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Question Number 28033 by abdo imad last updated on 18/Jan/18

$$1)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^2}dx$$$$2)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{xln\left(x\right)}{{(1+x^2)}^2}dx.$$

Commented by abdo imad last updated on 20/Jan/18

$$1)letputI={\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^2}dxthech.x=\frac{1}{t}give$$$$I=-{\int }_0^{\mathrm{\infty }}\frac{-ln\left(t\right)}{1+\frac{1}{t^2}}\frac{-dt}{t^2}=-{\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)}{t^2+1}=-I$$$$\Rightarrow 2I=0\Rightarrow I=0.$$$$2)letputJ={\int }_0^{\mathrm{\infty }}\frac{xln\left(x\right)}{{(1+x^2)}^2}dx$$$$x=tant\Rightarrow J={\int }_0^{\frac{\pi }{2}}\frac{tantln\left(tant\right)}{{({}^{}1+{tan}^{2}t)}^2}(1+{tan}^{2}t)dt$$$$={\int }_{0^{}}^{\frac{\pi }{2}}{cos}^{2}t\frac{sint}{cost}ln\left(\frac{sint}{cost}\right)dt={\int }_0^{\frac{\pi }{2}}costsintln\left(\frac{sint}{cost}\right)dt$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}sin\left(2t\right)(ln\left(sint\right)-ln\left(cost\right))dt$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}sin\left(2t\right)ln\left(sint\right)dt-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}sin\left(2t\right)ln\left(cost\right)dt$$$$thech.t=\frac{\pi }{2}-\alpha give$$$${\int }_0^{\frac{\pi }{2}}sin\left(2t\right)ln\left(cost\right)dt={\int }_0^{\frac{\pi }{2}}sin\left(2\alpha \right)ln\left(sin\alpha \right)d\alpha so$$$$J=0.$$$$$$

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