A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 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
I am looking for solution step by step for this integral, I know what the answer is. \[\int\limits_{0}^{\infty}\frac{x ^{3}}{e ^{x}1}dx=\frac{\pi ^{4}}{15}\]

This Question is Closed

JamesJ
 2 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.

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

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

cinar
 2 years ago
Best ResponseYou've already chosen the best response.2actually, 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..

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

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

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

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.2I see. Let me try and find one of the more elementary proofs.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.