Question Number 108692 by 150505R last updated on 18/Aug/20
Answered by Dwaipayan Shikari last updated on 18/Aug/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{log}\left({tan}\theta\right)}{{tan}^{\mathrm{2}} \theta+\mathrm{1}}{sec}^{\mathrm{2}} \theta{d}\theta\:\:\:\left({x}={tan}\theta,\:\mathrm{1}={sec}^{\mathrm{2}} \theta\frac{{d}\theta}{{dx}}\right) \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\theta\right)−{log}\left({sin}\theta\right)\:{d}\theta=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({cos}\theta\right)−{log}\left({sin}\theta\right)={I} \\ $$$$\mathrm{2}{I}=\mathrm{0} \\ $$$${I}=\mathrm{0} \\ $$$$ \\ $$
Commented by 150505R last updated on 18/Aug/20
$${thanks} \\ $$
Answered by mathmax by abdo last updated on 18/Aug/20
$$\mathrm{i}\:\mathrm{think}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{lnx}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$
Commented by 150505R last updated on 18/Aug/20
$${yes}\:{i}\:{has}\:{sent}\:{it}\:{again}. \\ $$
Commented by mnjuly1970 last updated on 18/Aug/20
![Answer is = (π^3 /8) hint put f(a)=∫_0 ^( ∞) (x^a /(1+x^(2 ) )) dx=^(x^2 =t) (1/2)∫_0 ^( ∞) (t^((a/2)−(1/2)) /(1+t))= =(1/2)β(((a+1)/2) , ((1−a)/2) )=(1/2)Γ(((a+1)/2) )Γ(((1−a)/2)) =(1/2) (π/(sin((((1+a)/2))π))) =(1/2) (π/(cos(((πa)/2)))) f^( ′′) (a) =(1/2)π(d/da)[(((π/2)sin(((πa)/2)))/(cos^2 (((πa)/2))=g(a)))]=(π^2 /4) (((π/2)cos(((πa)/2))g(a)−g^′ (a)sin(((πa)/2)))/(g^2 (a)))∣_(a=0) on the other hand :: f^( ′′) (a)=∫_0 ^( ∞) ((x^a ln^2 (x))/(1+x^2 )) dx goal:=f^( ′′) (0) =??? f^( ′′) (0)=(π^3 /(8 ))= goal.....](https://www.tinkutara.com/question/Q108726.png)
$$\mathbb{A}\mathrm{nswer}\:\mathrm{is}\:=\:\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\: \\ $$$$\mathrm{hint}\:\:\:\mathrm{put}\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\:\:\infty} \frac{{x}^{{a}} }{\mathrm{1}+{x}^{\mathrm{2}\:} }\:{dx}\overset{{x}^{\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\frac{{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{t}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\:,\:\frac{\mathrm{1}−{a}}{\mathrm{2}}\:\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\:\right)\Gamma\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi}{{sin}\left(\left(\frac{\mathrm{1}+{a}}{\mathrm{2}}\right)\pi\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi}{{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right)} \\ $$$${f}^{\:''} \left({a}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\pi\frac{{d}}{{da}}\left[\frac{\frac{\pi}{\mathrm{2}}{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}{{cos}^{\mathrm{2}} \left(\frac{\pi{a}}{\mathrm{2}}\right)={g}\left({a}\right)}\right]=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\frac{\frac{\pi}{\mathrm{2}}{cos}\left(\frac{\pi{a}}{\mathrm{2}}\right){g}\left({a}\right)−{g}^{'} \left({a}\right){sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}{{g}^{\mathrm{2}} \left({a}\right)}\mid_{{a}=\mathrm{0}} \\ $$$${on}\:{the}\:{other}\:{hand}\:\:::\:{f}^{\:''} \left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{a}} {ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\: \\ $$$${goal}:={f}^{\:''} \left(\mathrm{0}\right)\:=??? \\ $$$${f}^{\:''} \left(\mathrm{0}\right)=\frac{\pi^{\mathrm{3}} }{\mathrm{8}\:\:}=\:{goal}….. \\ $$$$\: \\ $$