Integral $\int_0^\infty \frac{dx}{x^5\left(e^{\frac{a}{x}}-1\right)}$

Asked Oct 01 '19 at 13:17

Active Oct 13 '19 at 22:48

Viewed 137 times

I was integrating the Planck distribution and came across this integral:

$$\int_0^\infty \frac {1}{x^5\cdot \left(e^{\frac{a}{x}}-1\right)} dx$$

$$\int_0^\infty \frac {1}{x^5\cdot \left(e^{\frac{1}{x}}-1\right)}dx = \frac {\pi^4}{15} $$

and I could analytically work out that the general solution is

$$\int_0^\infty \frac {1}{x^5\cdot \left(e^{\frac{a}{x}}-1\right)}dx = \frac {\pi^4}{15a^4} $$

I was wondering how one could prove this, and if my general solution is accurate or not.

edited Oct 01 '19 at 13:37

Zacky

asked Oct 01 '19 at 13:17

Neel Kulkarni

4

Did you not just compute it? Why do you think your computation might be wrong? – Paul Oct 01 '19 at 13:21
1

Missing $dx$ in equations. – StephenG - Help Ukraine Oct 01 '19 at 13:28

0 Answers0