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Question Number 53483 by dwdkswd last updated on 22/Jan/19

Commented by maxmathsup by imad last updated on 22/Jan/19

$$let{A}_s={\int }_0^{\mathrm{\infty }}\frac{x^s}{e^x-1}dx\Rightarrow A_s={\int }_0^{\mathrm{\infty }}\frac{e^{-x}{x}^s}{1-e^{-x}}dx={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^s\left(\sum _{n=0}^{\mathrm{\infty }}{e}^{-nx}\right)dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{x}^s{e}^{-(n+1)x}dx{=}_{(n+1)x=t}\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{t^s}{{(n+1)}^s}{e}^{-t}\frac{dt}{n+1}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+1)}^{s+1}}{\int }_0^{\mathrm{\infty }}{t}^s{e}^{-t}dtbutweknow\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt(x>0)\Rightarrow$$$${\int }_0^{\mathrm{\infty }}{t}^s{e}^{-t}dt=\mathrm{\Gamma }(s+1)\Rightarrow A_s=\xi (s+1)\mathrm{\Gamma }(s+1).$$

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

Commented by Tawa1 last updated on 22/Jan/19

$$\mathrm{Sir},\mathrm{is}\mathrm{this}\mathrm{advanced}\mathrm{algebra}??$$

Commented by Tawa1 last updated on 22/Jan/19

$$\mathrm{I}\mathrm{mean}\mathrm{the}\mathrm{textbook}\mathrm{you}\mathrm{snaped}$$

Commented by Tawa1 last updated on 22/Jan/19

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{name}\mathrm{sir}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

$$visitarchive.organdfreedownloadit..thebook$$

Commented by Tawa1 last updated on 22/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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