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Question Number 136425 by Ar Brandon last updated on 21/Mar/21

$$\mathrm{If}\alpha >0\mathrm{and}\beta >0,\mathrm{prove}$$ $${\int }_0^{\mathrm{\infty }}\frac{\ln \left(\alpha \mathrm{x}\right)}{{\beta }^2+{\mathrm{x}}^2}\mathrm{dx}=\frac{\pi }{2\beta }\ln \left(\alpha \beta \right)$$

Commented byDwaipayan Shikari last updated on 21/Mar/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{log\left(\alpha \right)}{{\beta }^2+x^2}+{\int }_0^{\mathrm{\infty }}\frac{log\left(x\right)}{x^2+{\beta }^2}dx$$ $$=\frac{\pi log\left(\alpha \right)}{2\beta }+{\tau }^{\prime }\left(0\right)$$ $$\tau \left(\eta \right)={\int }_0^{\mathrm{\infty }}\frac{x^{\eta }}{{\beta }^2+x^2}dx={\beta }^{\eta -1}{\int }_0^{\mathrm{\infty }}\frac{u^{\eta }}{u^2+1}dux=\beta u$$ $$=\frac{1}{2}{\beta }^{\eta -1}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{\eta +1}{2}-1}}{{(t+1)}^{\frac{\eta }{2}+\frac{1}{2}-\frac{\eta }{2}+\frac{1}{2}}}du=\frac{{\beta }^{\eta -1}}{2}.\frac{\mathrm{\Gamma }\left(\frac{\eta +1}{2}\right)\mathrm{\Gamma }(\frac{1}{2}-\frac{\eta }{2})}{\mathrm{\Gamma }\left(2\right)}$$ $$=\frac{{\beta }^{\eta -1}}{2}.\frac{\pi }{cos\left(\frac{\pi \eta }{2}\right)}$$ $${\tau }^{\prime }\left(\eta \right)=\frac{{\beta }^{\eta -1}log\left(\beta \right)}{2}.\frac{\pi }{cos\left(\frac{\pi \eta }{2}\right)}+\frac{{\beta }^{n-1}{\pi }^2}{4}sec\left(\frac{\pi \eta }{2}\right)tan\left(\frac{\pi \eta }{2}\right)$$ $${\tau }^{\prime }\left(0\right)=\frac{log\left(\beta \right)\pi }{2\beta }$$ $$\mathrm{\Phi }=\frac{\pi log\left(\alpha \right)}{2\beta }+\frac{\pi log\left(\beta \right)}{2\beta }=\frac{\pi log\left(\alpha \beta \right)}{2\beta }$$ $$$$

Commented byAr Brandon last updated on 21/Mar/21

Thanks

Commented byDwaipayan Shikari last updated on 21/Mar/21

$$$$ 😃\n

Answered by mathmax by abdo last updated on 21/Mar/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\log \left(\alpha \mathrm{x}\right)}{{\mathrm{x}}^2+{\beta }^2}\mathrm{dx}\Rightarrow \mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\log \alpha +\mathrm{logx}}{{\mathrm{x}}^2+{\beta }^2}\mathrm{dx}$$ $$=\log \alpha {\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+{\beta }^2}(\to \mathrm{x}=\beta \mathrm{u})+{\int }_0^{\mathrm{\infty }}\frac{\mathrm{logx}}{{\mathrm{x}}^2+{\beta }^2}\mathrm{dx}(\to \mathrm{x}=\beta \mathrm{u})$$ $$=\log \alpha {\int }_0^{\mathrm{\infty }}\frac{\beta \mathrm{du}}{{\beta }^2(1+{\mathrm{u}}^2)}+{\int }_0^{\mathrm{\infty }}\frac{\log \left(\beta \mathrm{u}\right)}{{\beta }^2(1+{\mathrm{u}}^2)}\beta \mathrm{du}$$ $$=\frac{\log \alpha }{\beta }.\frac{\pi }{2}+\frac{1}{\beta }{\int }_0^{\mathrm{\infty }}\frac{\log \beta +\mathrm{logu}}{1+{\mathrm{u}}^2}\mathrm{du}$$ $$=\frac{\pi \log \alpha }{2\beta }+\frac{\log \beta }{\beta }.\frac{\pi }{2}+\frac{1}{\beta }{\int }_0^{\mathrm{\infty }}\frac{\mathrm{logu}}{1+{\mathrm{u}}^2}\mathrm{du}(\to 0)$$ $$=\frac{\pi (\log \alpha +\log \beta )}{2\beta }=\frac{\pi }{2\beta }\log \left(\alpha \beta \right)$$

Commented byAr Brandon last updated on 21/Mar/21

Thanks

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