^{\infty} \frac{1}{k^{n}}= \frac{(-1)^{n-1}}{(n-1)!} \int_{0}^{1} \frac{\ln^{n-1} t}{1-t}\ dt$ (2)

... where $\zeta(*)$ is the so called 'Riemann Zeta Function' and $n>1$. Setting in (2) $\ln t=x$ You obtain the equivalent relation...

$\displaystyle \zeta(n)= \frac{1}{(n-1)!} \int_{0}^{\infty} \frac{x^{n-1}}{e^{x}-1}\ dx$ (3)

Kind regards

$\chi$ $\sigma$