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Question Number 104780 by Aziztisffola last updated on 23/Jul/20

Commented by Dwaipayan Shikari last updated on 23/Jul/20

$$\frac{{\pi }^2}{6}$$$$\mathrm{\Gamma }\left(s\right).\zeta \left(s\right)={\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x+1}dx$$$$\mathrm{\Gamma }\left(2\right).\zeta \left(2\right)={\int }_0^{\mathrm{\infty }}\frac{x^{2-1}}{e^x+1}dx$$$$(2-1)!\zeta \left(2\right)={\int }_0^{\mathrm{\infty }}\frac{x}{e^x+1}dx$$$$\zeta \left(2\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$$$so$$$${\int }_0^{\mathrm{\infty }}\frac{x}{e^x+1}=\frac{{\pi }^2}{6}$$

Commented by Aziztisffola last updated on 23/Jul/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{I}\mathrm{found}\mathrm{\Gamma }\left(2\right).\zeta \left(2\right)=\frac{{\pi }^2}{6}$$

Answered by abdomsup last updated on 23/Jul/20

$${\int }_0^{\mathrm{\infty }}\frac{x}{e^x-1}dx={\int }_0^{\mathrm{\infty }}\frac{{xe}^{-x}}{1-e^{-x}}dx$$$$={\int }_0^{\mathrm{\infty }}{xe}^{-x}\left(\sum _{n==0}^{\mathrm{\infty }}{e}^{-nx}\right)dx$$$$=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}x{e}^{-(n+1)x}dx$$$${=}_{(n+1)x=t}\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{t}{n+1}{e}^{-t}\frac{dt}{n+1}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+1)}^2}{\int }_0^{\mathrm{\infty }}t{e}^{-t}dt$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}.\mathrm{\Gamma }\left(2\right)$$$$=\xi \left(2\right).\mathrm{\Gamma }\left(2\right)=\frac{{\pi }^2}{6}\times 1!=\frac{{\pi }^2}{6}$$$$$$

Commented by Aziztisffola last updated on 23/Jul/20

$$\mathrm{Thanks}\mathrm{Sir}$$

Commented by mathmax by abdo last updated on 24/Jul/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}.$$

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