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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
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

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

cinarBest ResponseYou've already chosen the best response.2
tell me about it, it is really hard..
 2 years ago

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

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

cinarBest ResponseYou've already chosen the best response.2
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..
 2 years ago

cinarBest ResponseYou've already chosen the best response.2
I did not understand how passed this step \[= \int\limits_0^\infty \sum_{n=0}^\infty x^3 e^{(n+1)x} dx\]
 2 years ago

JamesJBest 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 \( 0 < e^{x} < 1 \), \[ \frac{e^{x}}{1e^{x}} = e^{x} ( 1 + e^{x} + e^{2x} + e^{3x} + ...) \] \[ = e^{x} + e^{2x} + e^{3x} + e^{4x} + ... \]
 2 years ago

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

cinarBest ResponseYou've already chosen the best response.2
yes it does, I am still waiting for \[\zeta(4)=\frac {\pi^4}{90}\]
 2 years ago

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