An Improper Riemann Integral

[gugo82]

Exercise
Evaluate:
\[
\intop_0^\infty \frac{x^3}{e^x - 1}\text{d} x\; .
\]

[gugo82]

"TeM":[quote="gugo82"]Abramowitz & Stegun?

Honestly I didn't know anything of this "Bible" (I peeked just now on the net...)!! [/quote]
It is a very useful reference for this kind of questions.
There is also a new online edition.

"TeM":The idea for the series came to me reviewing some similar example that our professor of
analysis mathematics showed us in his "insights for the most curious", while for all the rest
(as was ever mine wont) the source is mathworld.wolfram.com.

Just out of curiosity, who was your Mathematical Analysis prof.?

[gugo82]

@ TeM: Abramowitz & Stegun?

[gugo82]

@ TeM: Ok, your answer's correct!

***

Now a simple generalization...

Exercise
Evaluate:
\[
\intop_0^\infty \frac{x^\alpha}{e^x - 1}\ \text{d} x\; ,
\]
where \(\alpha \geq 1\).

[j18eos]

For clarity, one has:
\[
\frac{e^x}{1-e^x}=\sum_{n=1}^{+\infty}e^{nx}\,\text{for}\,x<0;
\]
formally, one has:
\[
\frac{e^{-x}}{1-e^{-x}}=\sum_{n=1}^{+\infty}\left(e^{-x}\right)^n\,\text{for}\,x>0.
\]

I think that you are wrong...