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Question Number 108692 by 150505R last updated on 18/Aug/20

Answered by Dwaipayan Shikari last updated on 18/Aug/20

$${\int }_0^{\frac{\pi }{2}}\frac{2log\left(tan\theta \right)}{{tan}^2\theta +1}{sec}^2\theta d\theta (x=tan\theta ,1={sec}^2\theta \frac{d\theta }{dx})$$$$2{\int }_0^{\frac{\pi }{2}}log\left(sin\theta \right)-log\left(sin\theta \right)d\theta =2{\int }_0^{\frac{\pi }{2}}log\left(cos\theta \right)-log\left(sin\theta \right)=I$$$$2I=0$$$$I=0$$$$$$

Commented by 150505R last updated on 18/Aug/20

$$thanks$$

Answered by mathmax by abdo last updated on 18/Aug/20

$$\mathrm{i}\mathrm{think}\mathrm{the}\mathrm{question}\mathrm{is}{\int }_0^{\mathrm{\infty }}\frac{{\left(\mathrm{lnx}\right)}^2}{1+{\mathrm{x}}^2}\mathrm{dx}$$

Commented by 150505R last updated on 18/Aug/20

$$yesihassentitagain.$$

Commented by mnjuly1970 last updated on 18/Aug/20

$$\mathbb{A}\mathrm{nswer}\mathrm{is}=\frac{{\pi }^3}{8}$$$$\mathrm{hint}\mathrm{put}f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^a}{1+x^2}dx\stackrel{x^2=t}{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{a}{2}-\frac{1}{2}}}{1+t}=$$$$=\frac{1}{2}\beta (\frac{a+1}{2},\frac{1-a}{2})=\frac{1}{2}\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{1-a}{2}\right)$$$$=\frac{1}{2}\frac{\pi }{sin\left(\left(\frac{1+a}{2}\right)\pi \right)}=\frac{1}{2}\frac{\pi }{cos\left(\frac{\pi a}{2}\right)}$$$$f^{{}^{″}}\left(a\right)=\frac{1}{2}\pi \frac{d}{da}\left[\frac{\frac{\pi }{2}sin\left(\frac{\pi a}{2}\right)}{{cos}^2\left(\frac{\pi a}{2}\right)=g\left(a\right)}\right]=\frac{{\pi }^2}{4}\frac{\frac{\pi }{2}cos\left(\frac{\pi a}{2}\right)g\left(a\right)-g^{{}^{\prime }}\left(a\right)sin\left(\frac{\pi a}{2}\right)}{g^2\left(a\right)}{\mid }_{a=0}$$$$ontheotherhand::f^{{}^{″}}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^a{ln}^2\left(x\right)}{1+x^2}dx$$$$goal:=f^{{}^{″}}\left(0\right)=???$$$$f^{{}^{″}}\left(0\right)=\frac{{\pi }^3}{8}=goal.....$$$$$$

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