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I know that it is possible to show that $\int_0^\infty\frac{x^2 {\rm d} x}{e^x+1} = \frac{3}{2}\zeta(3)$ and $\int_0^\infty\frac{x^2 {\rm d} x}{e^x-1} = 2\zeta(3)$ by rewriting the integrals to an expression containing the definition of the zeta function by doing a series expansion of the denominators.

However, given the apparent similarity of the integrals, I was wondering if it was also possible to show that $\int_0^\infty\frac{x^2 {\rm d} x}{e^x+1} = \frac{3}{4}\int_0^\infty\frac{x^2 {\rm d}x}{e^x-1}$ directly with integration techniques? I tried my best with some substitutions & contours but so far failed.

Ewoud
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  • did you try to expand $\frac{1}{e^x \pm 1}$ in geometric series in $e^{-kx}, k \ge 1$ do the even/odd trick in $k$ and substitute $x \to x/2$ in the even $k$ case etc...? – Conrad Oct 12 '20 at 13:35

2 Answers2

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Write $e^x-1=(e^{x/2}+1)(e^{x/2}-1)$ so $$\frac 2{e^x-1}=\frac1{e^{x/2}-1}-\frac1{e^{x/2}+1},$$ so $$2\int_0^\infty\frac {x^2}{e^x-1}\,dx=8\int_0^\infty\frac {(x/2)^2}{e^{x/2}-1}\frac {dx} 2- 8\int_0^\infty\frac {(x/2)^2}{e^{x/2}+1}\frac{dx}2$$ and so on.

kimchi lover
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  • I should have started with $e^{2x}-1=(e^x+1)(e^x-1)$ for a less messy version of the same change-of-variables calculation. – kimchi lover Oct 12 '20 at 14:00
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$$I=\int_0^\infty\frac{x^2}{e^x+1}dx=\int_0^\infty\frac{x^2(e^x-1)}{e^{2x}-1}dx$$ now $u=2x$ so $dx=du/2$ and so: $$I=\int_0^\infty\frac{x^2e^x}{e^{2x}-1}dx-\int_0^\infty\frac{x^2}{e^{2x}-1}dx$$ $$I=\int_0^\infty\frac{x^2e^x}{e^{2x}-1}dx-\frac14\int_0^\infty\frac{u^2}{e^u-1}du$$ now this left integral contains: $$\frac{e^x}{(e^x+1)(e^x-1)}$$ so you can easily split it up

Henry Lee
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