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Question Number 114681 by Dwaipayan Shikari last updated on 20/Sep/20

$$Provethat{\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x-1}dx=n!\zeta (n+1)$$

Answered by mnjuly1970 last updated on 20/Sep/20

$$proof::$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x-1}dx={\int }_0^{\mathrm{\infty }}\frac{x^n{e}^{-x}}{1-e^{-x}}dx$$$$={\int }_0^{\mathrm{\infty }}\underset{m=1}{\sum ^{\mathrm{\infty }}}{x}^n{e}^{-mx}dx=\underset{m=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{x}^n{e}^{-mx}dx$$$$\stackrel{⟨mx=t⟩}{=}\underset{m=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}\frac{t^{(n+1)-1}{e}^{-t}}{m^{n+1}}dt=\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+1)}{m^{n+1}}$$$$\mathrm{\Omega }=\mathrm{\Gamma }(n+1).\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{m^{n+1}}=n!\zeta (n+1)$$$$✓note\left(1\right)::\mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }\left(n\right)=n!$$$$✓note\left(2\right)::\zeta \left(s\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^s}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$$$

Commented by Dwaipayan Shikari last updated on 20/Sep/20

$$Greatsir!$$

Commented by mnjuly1970 last updated on 20/Sep/20

$$p.b.h.yyouarewelcom$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by mathdave last updated on 20/Sep/20

$$solution$$$$letI={\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x-1}dx={\int }_0^{\mathrm{\infty }}\frac{e^{-x}{x}^n}{1-e^{-x}}dx$$$$seriesof$$$$\frac{1}{1-e^{-x}}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{e}^{-kx}=\underset{k=1}{\sum ^{\mathrm{\infty }}}{e}^{-x(k-1)}=\underset{k=1}{\sum ^{\mathrm{\infty }}}{e}^{-kx+x}$$$$I=\underset{k=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^n{e}^{-kx+x}dx=\underset{k=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{x}^n{e}^{-kx}$$$$lety=kx,dx=\frac{dy}{k}$$$$I=\underset{k=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{e}^{-y}{\left(\frac{y}{k}\right)}^n\frac{dy}{k}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^{n+1}}{\int }_0^{\mathrm{\infty }}{e}^{-y}{y}^{n}dy$$$$I=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^{n+1}}{\int }_0^{\mathrm{\infty }}{y}^{(n+1)-1}{e}^{-y}dy$$$$I=\zeta (n+1)\mathrm{\Gamma }(n+1)but\mathrm{\Gamma }(n+1)=n!$$$$\because {\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x-1}=n!\zeta (n+1)Q.E.D$$$$bymathdave$$

Answered by Aziztisffola last updated on 20/Sep/20

$${\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x-1}dx={\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x}\frac{1}{1-e^{-x}}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{x^n}{e^x}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{e}^{-\mathrm{k}x}dx$$$$=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{x}^n{e}^{-kx-x}dx$$$$=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{x}^n{e}^{-(k+1)x}dx$$$$\mathrm{let}\mathrm{t}=(\mathrm{k}+1)x\Rightarrow \mathrm{dt}=(\mathrm{k}+1)\mathrm{d}x$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{t}}{\mathrm{k}+1}\right)}^{\mathrm{n}}{e}^{-\mathrm{t}}\frac{\mathrm{dt}}{\mathrm{k}+1}$$$$=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{{(\mathrm{k}+1)}^{\mathrm{n}+1}}{e}^{-\mathrm{t}}\mathrm{dt}$$$$=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{k}+1)}^{\mathrm{n}+1}}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}}{e}^{-\mathrm{t}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}}{e}^{-\mathrm{t}}\mathrm{dt}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{k}+1)}^{\mathrm{n}+1}}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{(\mathrm{n}+1)-1}{e}^{-\mathrm{t}}\mathrm{dt}\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(\mathrm{k}\right)}^{\mathrm{n}+1}}$$$$=\mathrm{\Gamma }(\mathrm{n}+1)\zeta (\mathrm{n}+1)$$$$=\mathrm{n}!\zeta (\mathrm{n}+1)$$

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