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Question Number 113766 by Riteshgoyal last updated on 15/Sep/20

$$$$$$I={\int }_0^{\mathrm{\infty }}(\frac{\pi }{1+{\pi }^2{x}^2}-\frac{1}{1+x^2})lnxdx$$$$put\pi x=tanA,x=tanB$$$$I={\int }_0^{\frac{\pi }{2}}(ln\left(tanA\right)-ln\pi )dA-{\int }_0^{\pi /2}ln\left(tanB\right)dB$$$$I=\frac{-\pi }{2}ln\pi$$$$$$$$$$$$$$

Answered by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}(\frac{\pi }{1+{\pi }^2{\mathrm{x}}^2}-\frac{1}{1+{\mathrm{x}}^2})\mathrm{lnx}\mathrm{dx}=\pi {\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{1+{\pi }^2{\mathrm{x}}^2}-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{1+{\mathrm{x}}^2}\mathrm{dx}=0(\mathrm{put}\mathrm{x}=\frac{1}{\mathrm{u}})$$$$\pi {\int }_0^{\mathrm{\infty }}\frac{\ln \left(\mathrm{x}\right)}{1+{\left(\pi \mathrm{x}\right)}^2}\mathrm{dx}{=}_{\pi \mathrm{x}=\mathrm{t}}\pi {\int }_0^{\mathrm{\infty }}\frac{\ln \left(\frac{\mathrm{t}}{\pi }\right)}{1+{\mathrm{t}}^2}\times \frac{\mathrm{dt}}{\pi }$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnt}-\ln \pi }{1+{\mathrm{t}}^2}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnt}}{1+{\mathrm{t}}^2}\mathrm{dt}-\ln \left(\pi \right)\times \frac{\pi }{2}=0-\frac{\pi }{2}\ln \left(\pi \right)\Rightarrow$$$$\mathrm{I}=-\frac{\pi }{2}\ln \left(\pi \right)$$

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