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Question Number 157175 by cortano last updated on 20/Oct/21

Commented by puissant last updated on 20/Oct/21

$$\mathrm{\Omega }=\frac{{\pi }^2}{4}$$

Answered by Mathspace last updated on 20/Oct/21

$$I={\int }_0^{\mathrm{\infty }}ln\left(\frac{1+e^{-x}}{1-e^{-x}}\right)dx$$$$={\int }_0^{\mathrm{\infty }}ln(1+e^{-x})dx-{\int }_0^{\mathrm{\infty }}ln(1-e^{-x)}dx$$$${ln}^{{}^{\prime }}(1+u)dx=\frac{1}{1+u}=\mathrm{\Sigma }{(-1)}^n{u}^n\Rightarrow$$$$ln(1+u)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{u^{n+1}}{n+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{u^n}{n}\Rightarrow$$$$ln(1+e^{-x})=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{e^{-nx}}{n}$$$$and{\int }_0^{\mathrm{\infty }}ln(1+e^{-x})dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^{\mathrm{\infty }}{e}^{-nx}dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}(-\frac{1}{n}){\left[e^{-nx}\right]}_0^{\mathrm{\infty }}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=-\delta \left(2\right)$$$$=-(2^{1-2}-1)\xi \left(2\right)$$$$=-(\frac{1}{2}-1)\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}$$$${ln}^{{}^{\prime }}(1-u)=-\frac{1}{1-u}=-\sum _{n=0}^{\mathrm{\infty }}{u}^n\Rightarrow$$$$ln(1-u)=-\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}\Rightarrow ln(1-e^{-x})$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-nx}}{n}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}ln(1-e^{-x})dx$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{\int }_0^{\mathrm{\infty }}{e}^{-nx}dx$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{[-\frac{1}{n}{e}^{-nx}]}_0^{\mathrm{\infty }}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=-\frac{{\pi }^2}{6}\Rightarrow$$$$I=\frac{{\pi }^2}{12}+\frac{{\pi }^2}{6}=\frac{3{\pi }^2}{12}\Rightarrow I=\frac{{\pi }^2}{4}$$

Answered by Mathspace last updated on 20/Oct/21

$$anotherway$$$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}ln\left(\frac{e^x+1}{e^x-1}\right)dx\Rightarrow$$$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}ln\left(\frac{1+e^{-x}}{1-e^{-x}}\right)dx$$$$={\int }_0^{\mathrm{\infty }}ln(1+e^{-x})dx-{\int }_0^{\mathrm{\infty }}ln(1-e^{-x)}dx$$$$wehave$$$${\int }_0^{\mathrm{\infty }}ln(1-e^{-x})dx{=}_{e^{-x}=t}{\int }_1^{0}ln(1-t)(-\frac{dt}{t})$$$$={\int }_0^1\frac{ln(1-t)}{t}dt$$$$={\int }_0^1\frac{1}{t}(-\sum _{n=1}^{\mathrm{\infty }}\frac{t^n}{n})dt$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{\int }_0^1{t}^{n-1}dt$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=-\frac{{\pi }^2}{6}$$$${\int }_0^{\mathrm{\infty }}ln(1+e^{-x})dx{=}_{e^{-x}=t}$$$${\int }_1^{0}ln(1+t)(-\frac{dt}{t})={\int }_0^1\frac{ln(1+t)}{t}dt$$$${ln}^{{}^{\prime }}(1+t)=\mathrm{\Sigma }{(-1)}^n{u}^n\Rightarrow$$$$ln(1+t)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{t}^{n+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{t}^n}{n}\Rightarrow$$$${\int }_0^1\frac{ln(1+t)}{t}dt={\int }_0^1\frac{1}{t}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{t}^n}{n}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^1{t}^{n-1}dt$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}=\frac{{\pi }^2}{12}\Rightarrow$$$$\mathrm{\Psi }=\frac{{\pi }^2}{12}-(-\frac{{\pi }^2}{6})=\frac{{\pi }^2}{12}+\frac{{\pi }^2}{6}=\frac{{\pi }^2}{4}$$$$$$

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