Question Number 27691 by abdo imad last updated on 12/Jan/18
$${let}\:{give}\:\:{A}=\int\int_{\mathrm{0}\leqslant{y}\leqslant{x}\leqslant\mathrm{1}} \:\:\:\:\:\:\frac{{dxdxy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:\:{and} \\ $$$${B}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \theta\right)}{\mathrm{2}{cos}\left(\mathrm{2}\theta\right)}{d}\theta\:\:{calculate}\:{A}\:{and}\:{prove}\:{that}\:{B}={A}. \\ $$
Commented by abdo imad last updated on 19/Jan/18
![A= ∫_0 ^1 (∫_0 ^x (dy/(1+y^2 ))) (dx/(1+x^2 )) = ∫_0 ^1 ((arctanx)/(1+x^2 ))dx by parts A= [actanx.arctanx]_0 ^1 − ∫_0 ^1 arctanx (dx/(1+x^2 )) 2A=(π^2 /(16)) ⇒ A= (π^2 /(32)) B= ∫_0 ^(π/4) ((ln(2((1+cos(2θ))/2)))/(2cos(2θ)))dθ= ∫_0 ^(π/4) ((ln(1+cos(2θ)))/(2cos(2θ)))dθ = ∫_0 ^(π/2) ((ln(1+cost))/(2cost)) (dt/2) = (1/4) ∫_0 ^(π/2) ((ln(1+cost))/(cost))dt let introduce F(x)= ∫_0 ^(π/2) ((ln(1+xcost))/(cost))dt with −1<x<1 F^′ (x)= ∫_0 ^(π/2) (dt/(1+xcost)) the ch. tan((t/2))=u give F^ (x)= ∫_0 ^(1 ) (1/(1+x((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 ))= ∫_0 ^1 ((2du)/(1+u^2 +x −xu^2 )) = ∫_0 ^1 ((2du)/((1−x)u^2 +1+x)) = (2/(1−x)) ∫_0 ^1 (du/(u^2 +((1+x)/(1−x)))) the ch. u=(√((1+x)/(1−x)))t give F^ (x)= (2/(1−x)) ∫_0 ^((√(1−x))/(√(1+x ))) (1/(((1+x)/(1−x))(1+u^2 ))) ((√(1+x))/(√(1−x)))dt = (2/(1−x)) ((1−x)/(1+x)) ((√(1+x))/(√(1−x))) ∫_0 ^((√(1−x))/(√(1+x))) (dt/(1+t^2 )) = (2/(√(1−x^2 ))) arctan((√((1−x)/(1+x)))) but with x=cosθ arctan((√((1−x)/(1+x))))=arctan(tan((θ/2)))= (θ/2) F^ (x)=(2/(√(1−x^2 ))) ((arcosx)/2)= ((arcosx)/(√(1−x^2 ))) F(x)= ∫_0 ^x ((arcost)/(√(1−t^2 )))dt +λ ( λ=F(0)=0) ⇒ F(x)= ∫_0 ^x ((arcost)/(√(1−t^2 )))dt B= (1/4) F(1)= (1/4) ∫_0 ^(1 ) ((arcost)/(√(1−t^2 )))dt by parts let put I= ∫_0 ^1 ((arcost)/(√(1−t^2 ))) dt by parts I= −arcost.arcost]_0 ^1 − ∫_0 ^1 (−arcost) ((−dt)/(√(1−t^2 ))) 2I= (π^2 /4) ⇒ I= (π^2 /8) ⇒ F(1)= (π^2 /8) and B= (π^2 /(32)) =A. for F(x)= ∫_0 ^x ((arcost)/(√(1−t^2 )))dt let integrate by parts F(x)= −arcot arcost]_0 ^x − ∫_0 ^x (−arcosx) ((−dt)/(√(1−x^2 ))) = (π^2 /4) −(arcosx)^2 −F(x) ⇒ F(x)= (π^2 /8) −(1/2) (arcosx)^2 so ∫_0 ^(π/2) ((ln(1+xcost))/(cost))dt = (π^2 /8) − (1/2) (arcosx)^2 .](Q28046.png)
$${A}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts} \\ $$$${A}=\:\left[{actanx}.{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctanx}\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{A}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:\:\Rightarrow\:{A}=\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{32}} \\ $$$${B}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{ln}\left(\mathrm{2}\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right)}{\mathrm{2}{cos}\left(\mathrm{2}\theta\right)}{d}\theta=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{ln}\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right)}{\mathrm{2}{cos}\left(\mathrm{2}\theta\right)}{d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{ln}\left(\mathrm{1}+{cost}\right)}{\mathrm{2}{cost}}\:\frac{{dt}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{ln}\left(\mathrm{1}+{cost}\right)}{{cost}}{dt}\:{let}\:{introduce} \\ $$$${F}\left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{ln}\left(\mathrm{1}+{xcost}\right)}{{cost}}{dt}\:\:{with}\:\:−\mathrm{1}<{x}<\mathrm{1} \\ $$$${F}^{'} \left({x}\right)=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{dt}}{\mathrm{1}+{xcost}}\:\:{the}\:{ch}.