Question Number 137937 by benjo_mathlover last updated on 08/Apr/21
$$\:\underset{−\infty} {\overset{\:\:\:\infty} {\int}}\frac{\mathrm{ln}\left(\:\mid{x}\mid\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:=? \\ $$
Answered by Ñï= last updated on 14/Apr/21
$$\int_{−\infty} ^{+\infty} \frac{{ln}\mid{x}\mid}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${I}\left({a}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}=\mathrm{2}{a}\int_{\mathrm{0}} ^{\infty} \frac{{lna}+{lnx}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{lna}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\pi}{{a}}{lna}+\left\{\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{−{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{0}\right\} \\ $$$$=\frac{\pi}{{a}}{lna} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \frac{{ln}\mid{x}\mid}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{{I}\left({a}\right)'}{\mathrm{2}{a}}=\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }\left({lna}−\mathrm{1}\right) \\ $$