Question Number 167965 by mnjuly1970 last updated on 30/Mar/22
Answered by mindispower last updated on 01/Apr/22
![Ω=∫_0 ^∞ 2((1−e^(−2x) )/((1+e^(−2x) )^2 x))e^(−x) =−2∫_0 ^1 ((1−t^2 )/((1+t^2 )^2 ln(t))) =−2∫_∞ ^1 ((1−(1/t^2 ))/((1+(1/t^2 ))^2 ln((1/t)))).−(dt/t^2 ) =−2∫_1 ^∞ ((1−t^2 )/((t^2 +1)^2 ln(t)))⇒2Ω=−2∫_0 ^∞ ((1−t^2 )/((1+t^2 )ln(t)))dt f(a)=−2∫_0 ^∞ ((1−t^a )/((1+t^2 )^2 ln(t))),a∈[0,2] f(0)=0,f(2)=2Ω f′(a)=2∫_0 ^∞ (t^a /((1+t^2 )^2 ))dr =2∫_0 ^(π/2) sin^a (t)cos^(2−a) (t)dt =β(((a+1)/2),((3−a)/2))=((Γ(((a+1)/2))Γ(((3−a)/2)))/(Γ(2))) =(((1−a)/2)).(π/(sin((π/2)(1+a))))=((π(1−a))/(2cos(((πa)/2)))) 2Ω=∫_0 ^2 f′(a)da (π/2)∫_0 ^1 ((1−a)/(cos(((πa)/2))))+∫_1 ^2 ((1−a)/(cos(((πa)/2)))) =(π/2)(∫_0 ^1 ((ada)/(sin(((πa)/2))))+∫_0 ^1 ((−udu)/(cos((π/2)(1+u)))) =π∫_0 ^1 ((daa)/(sin(((πa)/2))))=(4/π)∫_0 ^(π/2) (u/(sin(u)))du ∫_0 ^(π/2) (u/(sin(u)))du=∫_0 ^(π/2) ((usin(u))/((1−cos(u))(1+cos(u))))du =(1/2)[uln(((1−cosx)/(1+cos(x))))]−(1/2)∫_0 ^(π/2) ln(((1−cos(x))/(1+cos(x)))) =−∫_0 ^(π/2) ln(tg((u/2)))du=−2∫_0 ^1 ((ln(x))/(1+x^2 ))=2G ⇒2Ω=(4/π).2G Ω=∫_0 ^∞ ((tanh (x))/x).sech (x)dx=((4G)/π)](Q168038.png)
$$\Omega=\int_{\mathrm{0}} ^{\infty} \mathrm{2}\frac{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{\left(\mathrm{1}+{e}^{−\mathrm{2}\boldsymbol{{x}}} \right)^{\mathrm{2}} \boldsymbol{{x}}}{e}^{−{x}} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} {ln}\left({t}\right)}\: \\ $$$$=−\mathrm{2}\int_{\infty} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{2}} {ln}\left(\frac{\mathrm{1}}{{t}}\right)}.−\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$=−\mathrm{2}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {ln}\left({t}\right)}\Rightarrow\mathrm{2}\Omega=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right){ln}\left({t}\right)}{dt} \\ $$$${f}\left({a}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{t}^{{a}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} {ln}\left({t}\right)},{a}\in\left[\mathrm{0},\mathrm{2}\right] \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0},{f}\left(\mathrm{2}\right)=\mathrm{2}\Omega \\ $$$${f}'\left({a}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dr} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{a}} \left({t}\right){cos}^{\mathrm{2}−{a}} \left({t}\right){dt} \\ $$$$=\beta\left(\frac{{a}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}−{a}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}−{a}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}\right)} \\ $$$$=\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right).\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{a}\right)\right)}=\frac{\pi\left(\mathrm{1}−{a}\right)}{\mathrm{2}{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)} \\ $$$$\mathrm{2}\Omega=\int_{\mathrm{0}} ^{\mathrm{2}} {f}'\left({a}\right){da} \\ $$$$\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{a}}{{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}−{a}}{{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ada}}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{udu}}{{cos}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{u}\right)\right.}\right) \\ $$$$ \\ $$$$=\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{daa}}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}=\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{u}}{{sin}\left({u}\right)}{du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{u}}{{sin}\left({u}\right)}{du}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{usin}\left({u}\right)}{\left(\mathrm{1}−{cos}\left({u}\right)\right)\left(\mathrm{1}+{cos}\left({u}\right)\right)}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{uln}\left(\frac{\mathrm{1}−{cosx}}{\mathrm{1}+{cos}\left({x}\right)}\right)\right]−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}−{cos}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}\right) \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({tg}\left(\frac{{u}}{\mathrm{2}}\right)\right){du}=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }=\mathrm{2}{G} \\ $$$$\Rightarrow\mathrm{2}\Omega=\frac{\mathrm{4}}{\pi}.\mathrm{2}{G} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tanh}\:\left({x}\right)}{{x}}.\mathrm{sech}\:\left({x}\right){dx}=\frac{\mathrm{4}{G}}{\pi} \\ $$$$ \\ $$$$ \\ $$