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Question Number 33119 by abdo imad last updated on 10/Apr/18

$$find{\int }_0^{\mathrm{\infty }}\frac{t^n}{e^t-1}dtbyusing\xi \left(x\right)fornintegr$$$$\xi \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x}withx>1.$$

Commented by prof Abdo imad last updated on 15/Apr/18

$$letput{A}_n={\int }_0^{\mathrm{\infty }}\frac{t^n}{e^t-1}dt$$$$A_n={\int }_0^{\mathrm{\infty }}\frac{e^{-t}{t}^n}{1-e^{-t}}dt={\int }_0^{\mathrm{\infty }}{t}^n{e}^{-t}\left(\sum _{p=0}^{\mathrm{\infty }}{e}^{-pt}\right)dt$$$$=\sum _{p=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{t}^n{e}^{-(p+1)t}dtthech(p+1)t=ugive$$$$A_n=\sum _{p=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{u^n}{{(p+1)}^n}{e}^{-u}\frac{du}{(p+1)}$$$$=\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(p+1)}^{n+1}}{\int }_0^{\mathrm{\infty }}{u}^n{e}^{-u}dubutweknowthat$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dtforx>0\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{u}^n{e}^{-u}du=\mathrm{\Gamma }(n+1)and\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(p+1)}^{n+1}}$$$$=\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^{n+1}}=\xi (n+1)\Rightarrow$$$$A_n=\mathrm{\Gamma }(n+1).\xi (n+1).$$

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