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Question Number 137937 by benjo_mathlover last updated on 08/Apr/21

 ∫_(−∞) ^(   ∞) ((ln( ∣x∣))/((x^2 +a^2 )^2 )) dx =?

$$\:\underset{−\infty} {\overset{\:\:\:\infty} {\int}}\frac{\mathrm{ln}\left(\:\mid{x}\mid\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:=? \\ $$

Answered by Ñï= last updated on 14/Apr/21

∫_(−∞) ^(+∞) ((ln∣x∣)/((x^2 +a^2 )^2 ))dx=2∫_0 ^∞ ((lnx)/((x^2 +a^2 )^2 ))dx  I(a)=2∫_0 ^∞ ((lnx)/(x^2 +a^2 ))dx=2a∫_0 ^∞ ((lna+lnx)/(a^2 x^2 +a^2 ))dx  =(2/a)∫_0 ^∞ ((lna)/(x^2 +1))dx+(2/a)∫_0 ^∞ ((lnx)/(x^2 +1))dx  =(π/a)lna+{(2/a)∫_0 ^∞ ((−lnx)/(1+x^2 ))dx=0}  =(π/a)lna  ⇒∫_(−∞) ^(+∞) ((ln∣x∣)/((x^2 +a^2 )^2 ))dx=−((I(a)′)/(2a))=(π/(2a^3 ))(lna−1)

$$\int_{−\infty} ^{+\infty} \frac{{ln}\mid{x}\mid}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${I}\left({a}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}=\mathrm{2}{a}\int_{\mathrm{0}} ^{\infty} \frac{{lna}+{lnx}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{lna}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\pi}{{a}}{lna}+\left\{\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{−{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{0}\right\} \\ $$$$=\frac{\pi}{{a}}{lna} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \frac{{ln}\mid{x}\mid}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{{I}\left({a}\right)'}{\mathrm{2}{a}}=\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }\left({lna}−\mathrm{1}\right) \\ $$

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