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Question Number 53599 by maxmathsup by imad last updated on 23/Jan/19

$$1)calculate{A}_n={\int }_0^{\mathrm{\infty }}\frac{x^{n-1}}{e^x+1}dxwithnintegrnatural(n⩾2)$$$$2)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{x}{e^x+1}dx$$

Commented by maxmathsup by imad last updated on 24/Jan/19

$$1)wehave{A}_n={\int }_0^{\mathrm{\infty }}\frac{x^{n-1}{e}^{-x}}{1+e^{-x}}dx={\int }_0^{\mathrm{\infty }}{x}^{n-1}{e}^{-x}\left(\sum _{p=0}^{\mathrm{\infty }}{(-1)}^p{e}^{-px}\right)dx$$$$=\sum _{p=0}^{\mathrm{\infty }}{(-1)}^p{\int }_0^{\mathrm{\infty }}{x}^{n-1}{e}^{-(p+1)x}dx=\sum _{p=0}^{\mathrm{\infty }}{(-1)}^p{A}_p$$$$A_p={\int }_0^{\mathrm{\infty }}{x}^{n-1}{e}^{-(p+1)x}dx{=}_{(p+1)x=t}{\int }_0^{\mathrm{\infty }}{\left(\frac{t}{p+1}\right)}^{n-1}{e}^{-t}\frac{dt}{p+1}$$$$=\frac{1}{{(p+1)}^n}{\int }_0^{\mathrm{\infty }}{t}^{n-1}{e}^{-t}dt=\frac{\mathrm{\Gamma }\left(n\right)}{{(p+1)}^n}\Rightarrow A_n=\mathrm{\Gamma }\left(n\right)\sum _{p=0}^{\mathrm{\infty }}\frac{{(-1)}^p}{{(p+1)}^n}$$$$=\mathrm{\Gamma }\left(n\right).\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^{p-1}}{p^n}letremember\xi \left(x\right)=\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^x}withx>1$$$$wehave\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^{p-1}}{p^n}=-\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{\left(2k\right)}^n}+\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(2k+1)}^n}$$$$=-\frac{1}{2^n}\xi \left(n\right)+\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(2k+1)}^n}but$$$$\xi \left(n\right)=\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^n}=\sum _{k=1}^n\frac{1}{{\left(2k\right)}^n}+\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(2k+1)}^n}=\frac{1}{2^n}\xi \left(n\right)+\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(2k+1)}^n}\Rightarrow$$$$\sum _{k=0}^n\frac{1}{{(2k+1)}^n}=(1-2^{-n})\xi \left(n\right)\Rightarrow \sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^{p-1}}{p^n}=-2^{-n}\xi \left(n\right)+(1-2^{-n})\xi \left(n\right)$$$$=(1-2^{1-n})\xi \left(n\right)\Rightarrow A_n=(1-2^{1-n})\xi \left(n\right)\mathrm{\Gamma }\left(n\right)$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt\Rightarrow \mathrm{\Gamma }\left(n\right)={\int }_0^{\mathrm{\infty }}{t}^{n-1}{e}^{-t}dt={\left[\frac{1}{n}{t}^n{e}^{-t}\right]}_0^{+\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{t^n}{n}{e}^{-t}dt$$$$=\frac{1}{n}\mathrm{\Gamma }(n+1)\Rightarrow \mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }\left(n\right)\Rightarrow \mathrm{\Gamma }(n+1)=n!\Rightarrow$$$$A_n=(1-2^{1-n})\xi \left(n\right)(n-1)!$$$$2)weseethat{\int }_0^{\mathrm{\infty }}\frac{x}{e^x+1}dx=A_2=(1-\frac{1}{2})\xi \left(2\right)=\frac{1}{2}.\frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}.$$

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