Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 192407 by mnjuly1970 last updated on 17/May/23

$$$$$$findthevalueofthefollowingintegral$$$$$$$$\chi ={\int }_0^{\mathrm{\infty }}\frac{{\ln }^2\left(x\right)}{1+x^2}dx=?$$$$---------$$

Answered by ARUNG_Brandon_MBU last updated on 17/May/23

$$\chi ={\int }_0^{\mathrm{\infty }}\frac{{\ln }^{2}x}{1+x^2}dx=\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{t^{-\frac{1}{2}}{\ln }^{2}t}{1+t}dt$$$$=\frac{{\partial }^2}{\partial {\alpha }^2}{\mid }_{\alpha =\frac{1}{2}}\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{t^{\alpha -1}}{1+t}dt=\frac{{\partial }^2}{\partial {\alpha }^2}{\mid }_{\alpha =\frac{1}{2}}\frac{1}{8}\beta (\alpha ,1-\alpha )$$$$=\frac{{\partial }^2}{\partial {\alpha }^2}{\mid }_{\alpha =\frac{1}{2}}\frac{1}{8}\cdot \frac{\pi }{\sin \left(\pi \alpha \right)}=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =\frac{1}{2}}-\frac{1}{8}\cdot \frac{{\pi }^2\cos \left(\pi \alpha \right)}{{\sin }^2\left(\pi \alpha \right)}$$$$=-\frac{{\pi }^3}{8}{\mid }_{\alpha =\frac{1}{2}}(-\frac{1}{\sin \left(\pi \alpha \right)}-\frac{{2\cos }^2\left(\pi \alpha \right)}{{\sin }^3\left(\pi \alpha \right)})=\frac{{\pi }^3}{8}$$

Commented by mnjuly1970 last updated on 17/May/23

$$sonicesolutionSir$$$$thankyouverymuch.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com