\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give} \\ $$$${F}^{} \left({x}\right)=\:\:\int_{\mathrm{0}} ^{\mathrm{1}\:\:\:\:\:\:} \:\frac{\mathrm{1}}{\mathrm{1}+{x}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+{x}\:−{xu}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}−{x}\right){u}^{\mathrm{2}} +\mathrm{1}+{x}}\:=\:\frac{\mathrm{2}}{\mathrm{1}−{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{the}\:{ch}. \\ $$$${u}=\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{t}\:{give} \\ $$$${F}^{} \left({x}\right)=\:\:\frac{\mathrm{2}}{\mathrm{1}−{x}}\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{1}−{x}}}{\sqrt{\mathrm{1}+{x}\:\:}}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{\sqrt{\mathrm{1}+{x}}}{\sqrt{\mathrm{1}−{x}}}{dt} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{1}−{x}}\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:\frac{\sqrt{\mathrm{1}+{x}}}{\sqrt{\mathrm{1}−{x}}}\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{1}−{x}}}{\sqrt{\mathrm{1}+{x}}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\right)\:{but}\:{with}\:\:{x}={cos}\theta \\ $$$${arctan}\left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\right)={arctan}\left({tan}\left(\frac{\theta}{\mathrm{2}}\right)\right)=\:\frac{\theta}{\mathrm{2}} \\ $$$${F}^{} \left({x}\right)=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\frac{{arcosx}}{\mathrm{2}}=\:\frac{{arcosx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${F}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{{arcost}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:+\lambda\:\:\:\:\left(\:\lambda={F}\left(\mathrm{0}\right)=\mathrm{0}\right) \\ $$$$\Rightarrow\:\:\:\:{F}\left({x}\right)=\:\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{{arcost}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$${B}=\:\frac{\mathrm{1}}{\mathrm{4}}\:{F}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{1}\:\:} \:\:\:\frac{{arcost}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:{by}\:{parts} \\ $$$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arcost}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt}\:\:\:\:{by}\:{parts} \\ $$$$\left.{I}=\:\:−{arcost}.{arcost}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(−{arcost}\right)\:\frac{−{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$\mathrm{2}{I}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow\:{I}=\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:\:\Rightarrow\:{F}\left(\mathrm{1}\right)=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:{and}\:{B}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\:\:={A}. \\ $$$${for}\:\:{F}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{arcost}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:\:\:{let}\:{integrate}\:{by}\:{parts} \\ $$$$\left.{F}\left({x}\right)=\:−{arcot}\:{arcost}\right]_{\mathrm{0}} ^{{x}} \:\:−\:\int_{\mathrm{0}} ^{{x}} \:\left(−{arcosx}\right)\:\frac{−{dt}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:−\left({arcosx}\right)^{\mathrm{2}} \:−{F}\left({x}\right) \\ $$$$\Rightarrow\:{F}\left({x}\right)=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\left({arcosx}\right)^{\mathrm{2}} \:\:{so} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{xcost}\right)}{{cost}}{dt}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({arcosx}\right)^{\mathrm{2}} \:. \\ $$