Definite integral Problem

Calculus Level 4

Compute , 0 x 3 e x 1 d x \int_{0}^{\infty }{\frac{{{x}^{3}}}{{{e}^{x}}-1}dx}


The answer is 6.49394.

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1 solution

Akiva Weinberger
May 9, 2015

We have:

1 e x 1 = e x 1 e x \dfrac1{e^x-1}=\dfrac{e^{-x}}{1-e^{-x}}

1 e x 1 = e x ( 1 + e x + e 2 x + ) \phantom{\dfrac1{e^x-1}}=e^{-x}(1+e^{-x}+e^{-2x}+\dotsb)

1 e x 1 = e x + e 2 x + e 3 x + \phantom{\dfrac1{e^x-1}}=e^{-x}+e^{-2x}+e^{-3x}+\dotsb

Thus:

0 x 3 e x 1 d x = 0 ( x 3 e x + x 3 e 2 x + x 3 e 3 x + ) d x \displaystyle\int_0^\infty\frac{x^3}{e^x-1}\operatorname d\!x=\int_0^\infty\left(x^3e^{-x}+x^3e^{-2x}+x^3e^{-3x}+\dotsb\right)\operatorname d\!x

It's easy to show that 0 x 3 e a x d x = 6 a 4 \displaystyle\int_0^\infty x^3e^{-ax}\operatorname d\!x=\frac6{a^4} . (Hint: Substitute u = a x u=ax and use Gamma... or just integrate by parts a bunch of times.) Thus:

0 x 3 e x 1 d x = 0 ( x 3 e x + x 3 e 2 x + x 3 e 3 x + ) d x \displaystyle\int_0^\infty\frac{x^3}{e^x-1}\operatorname d\!x=\int_0^\infty\left(x^3e^{-x}+x^3e^{-2x}+x^3e^{-3x}+\dotsb\right)\operatorname d\!x

0 x 3 e x 1 d x = 6 1 4 + 6 2 4 + 6 3 4 + \displaystyle\phantom{\int_0^\infty\frac{x^3}{e^x-1}\operatorname d\!x}=\frac6{1^4}+\frac6{2^4}+\frac6{3^4}+\dotsb

0 x 3 e x 1 d x = 6 ζ ( 4 ) = 6 π 4 90 = π 4 15 \displaystyle\phantom{\int_0^\infty\frac{x^3}{e^x-1}\operatorname d\!x}=6\zeta(4)=6\frac{\pi^4}{90}=\boxed{\dfrac{\pi^4}{15}}

Nice solution thank you very much

Ramez Hindi - 6 years, 1 month ago

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