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cinar

  • 2 years ago

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

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  1. JamesJ
    • 2 years ago
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    *bookmark. Going to bed now, but I'll write this out for you tomorrow. Nice question.

  2. cinar
    • 2 years ago
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    tell me about it, it is really hard..

  3. JamesJ
    • 2 years ago
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    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
    • 2 years ago
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    We can now generalize this to \[ \int_0^\infty \frac{x^t}{e^x - 1} dx = \Gamma(t+1)\zeta(t+1) \]

  5. cinar
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    \[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + ... \] 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
    • 2 years ago
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    does this make sense now?

  9. cinar
    • 2 years ago
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    yes it does, I am still waiting for \[\zeta(4)=\frac {\pi^4}{90}\]

  10. JamesJ
    • 2 years ago
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    I see. Let me try and find one of the more elementary proofs.

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