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Question Number 196408 by Erico last updated on 24/Aug/23

$$\mathrm{Calcul}{\underset{0}{\int }}^{+\mathrm{\infty }}\frac{\mathrm{lnt}}{\sqrt{\mathrm{t}}(1+{\mathrm{t}}^2)}\mathrm{dt}$$

Answered by qaz last updated on 24/Aug/23

$${\int }_0^{\mathrm{\infty }}\frac{lnt}{\sqrt{t}(1+t^2)}dt=\frac{\partial }{\partial a}{\mid }_{a=-1/2}{\int }_0^{\mathrm{\infty }}\frac{t^a}{1+t^2}dt$$$$=-\frac{\partial }{\partial a}{\mid }_{a=-1/2}\frac{\pi }{2}csc\left(\frac{a+1}{2}\pi \right)=\frac{{\pi }^2}{4}\cot \left(\frac{a+1}{2}\pi \right)csc\left(\frac{a+1}{2}\pi \right){\mid }_{a=-1/2}$$$$=\frac{\sqrt{2}{\pi }^2}{4}$$

Answered by Mathspace last updated on 25/Aug/23

$$I={\int }_0^{\mathrm{\infty }}\frac{lnt}{\sqrt{t}(1+t^2)}dt(\sqrt{t}=u)$$$$={\int }_0^{\mathrm{\infty }}\frac{2lnu}{u(1+u^4)}\left(2u\right)du$$$$=4{\int }_0^{\mathrm{\infty }}\frac{lnu}{1+u^4}du(u^4=x)$$$$=4{\int }_0^{\mathrm{\infty }}\frac{ln\left(x^{\frac{1}{4}}\right)}{1+x}\times \frac{1}{4}{x}^{\frac{1}{4}-1}dx$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{4}-1}lnx}{1+x}dx$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x}dx-1<a<1$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x}lnxdx\Rightarrow$$$$4I=f^{{}^{\prime }}\left(\frac{1}{4}\right)butf\left(a\right)=\frac{\pi }{sin\left(\pi a\right)}$$$$\Rightarrow f^{{}^{\prime }}\left(a\right)=\pi .\frac{-\pi cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}$$$$=-{\pi }^2\frac{cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}\Rightarrow$$$$f^{{}^{\prime }}\left(\frac{1}{4}\right)=-{\pi }^2\frac{cos\left(\frac{\pi }{4}\right)}{{sin}^2\left(\frac{\pi }{4}\right)}$$$$=-{\pi }^2.\frac{1}{\sqrt{2}.\left(\frac{1}{2}\right)}=-{\pi }^2\sqrt{2}$$$$I=\frac{1}{4}{f}^{{}^{\prime }}\left(\frac{1}{4}\right)\Rightarrow I=-\frac{\sqrt{2}}{4}{\pi }^2$$

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