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

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2*bookmark. Going to bed now, but I'll write this out for you tomorrow. Nice question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tell me about it, it is really hard..

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Ok... \[ \int_0^\infty \frac{x^3}{e^x  1} dx = \int_0^\infty \frac{x^3e^{x}}{1e^{x}} \] \[ = \int_0^\infty \sum_{n=0}^\infty x^3 e^{(n+1)x} dx \] \[ = \sum_{n=0}^\infty \int_0^\infty x^3 e^{(n+1)x} dx \] Changing variable \( u = (n+1)x \), this is equal to \[ \sum_{n=0}^\infty \int_0^\infty \frac{u^3}{(n+1)^4} e^{u} du \] \[ = \sum_{n=0}^\infty \frac{1}{(n+1)^4} \Gamma(4) \ \hbox{, by definition of the Gamma function }\] \[ = 6 \ \sum_{n=1}^\infty \frac{1}{n^4} \ \ \ \ \ \hbox{, as } \Gamma(4) = 3! = 6 \] \[ = 6 \ \frac{\pi^4}{90} \ = \ \frac{\pi^4}{15} \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2We can now generalize this to \[ \int_0^\infty \frac{x^t}{e^x  1} dx = \Gamma(t+1)\zeta(t+1) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually, 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..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did not understand how passed this step \[= \int\limits_0^\infty \sum_{n=0}^\infty x^3 e^{(n+1)x} dx\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2\[ \frac{1}{1x} = 1 + x + x^2 + x^3 + ... \] provided x < 1. As for \( x \in (0, \infty) \) we have \( 0 < e^{x} < 1 \), \[ \frac{e^{x}}{1e^{x}} = e^{x} ( 1 + e^{x} + e^{2x} + e^{3x} + ...) \] \[ = e^{x} + e^{2x} + e^{3x} + e^{4x} + ... \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2does this make sense now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes it does, I am still waiting for \[\zeta(4)=\frac {\pi^4}{90}\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2I see. Let me try and find one of the more elementary proofs.
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