How do we calculate improper integrals of the form $ \int\limits_0^\infty \frac{x^{n}}{e^{x} \pm 1} \mathrm{d}x$?

Asked Oct 04 '13 at 12:38

Active Dec 13 '13 at 00:31

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It follows that $ \int\limits_0^\infty \frac{x^{3}}{e^{x} - 1} \mathrm{d}x = \frac{\pi^{4}}{15},~ \int\limits_0^\infty \frac{x^{3}}{e^{x} + 1} \mathrm{d}x = \frac{7 \pi^{4}}{120},~ \int\limits_0^\infty \frac{x^{5}}{e^{x} - 1} \mathrm{d}x = \frac{8 \pi^{6}}{63},~ \int\limits_0^\infty \frac{x^{2}}{e^{x} - 1} \mathrm{d}x = 2 \zeta (3).$

How did we get those results?

edited Dec 13 '13 at 00:31

Potato

41,411

asked Oct 04 '13 at 12:38

jacie

1

See Ron Gordon's answer here. – user1337 Oct 04 '13 at 12:41
Here is the technique. – Mhenni Benghorbal Oct 04 '13 at 13:01
Thanks a lot. And it also works when the denominator is of the form $e^{u}-1$. – jacie Oct 04 '13 at 19:31
The "$\large +$" integrals can be expressed in terms of the "$\large -$" integrals. So, you just need the "$\large -$" integrals. – Felix Marin Oct 05 '13 at 09:11

How do we calculate improper integrals of the form $ \int\limits_0^\infty \frac{x^{n}}{e^{x} \pm 1} \mathrm{d}x$?

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