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Question Number 167965 by mnjuly1970 last updated on 30/Mar/22

Answered by mindispower last updated on 01/Apr/22

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}2\frac{1-e^{-2x}}{{(1+e^{-2\mathit{x}})}^2\mathit{x}}{e}^{-x}$$$$=-2{\int }_0^1\frac{1-t^2}{{(1+t^2)}^{2}ln\left(t\right)}$$$$=-2{\int }_{\mathrm{\infty }}^1\frac{1-\frac{1}{t^2}}{{(1+\frac{1}{t^2})}^{2}ln\left(\frac{1}{t}\right)}.-\frac{dt}{t^2}$$$$$$$$$$$$=-2{\int }_1^{\mathrm{\infty }}\frac{1-t^2}{{(t^2+1)}^{2}ln\left(t\right)}\Rightarrow 2\mathrm{\Omega }=-2{\int }_0^{\mathrm{\infty }}\frac{1-t^2}{(1+t^2)ln\left(t\right)}dt$$$$f\left(a\right)=-2{\int }_0^{\mathrm{\infty }}\frac{1-t^a}{{(1+t^2)}^{2}ln\left(t\right)},a\in [0,2]$$$$f\left(0\right)=0,f\left(2\right)=2\mathrm{\Omega }$$$$f^{\prime }\left(a\right)=2{\int }_0^{\mathrm{\infty }}\frac{t^a}{{(1+t^2)}^2}dr$$$$=2{\int }_0^{\frac{\pi }{2}}{sin}^a\left(t\right){cos}^{2-a}\left(t\right)dt$$$$=\beta (\frac{a+1}{2},\frac{3-a}{2})=\frac{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{3-a}{2}\right)}{\mathrm{\Gamma }\left(2\right)}$$$$=\left(\frac{1-a}{2}\right).\frac{\pi }{sin\left(\frac{\pi }{2}(1+a)\right)}=\frac{\pi (1-a)}{2cos\left(\frac{\pi a}{2}\right)}$$$$2\mathrm{\Omega }={\int }_0^2{f}^{\prime }\left(a\right)da$$$$\frac{\pi }{2}{\int }_0^1\frac{1-a}{cos\left(\frac{\pi a}{2}\right)}+{\int }_1^2\frac{1-a}{cos\left(\frac{\pi a}{2}\right)}$$$$=\frac{\pi }{2}({\int }_0^1\frac{ada}{sin\left(\frac{\pi a}{2}\right)}+{\int }_0^1\frac{-udu}{cos(\frac{\pi }{2}(1+u)})$$$$$$$$=\pi {\int }_0^1\frac{daa}{sin\left(\frac{\pi a}{2}\right)}=\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{u}{sin\left(u\right)}du$$$${\int }_0^{\frac{\pi }{2}}\frac{u}{sin\left(u\right)}du={\int }_0^{\frac{\pi }{2}}\frac{usin\left(u\right)}{(1-cos\left(u\right))(1+cos\left(u\right))}du$$$$=\frac{1}{2}\left[uln\left(\frac{1-cosx}{1+cos\left(x\right)}\right)\right]-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}ln\left(\frac{1-cos\left(x\right)}{1+cos\left(x\right)}\right)$$$$=-{\int }_0^{\frac{\pi }{2}}ln\left(tg\left(\frac{u}{2}\right)\right)du=-2{\int }_0^1\frac{ln\left(x\right)}{1+x^2}=2G$$$$\Rightarrow 2\mathrm{\Omega }=\frac{4}{\pi }.2G$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\tanh \left(x\right)}{x}.\mathrm{sech}\left(x\right)dx=\frac{4G}{\pi }$$$$$$$$$$

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