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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 byprof 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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