Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 142430 by Mathspace last updated on 31/May/21

$$calculate{\int }_0^{\mathrm{\infty }}\frac{{log}^{2}x}{1+x^2}dx$$

Answered by Dwaipayan Shikari last updated on 31/May/21

$${\int }_0^{\mathrm{\infty }}\frac{x^a}{1+x^2}dx=\frac{\pi }{2sin\left(\frac{\pi (1-a)}{2}\right)}$$$${\int }_0^{\mathrm{\infty }}\frac{x^a{log}^2\left(x\right)}{1+x^2}dx=\frac{{\partial }^2}{\partial a^2}\left(\frac{\pi }{2}cosec\left(\frac{\pi }{2}(1-a)\right)\right)$$$${\int }_0^{\mathrm{\infty }}\frac{{log}^2\left(x\right)}{1+x^2}dx=\frac{{\partial }^2}{\partial a^2}{\mid }_{a=0}\frac{\pi }{2}cosec\left(\frac{\pi }{2}(1-a)\right)$$

Answered by mathmax by abdo last updated on 01/Jun/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}={\mathrm{t}}^{\frac{1}{2}}[\mathrm{give}$$$$\mathrm{\Phi }=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{{\log }^2\mathrm{t}}{1+\mathrm{t}}\frac{1}{2}{\mathrm{t}}^{-\frac{1}{2}}\mathrm{dt}=\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{-\frac{1}{2}}{\log }^2\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial \mathrm{a}}\left(\frac{{\mathrm{e}}^{(\mathrm{a}-1)\mathrm{logt}}}{1+\mathrm{t}}\right)\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}\mathrm{logt}}{1+\mathrm{t}}\Rightarrow {\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}{\log }^2\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow$$$${\mathrm{f}}^{\left(2\right)}\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{-\frac{1}{2}}{\log }^2\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}=8\mathrm{\Phi }$$$$\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{{\pi }^2\cos \left(\pi \mathrm{a}\right)}{{\sin }^2\left(\pi \mathrm{a}\right)}(0<\mathrm{a}<1)$$$$\mathrm{and}{\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)=-{\pi }^2.\frac{-\pi \sin \left(\pi \mathrm{a}\right){\sin }^2\left(\pi \mathrm{a}\right)-\cos \left(\pi \mathrm{a}\right).2\sin \left(\pi \mathrm{a}\right)\pi \cos \left(\pi \mathrm{a}\right)}{{\sin }^4\left(\pi \mathrm{a}\right)}$$$$8\mathrm{\Phi }={\mathrm{f}}^{\left(2\right)}\left(\frac{1}{2}\right)=-{\pi }^2\times \frac{-\pi }{1}={\pi }^3\Rightarrow \mathrm{\Phi }=\frac{{\pi }^3}{8}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com