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Question Number 40883 by prof Abdo imad last updated on 28/Jul/18

$$find{\int }_0^{\mathrm{\infty }}\frac{t^p}{e^t-1}dtwithp\in N^{★}$$

Answered by maxmathsup by imad last updated on 31/Jul/18

$$let{A}_p={\int }_0^{\mathrm{\infty }}\frac{t^p}{e^t-1}\Rightarrow A_p={\int }_0^{\mathrm{\infty }}\frac{e^{-t}{t}^p}{1-e^{-t}}$$$$={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^p\left(\sum _{n=0}^{\mathrm{\infty }}{e}^{-nt}\right)=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{t}^p{e}^{-(n+1)t}dtchangement(n+1)t=x$$$$give$$$$A_p=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{x}{n+1}\right)}^p{e}^{-x}\frac{dx}{n+1}=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{x^p{e}^{-x}}{{(n+1)}^{p+1}}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+1)}^{p+1}}{\int }_0^{\mathrm{\infty }}{x}^p{e}^{-x}dxbyparts$$$$w_p={\int }_0^{\mathrm{\infty }}{x}^p{e}^{-x}dx={[-x^p{e}^{-x}]}_0^{+\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}{px}^{p-1}{e}^{-p}dx$$$$={pw}_{p-1}\Rightarrow w_p=p!w_{0}and{w}_0={\int }_0^{\mathrm{\infty }}{e}^{-x}dx=1\Rightarrow w_p=p!\Rightarrow$$$$A_p=p!\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{p+1}}=p!\xi (p+1)with\xi \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x},x>1$$$$$$

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