Question

I am looking for solution step by step for this integral, I know what the answer is.

\[\int\limits_{0}^{\infty}\frac{x ^{3}}{e ^{x}-1}dx=\frac{\pi ^{4}}{15}\]

Answer 1

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

Answer 2

tell me about it, it is really hard..

Answer 3

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
\[ \sum_{n=0}^\infty \int_0^\infty \frac{u^3}{(n+1)^4} e^{-u} du \]
\[ = \sum_{n=0}^\infty \frac{1}{(n+1)^4} \Gamma(4) \ \hbox{, by definition of the Gamma function }\]
\[ = 6 \ \sum_{n=1}^\infty \frac{1}{n^4} \ \ \ \ \ \hbox{, as } \Gamma(4) = 3! = 6 \]
\[ = 6 \ \frac{\pi^4}{90} \ = \ \frac{\pi^4}{15} \]

Answer 4

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

Answer 5

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..

Answer 6

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

Answer 7

\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + ... \]
provided |x| < 1.

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

Answer 8

does this make sense now?

Answer 9

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

Answer 10

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