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Question Number 140221 by qaz last updated on 05/May/21

$${\int }_0^{\mathrm{\infty }}{x}^2[ln(1+e^x)-x]dx=\frac{7{\pi }^4}{360}$$

Answered by Dwaipayan Shikari last updated on 05/May/21

$${\int }_0^{\mathrm{\infty }}{x}^{2}log\left(\frac{1+e^x}{e^x}\right)dx$$$$=\mathrm{\Sigma }{\int }_0^{\mathrm{\infty }}{(-1)}^n{x}^2\frac{e^{-nx}}{n}dx$$$$=\mathrm{\Sigma }\frac{{(-1)}^{n+1}}{n^4}\mathrm{\Gamma }\left(3\right)=2\zeta \left(4\right)(1-\frac{1}{2^3})=\frac{7}{4}\zeta \left(4\right)=\frac{7{\pi }^4}{360}$$

Commented by qaz last updated on 05/May/21

$$thankyousir$$

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