Question Number 167965 by mnjuly1970 last updated on 30/Mar/22
Answered by mindispower last updated on 01/Apr/22
$$\Omega=\int_{\mathrm{0}} ^{\infty} \mathrm{2}\frac{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{\left(\mathrm{1}+{e}^{−\mathrm{2}\boldsymbol{{x}}} \right)^{\mathrm{2}} \boldsymbol{{x}}}{e}^{−{x}} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} {ln}\left({t}\right)}\: \\ $$$$=−\mathrm{2}\int_{\infty} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{2}} {ln}\left(\frac{\mathrm{1}}{{t}}\right)}.−\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$=−\mathrm{2}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {ln}\left({t}\right)}\Rightarrow\mathrm{2}\Omega=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right){ln}\left({t}\right)}{dt} \\ $$$${f}\left({a}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{t}^{{a}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} {ln}\left({t}\right)},{a}\in\left[\mathrm{0},\mathrm{2}\right] \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0},{f}\left(\mathrm{2}\right)=\mathrm{2}\Omega \\ $$$${f}'\left({a}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dr} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{a}} \left({t}\right){cos}^{\mathrm{2}−{a}} \left({t}\right){dt} \\ $$$$=\beta\left(\frac{{a}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}−{a}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}−{a}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}\right)} \\ $$$$=\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right).\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{a}\right)\right)}=\frac{\pi\left(\mathrm{1}−{a}\right)}{\mathrm{2}{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)} \\ $$$$\mathrm{2}\Omega=\int_{\mathrm{0}} ^{\mathrm{2}} {f}'\left({a}\right){da} \\ $$$$\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{a}}{{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}−{a}}{{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ada}}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{udu}}{{cos}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{u}\right)\right.}\right) \\ $$$$ \\ $$$$=\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{daa}}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}=\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{u}}{{sin}\left({u}\right)}{du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{u}}{{sin}\left({u}\right)}{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{usin}\left({u}\right)}{\left(\mathrm{1}−{cos}\left({u}\right)\right)\left(\mathrm{1}+{cos}\left({u}\right)\right)}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{uln}\left(\frac{\mathrm{1}−{cosx}}{\mathrm{1}+{cos}\left({x}\right)}\right)\right]−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}−{cos}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}\right) \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({tg}\left(\frac{{u}}{\mathrm{2}}\right)\right){du}=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }=\mathrm{2}{G} \\ $$$$\Rightarrow\mathrm{2}\Omega=\frac{\mathrm{4}}{\pi}.\mathrm{2}{G} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tanh}\:\left({x}\right)}{{x}}.\mathrm{sech}\:\left({x}\right){dx}=\frac{\mathrm{4}{G}}{\pi} \\ $$$$ \\ $$$$ \\ $$