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Question Number 134291 by Lordose last updated on 02/Mar/21

$$$$$$\mathbf{Prove}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{a}}}{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=(1-2^{-\mathrm{a}})\mathit{\zeta }(\mathrm{a}+1)\mathbf{\Gamma }(\mathrm{a}+1)$$$$$$

Answered by Dwaipayan Shikari last updated on 02/Mar/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^a}{1+e^x}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^{\mathrm{\infty }}{e}^{-nx}{x}^{a}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\frac{1}{n^{a+1}}\mathrm{\Gamma }(a+1)$$$$=\mathrm{\Gamma }(a+1)\zeta (a+1)(1-\frac{1}{2^{a+1-1}})=a!\zeta (a+1)(1-2^{-a})$$

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