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Question Number 111024 by mathmax by abdo last updated on 01/Sep/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\mathrm{lnx}}{{(1+{\mathrm{x}}^2)}^3}\mathrm{dx}$$

Answered by mathdave last updated on 01/Sep/20

$$solution$$$$letI={\int }_0^{\mathrm{\infty }}\frac{x^2\ln x}{{(1+x^2)}^3}dxthen$$$$\frac{\partial }{\partial a}{\mid }_{a=0}I\left(a\right)=\frac{1}{4}\frac{\partial }{\partial a}{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{2}+a}}{{(1+x)}^3}dx$$$$I^1\left(a\right)=\frac{1}{4}\frac{\partial }{\partial a}\beta (\frac{3}{2}+a,\frac{3}{2}-a)=\frac{1}{4}\frac{\partial }{\partial a}\left[\frac{\mathrm{\Gamma }(\frac{3}{2}+a)\mathrm{\Gamma }(\frac{3}{2}-a)}{\mathrm{\Gamma }\left(3\right)}\right]$$$$I^{{}^{\prime }}\left(0\right)=\frac{1}{4}\bullet \frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(3\right)}[\psi \left(\frac{3}{2}\right)-\psi \left(\frac{3}{2}\right)]=0$$$${\int }_0^{\mathrm{\infty }}\frac{x^2\ln x}{{(1+x^2)}^3}dx=0$$$$mathdave\left(01/09/2020\right)$$$$$$

Commented by abdomsup last updated on 01/Sep/20

$$thankyou$$

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