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Question Number 140310 by liberty last updated on 06/May/21

prove that ∫_0 ^∞ ((ln x)/(x^2 +1)) dx = 0

$$\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx}\:=\:\mathrm{0} \\ $$

Answered by qaz last updated on 06/May/21

∫_0 ^∞ ((lnx)/(1+x^2 ))dx=∫_∞ ^0 ((−lnx)/(1+x^2 ))(−1)dx =−∫_0 ^∞ ((lnx)/(1+x^2 ))dx=0

$$\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\infty} ^{\mathrm{0}} \frac{−{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }\left(−\mathrm{1}\right){dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$

Commented by TheSupreme last updated on 06/May/21

t=(1/x) dx=−(1/t^2 )dt ∫_0 ^∞ ((ln(x))/(1+x^2 ))dx=∫_∞ ^0 ((ln((1/t)))/(1+(1/t^2 )))(−(1/t^2 ))dt=∫_∞ ^0 ((ln(t))/(t^2 +1))dt I=−I=0

$${t}=\frac{\mathrm{1}}{{x}} \\ $$$${dx}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\infty} ^{\mathrm{0}} \frac{{ln}\left(\frac{\mathrm{1}}{{t}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\left(−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}=\int_{\infty} ^{\mathrm{0}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$${I}=−{I}=\mathrm{0} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 06/May/21

∫_0 ^∞ ((lnx)/(x^2 +1))dx =−(1/2)Re(Σ Res(f a_i )) with f(z)=((log^2 z)/(z^2 +1)) f(z)=((log^2 z)/((z−i)(z+i))) ⇒Res(f,i)=((log^2 i)/(2i)) =(((log(e^((iπ)/2) ))^2 )/(2i))=(((((iπ)/2))^2 )/(2i))=((−π^2 )/(8i)) Res(f,−i)=((log^2 (−i))/(2i)) =((log^2 (e^(−((iπ)/2)) ))/(2i)) =−(π^2 /(8i)) ⇒ Σ Res=((iπ^2 )/8)+((iπ^2 )/8) =((iπ^2 )/4) ⇒Re(Σ Res)=0 ⇒∫_0 ^∞ ((logx)/(x^2 +1))dx=0

$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\Sigma\:\mathrm{Res}\left(\mathrm{f}\:\mathrm{a}_{\mathrm{i}} \right)\right)\:\mathrm{with}\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{log}^{\mathrm{2}} \mathrm{z}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{log}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)}\:\Rightarrow\mathrm{Res}\left(\mathrm{f},\mathrm{i}\right)=\frac{\mathrm{log}^{\mathrm{2}} \mathrm{i}}{\mathrm{2i}}\:=\frac{\left(\mathrm{log}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)\right)^{\mathrm{2}} }{\mathrm{2i}}=\frac{\left(\frac{\mathrm{i}\pi}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2i}}=\frac{−\pi^{\mathrm{2}} }{\mathrm{8i}} \\ $$$$\mathrm{Res}\left(\mathrm{f},−\mathrm{i}\right)=\frac{\mathrm{log}^{\mathrm{2}} \left(−\mathrm{i}\right)}{\mathrm{2i}}\:=\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)}{\mathrm{2i}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{8i}}\:\Rightarrow \\ $$$$\Sigma\:\mathrm{Res}=\frac{\mathrm{i}\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{i}\pi^{\mathrm{2}} }{\mathrm{8}}\:=\frac{\mathrm{i}\pi^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\mathrm{Re}\left(\Sigma\:\mathrm{Res}\right)=\mathrm{0}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{logx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}=\mathrm{0} \\ $$

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