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Question Number 136555 by mnjuly1970 last updated on 23/Mar/21

$$....nice......calculus....$$$$provethat$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{ln(1+x^2)}{{(1+x^2)}^2}dx=\frac{\pi }{2}ln\left(2\right)-\frac{\pi }{4}$$$$::::::::$$

Answered by mnjuly1970 last updated on 23/Mar/21

$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{{(1+x^2)}^a}...(a>\frac{1}{2})$$$$f\left(a\right)\stackrel{x^2=y}{=}\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{y^{\frac{-1}{2}}}{{(1+y)}^a}dy=\beta (\frac{1}{2},a-\frac{1}{2})$$$$f\left(a\right)=\frac{1}{2}\left(\frac{\sqrt{\pi }\mathrm{\Gamma }(a-\frac{1}{2})}{\mathrm{\Gamma }\left(a\right)}\right)$$$$goal::\mathit{\varphi }=-f{}^{\prime }\left(2\right)$$$$f{}^{\prime }\left(a\right)=\frac{\sqrt{\pi }}{2}\left(\frac{{\mathrm{\Gamma }}^{\prime }(a-\frac{1}{2}).\mathrm{\Gamma }\left(a\right)-{\mathrm{\Gamma }}^{\prime }\left(a\right).\mathrm{\Gamma }(a-\frac{1}{2})}{{\mathrm{\Gamma }}^2\left(a\right)}\right)$$$$f{}^{\prime }\left(2\right)=\frac{\sqrt{\pi }}{2}\left(\frac{\psi \left(\frac{3}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)-\psi \left(2\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{1^2}\right)$$$$=\frac{\sqrt{\pi }}{2}\left(\frac{\sqrt{\pi }}{2}(2-\gamma -2ln\left(2\right)-(1-\gamma ))\right)$$$$\therefore -\mathit{\varphi }=\frac{\pi }{4}(2-\gamma -2ln\left(2\right)-1+\gamma )$$$$=\frac{\pi }{4}(1-2ln\left(2\right))=\frac{\pi }{4}-\frac{\pi }{2}ln\left(2\right)..✓$$$$\therefore \mathit{\varphi }=\frac{\pi }{2}ln\left(2\right)-\frac{\pi }{4}....$$$$$$$$m.n$$

Answered by mindispower last updated on 23/Mar/21

$$={\int }_0^{\frac{\pi }{2}}\frac{ln\left(\frac{1}{{cos}^2\left(x\right)}\right)}{{(1+{tg}^2\left(x\right))}^2}dtg\left(x\right)$$$$={\int }_0^{\frac{\pi }{2}}-2{cos}^2\left(x\right)ln\left(cos\left(x\right)\right)dx..A$$$$\beta (a,b)=2{\int }_0^{\frac{\pi }{2}}{cos}^{2a-1}\left(x\right).{sin}^{2b-1}\left(x\right)dx$$$$A=-\frac{1}{2}{\partial }^a\beta (a,b)\mid (a,b)=(\frac{3}{2},\frac{1}{2})$$$$=-\frac{1}{2}\beta (a,b)(\mathrm{\Psi }\left(a\right)-\mathrm{\Psi }(a+b))$$$$=-\frac{1}{2}.\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(2\right)}(\mathrm{\Psi }\left(\frac{3}{2}\right)-\mathrm{\Psi }\left(2\right))$$$$=-\frac{1}{2}.\frac{1}{2}.\pi (2+\mathrm{\Psi }\left(\frac{1}{2}\right)-(1-\gamma ))$$$$=-\frac{\pi }{4}(1-2ln\left(2\right))=\frac{\pi ln\left(2\right)}{4}-\frac{\pi }{4}$$$$$$$$$$

Commented by mnjuly1970 last updated on 23/Mar/21

$$thanksalot....$$

Answered by Dwaipayan Shikari last updated on 23/Mar/21

$$\vartheta \left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{{(1+x^2)}^{\alpha }}{{(1+x^2)}^2}dx={\int }_0^{\mathrm{\infty }}\frac{1}{{(1+x^2)}^{2-\alpha }}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2}-1}}{{(1+u)}^{\frac{1}{2}+\frac{3}{2}-\alpha }}du=\frac{1}{2}.\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(\frac{3}{2}-\alpha )}{\mathrm{\Gamma }(2-\alpha )}$$$${\vartheta }^{\prime }\left(\alpha \right)=\frac{-\sqrt{\pi }\mathrm{\Gamma }(2-\alpha ){\mathrm{\Gamma }}^{\prime }(\frac{3}{2}-\alpha )+\sqrt{\pi }\mathrm{\Gamma }(2-\alpha )\psi (2-\alpha )\mathrm{\Gamma }(\frac{3}{2}-\alpha )}{2{\mathrm{\Gamma }}^2(2-\alpha )}$$$${\vartheta }^{\prime }\left(0\right)=\frac{-\frac{\pi }{2}\psi \left(\frac{3}{2}\right)+\frac{\pi }{2}\psi \left(2\right)}{2}=\frac{\pi }{4}(-\gamma +1+\gamma -2log\left(2\right))$$$$=\frac{\pi }{4}-\frac{\pi }{2}log\left(2\right)$$$$-{\vartheta }^{\prime }\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{log(x^2+1)}{{(x^2+1)}^2}dx=\frac{\pi }{2}log\left(2\right)-\frac{\pi }{4}$$

Commented by mnjuly1970 last updated on 23/Mar/21

$$thanking...$$

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