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Question Number 113766 by Riteshgoyal last updated on 15/Sep/20
I=∫_0 ^∞ ((π/(1+π^2 x^2 ))−(1/(1+x^2 )))lnx dx put πx=tanA, x =tanB I=∫_0 ^(π/2) (ln(tanA)−lnπ)dA−∫_0 ^(π/2) ln(tanB)dB I=((−π)/2)lnπ
$$ \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\pi}{\mathrm{1}+\pi^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){lnx}\:{dx} \\ $$$${put}\:\pi{x}={tanA},\:{x}\:={tanB} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({ln}\left({tanA}\right)−{ln}\pi\right){dA}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({tanB}\right){dB} \\ $$$${I}=\frac{−\pi}{\mathrm{2}}{ln}\pi \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 16/Sep/20
I =∫_0 ^∞ ((π/(1+π^2 x^2 ))−(1/(1+x^2 )))lnx dx =π∫_0 ^∞ ((lnx)/(1+π^2 x^2 )) −∫_0 ^∞ ((lnx)/(1+x^2 )) dx ∫_0 ^∞ ((lnx)/(1+x^2 )) dx =0 (put x=(1/u)) π∫_0 ^∞ ((ln(x))/(1+(πx)^2 )) dx =_(πx =t) π ∫_0 ^∞ ((ln((t/π)))/(1+t^2 ))×(dt/π) =∫_0 ^∞ ((lnt−lnπ)/(1+t^2 )) dt =∫_0 ^∞ ((lnt)/(1+t^2 ))dt −ln(π)×(π/2) =0−(π/2)ln(π) ⇒ I =−(π/2)ln(π)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\pi}{\mathrm{1}+\pi^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\mathrm{lnx}\:\mathrm{dx}\:=\pi\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnx}}{\mathrm{1}+\pi^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }\:−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\mathrm{0}\:\:\:\left(\mathrm{put}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{u}}\right) \\ $$$$\pi\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{1}+\left(\pi\mathrm{x}\right)^{\mathrm{2}} }\:\mathrm{dx}\:=_{\pi\mathrm{x}\:=\mathrm{t}} \pi\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{ln}\left(\frac{\mathrm{t}}{\pi}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }×\frac{\mathrm{dt}}{\pi} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnt}−\mathrm{ln}\pi}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:−\mathrm{ln}\left(\pi\right)×\frac{\pi}{\mathrm{2}}\:=\mathrm{0}−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\pi\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\pi\right) \\ $$

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