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

\(\displaystyle \int_{0}^\infty \frac{x^3}{e^x-1} \text{d}x\)

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 \(\displaystyle \int_{-\infty}^\infty e^x \text{d}x\)). 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 \(\displaystyle \frac{\pi^4}{15}\).) 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

\(\displaystyle \int_{0}^\infty \frac{x^3}{e^x-1} \text{d}x\)

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 \(\displaystyle \int_{-\infty}^\infty e^x \text{d}x\)). 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 \(\displaystyle \frac{\pi^4}{15}\).) 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 \(\displaystyle \frac{1}{e^x-1}\) 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

$\chi$ $\sigma$

 

[blue_raver22]

Hello! Evaluating integrals using contours can be a powerful tool, but it may not always be the most straightforward approach. In this case, it seems like the integral may be more easily evaluated using techniques from real analysis.

One possible approach is to use integration by parts. Let u = x^3 and dv = (e^x-1)^{-1} dx. Then du = 3x^2 dx and v = -ln(e^x-1). Using the integration by parts formula, we have:

\int_{0}^\infty \frac{x^3}{e^x-1} \text{d}x = \left[-x^3 \ln(e^x-1)\right]_{0}^\infty - \int_{0}^\infty -3x^2 \ln(e^x-1) \text{d}x

The first term evaluates to 0, and the second term can be simplified using the substitution u = e^x-1. This gives:

\int_{0}^\infty \frac{x^3}{e^x-1} \text{d}x = 3\int_{0}^\infty \frac{x^2}{e^x} \text{d}x = 3\int_{0}^\infty x^2 e^{-x} \text{d}x

This integral can then be evaluated using techniques from real analysis, such as integration by parts or the substitution u = -x. With some algebraic manipulation, you should be able to arrive at the result \frac{\pi^4}{15}.

I hope this helps and provides a more elementary approach for evaluating this integral. Good luck!