## cinar 3 years ago I am looking for solution step by step for this integral, I know what the answer is. $\int\limits_{0}^{\infty}\frac{x ^{3}}{e ^{x}-1}dx=\frac{\pi ^{4}}{15}$

1. JamesJ

*bookmark. Going to bed now, but I'll write this out for you tomorrow. Nice question.

2. cinar

tell me about it, it is really hard..

3. JamesJ

Ok... $\int_0^\infty \frac{x^3}{e^x - 1} dx = \int_0^\infty \frac{x^3e^{-x}}{1-e^{-x}}$ $= \int_0^\infty \sum_{n=0}^\infty x^3 e^{-(n+1)x} dx$ $= \sum_{n=0}^\infty \int_0^\infty x^3 e^{-(n+1)x} dx$ Changing variable $$u = (n+1)x$$, this is equal to $\sum_{n=0}^\infty \int_0^\infty \frac{u^3}{(n+1)^4} e^{-u} du$ $= \sum_{n=0}^\infty \frac{1}{(n+1)^4} \Gamma(4) \ \hbox{, by definition of the Gamma function }$ $= 6 \ \sum_{n=1}^\infty \frac{1}{n^4} \ \ \ \ \ \hbox{, as } \Gamma(4) = 3! = 6$ $= 6 \ \frac{\pi^4}{90} \ = \ \frac{\pi^4}{15}$

4. JamesJ

We can now generalize this to $\int_0^\infty \frac{x^t}{e^x - 1} dx = \Gamma(t+1)\zeta(t+1)$

5. cinar

actually, my question contains $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n ^{s}}$ $\zeta(4)=1+\frac1{2^{4}}+\frac1{3^{4}}+\frac1{3^{4}}...=\frac {\pi ^{4}}{90}$ how it is solution..

6. cinar

I did not understand how passed this step $= \int\limits_0^\infty \sum_{n=0}^\infty x^3 e^{-(n+1)x} dx$

7. JamesJ

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$ provided |x| < 1. As for $$x \in (0, \infty)$$ we have $$0 < e^{-x} < 1$$, $\frac{e^{-x}}{1-e^{-x}} = e^{-x} ( 1 + e^{-x} + e^{-2x} + e^{-3x} + ...)$ $= e^{-x} + e^{-2x} + e^{-3x} + e^{-4x} + ...$

8. JamesJ

does this make sense now?

9. cinar

yes it does, I am still waiting for $\zeta(4)=\frac {\pi^4}{90}$

10. JamesJ

I see. Let me try and find one of the more elementary proofs.