## cinar Group Title I am looking for solution step by step for this integral, I know what the answer is. $\int\limits_{0}^{\infty}\frac{x ^{3}}{e ^{x}-1}dx=\frac{\pi ^{4}}{15}$ 2 years ago 2 years ago

1. JamesJ Group Title

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

2. cinar Group Title

tell me about it, it is really hard..

3. JamesJ Group Title

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 Group Title

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

5. cinar Group Title

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 Group Title

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 Group Title

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$ 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 Group Title

does this make sense now?

9. cinar Group Title

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

10. JamesJ Group Title

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