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Question Number 213662 by Spillover last updated on 13/Nov/24

Answered by mathmax last updated on 14/Nov/24

$$I={\int }_0^{\mathrm{\infty }}\frac{x^2}{1+e^x}dx\Rightarrow I={\int }_0^{\mathrm{\infty }}\frac{e^{-x}{x}^2}{1+e^{-x}}dx$$$$={\int }_0^{\mathrm{\infty }}{x}^2{e}^{-x}\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-nx}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{x}^2{e}^{-(n+1)x}dx(n+1)x=t$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}\frac{t^2}{{(n+1)}^2}{e}^{-t}\frac{dt}{n+1}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^3}{\int }_0^{\mathrm{\infty }}{t}^{3-1}{e}^{-t}dt$$$$=\mathrm{\Gamma }\left(3\right).\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^3}$$$$wehave\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^3}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^3}$$$$=eta\left(3\right)=(1-2^{1-3})\zeta \left(3\right)=(1-\frac{1}{4})\zeta \left(3\right)$$$$=\frac{3}{4}\zeta \left(3\right)\mathrm{\Gamma }\left(3\right)=2!=2\Rightarrow I=\frac{3}{2}\zeta \left(3\right)$$$$$$

Commented by Spillover last updated on 16/Nov/24

$$greatsolution$$

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