Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 111025 by mathmax by abdo last updated on 01/Sep/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2\ln \left(\mathrm{x}\right)}{{(1+\mathrm{x})}^4}\mathrm{dx}$$

Answered by mathdave last updated on 01/Sep/20

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com