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Question Number 31102 by abdo imad last updated on 02/Mar/18

$$find{\int }_0^{+\mathrm{\infty }}\frac{lnx}{x^2+a^2}dx$$$$2)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+a^2)}^3}.$$

Commented by abdo imad last updated on 05/Mar/18

$$letputf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{lnx}{x^2+a^2}dxch.x=atwitha>0give$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{lna+lnt}{a^2(1+t^2)}dt=\frac{lna}{a^2}{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}+\frac{1}{a^2}{\int }_0^{\mathrm{\infty }}\frac{lnt}{1+t^2}dtbut$$$$wehaveprovedthat{\int }_0^{\mathrm{\infty }}\frac{lnt}{1+t^2}dt=0\Rightarrow$$$$f\left(a\right)=\frac{\pi lna}{2a^2}$$$$2)wehave{f}^{{}^{\prime }}\left(a\right)=-2a{\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+a^2)}^2}dx\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+a^2)}^2}=\frac{-1}{2a}{f}^{{}^{\prime }}\left(a\right)=\frac{-\pi lna}{4a^3}.$$

Commented by abdo imad last updated on 05/Mar/18

$$\frac{d}{da}\left(\frac{-f^{{}^{\prime }}\left(a\right)}{2a}\right)={\int }_0^{\mathrm{\infty }}\frac{-22a(x^2+a^2)}{{(x^2+a^2)}^4}lnxdx=-4a{\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+a^2)}^3}dx$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{lnx}{{(x^2+a^2)}^3}dx=\frac{-1}{4a}\frac{\partial }{\partial a}\left(\frac{-\pi lna}{4a^3}\right)$$$$=\frac{\pi }{16a}{\left(\frac{lna}{a^3}\right)}^{{}^{\prime }}=\frac{\pi }{16a}\left(\frac{a^2-3a^{2}lna}{a^6}\right)=\frac{\pi a^2(1-3lna)}{16a^7}$$$$=\frac{\pi (1-3lna)}{16a^4}.$$$$$$

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