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Question Number 28033 by abdo imad last updated on 18/Jan/18
1) find the value of ∫_0 ^∞ ((ln(x))/(1+x^2 )) dx 2) find the value of ∫_0 ^∞ ((xln(x))/((1+x^2 )^2 ))dx .
$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$\left.\mathrm{2}\right)\:\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo imad last updated on 20/Jan/18
1) let put I= ∫_0 ^∞ ((ln(x))/(1+x^2 ))dx the ch. x=(1/t) give I= −∫_0 ^∞ ((−ln(t))/(1+(1/t^2 ))) ((−dt)/t^2 )= −∫_0 ^∞ ((ln(t))/(t^2 +1)) =−I ⇒2I=0⇒ I=0. 2) let put J= ∫_0 ^∞ ((xln(x))/((1+x^2 )^2 ))dx x=tant⇒ J= ∫_0 ^(π/2) ((tantln(tant))/((^ 1+tan^2 t)^2 ))(1+tan^2 t)dt = ∫_0^ ^(π/2) cos^2 t ((sint)/(cost))ln(((sint)/(cost)))dt= ∫_0 ^(π/2) cost sint ln(((sint)/(cost)))dt = (1/2) ∫_0 ^(π/2) sin(2t)(ln(sint) −ln(cost))dt = (1/2)∫_0 ^(π/2) sin(2t)ln(sint)dt −(1/2)∫_0 ^(π/2) sin(2t)ln(cost)dt the ch. t= (π/2) −α give ∫_0 ^(π/2) sin(2t)ln(cost)dt=∫_0 ^(π/2) sin(2α)ln(sinα)dα so J =0 .
$$\left.\mathrm{1}\right)\:{let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{the}\:{ch}.\:{x}=\frac{\mathrm{1}}{{t}}\:{give} \\ $$$${I}=\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{−{ln}\left({t}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\:\frac{−{dt}}{{t}^{\mathrm{2}} }=\:−\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}\:=−{I} \\ $$$$\Rightarrow\mathrm{2}{I}=\mathrm{0}\Rightarrow\:\:{I}=\mathrm{0}. \\ $$$$\left.\mathrm{2}\right)\:\:{let}\:{put}\:{J}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{xln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\: \\ $$$${x}={tant}\Rightarrow\:\:{J}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{tantln}\left({tant}\right)}{\left(^{} \mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\:\int_{\mathrm{0}^{} } ^{\frac{\pi}{\mathrm{2}}} \:\:\:{cos}^{\mathrm{2}} {t}\:\frac{{sint}}{{cost}}{ln}\left(\frac{{sint}}{{cost}}\right){dt}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cost}\:{sint}\:{ln}\left(\frac{{sint}}{{cost}}\right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}\left(\mathrm{2}{t}\right)\left({ln}\left({sint}\right)\:−{ln}\left({cost}\right)\right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\left(\mathrm{2}{t}\right){ln}\left({sint}\right){dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left(\mathrm{2}{t}\right){ln}\left({cost}\right){dt} \\ $$$${the}\:{ch}.\:{t}=\:\frac{\pi}{\mathrm{2}}\:−\alpha\:\:{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\left(\mathrm{2}{t}\right){ln}\left({cost}\right){dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\left(\mathrm{2}\alpha\right){ln}\left({sin}\alpha\right){d}\alpha\:\:{so} \\ $$$${J}\:=\mathrm{0}\:\:. \\ $$$$ \\ $$

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