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Question Number 108821 by mathmax by abdo last updated on 19/Aug/20
find ∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))dx
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$
Answered by mathmax by abdo last updated on 20/Aug/20
for this type of integral i use the formulae ∫_0 ^∞ q(x)ln(x)dx =−(1/2)Re(Σ_i Res(q(z)ln^2 z ,a_i ) with q(z)=(1/((z^2 +1)^2 )) let w(z) =((ln^2 z)/((z^2 +1)^2 )) poles of w? w(z) =((ln^2 z)/((z−i)^2 (z+i)^2 )) so the poles are +^− i (doubles) Res(w,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 w(z)}^((1)) =lim_(z→i) {((ln^2 z)/((z+i)^2 ))}^((1)) =lim_(z→i) ((2((lnz)/z)(z+i)^2 −2(z+i)ln^2 z)/((z+i)^4 )) =lim_(z→i) ((2((lnz)/z)(z+i)−2ln^2 z)/((z+i)^3 )) =((((4i)/i)ln(i)−2ln^2 (i))/((2i)^3 )) =((4(((iπ)/2))−2(((iπ)/2))^2 )/(−8i)) =((2iπ+(π^2 /2))/(−8i)) =−(π/4)+(π^2 /(16))i Res(w,−i) =lim_(z→−i) {((ln^2 z)/((z−i)^2 ))}^((1)) =lim_(z→−i) ((2((lnz)/z)(z−i)^2 −2(z−i)ln^2 z)/((z−i)^4 )) =lim_(z→−i) ((2((lnz)/z)(z−i)−2ln^2 z)/((z−i)^3 )) =((((−4i)/(−i))ln(−i)−2ln^2 (−i))/((−2i)^3 )) =((4(−((iπ)/2))−2(−((iπ)/2))^2 )/(8i)) =((−2iπ +(π^2 /2))/(8i)) =−(π/4) −(π^2 /(16))i ⇒ Σ Re(...) =−(π/2) ⇒∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))dx =−(1/2)(−(π/2)) =(π/4)
$$\mathrm{for}\:\mathrm{this}\:\mathrm{type}\:\mathrm{of}\:\mathrm{integral}\:\:\mathrm{i}\:\mathrm{use}\:\mathrm{the}\:\mathrm{formulae} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{q}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\sum_{\mathrm{i}} \:\mathrm{Res}\left(\mathrm{q}\left(\mathrm{z}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{z}\:,\mathrm{a}_{\mathrm{i}} \right)\right. \\ $$$$\mathrm{with}\:\mathrm{q}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{let}\:\mathrm{w}\left(\mathrm{z}\right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\mathrm{poles}\:\mathrm{of}\:\mathrm{w}? \\ $$$$\mathrm{w}\left(\mathrm{z}\right)\:=\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} }\:\:\mathrm{so}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{are}\:\overset{−} {+}\mathrm{i}\:\left(\mathrm{doubles}\right) \\ $$$$\mathrm{Res}\left(\mathrm{w},\mathrm{i}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} \mathrm{w}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\left\{\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\frac{\mathrm{2}\frac{\mathrm{lnz}}{\mathrm{z}}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}+\mathrm{i}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\frac{\mathrm{2}\frac{\mathrm{lnz}}{\mathrm{z}}\left(\mathrm{z}+\mathrm{i}\right)−\mathrm{2ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\:=\frac{\frac{\mathrm{4i}}{\mathrm{i}}\mathrm{ln}\left(\mathrm{i}\right)−\mathrm{2ln}^{\mathrm{2}} \left(\mathrm{i}\right)}{\left(\mathrm{2i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{4}\left(\frac{\mathrm{i}\pi}{\mathrm{2}}\right)−\mathrm{2}\left(\frac{\mathrm{i}\pi}{\mathrm{2}}\right)^{\mathrm{2}} }{−\mathrm{8i}}\:=\frac{\mathrm{2i}\pi+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}}{−\mathrm{8i}}\:=−\frac{\pi}{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\mathrm{i} \\ $$$$\mathrm{Res}\left(\mathrm{w},−\mathrm{i}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{i}} \:\:\left\{\frac{\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{i}} \:\:\:\:\frac{\mathrm{2}\frac{\mathrm{lnz}}{\mathrm{z}}\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}−\mathrm{i}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{i}} \:\:\:\:\frac{\mathrm{2}\frac{\mathrm{lnz}}{\mathrm{z}}\left(\mathrm{z}−\mathrm{i}\right)−\mathrm{2ln}^{\mathrm{2}} \mathrm{z}}{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} }\:=\frac{\frac{−\mathrm{4i}}{−\mathrm{i}}\mathrm{ln}\left(−\mathrm{i}\right)−\mathrm{2ln}^{\mathrm{2}} \left(−\mathrm{i}\right)}{\left(−\mathrm{2i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{4}\left(−\frac{\mathrm{i}\pi}{\mathrm{2}}\right)−\mathrm{2}\left(−\frac{\mathrm{i}\pi}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{8i}}\:=\frac{−\mathrm{2i}\pi\:+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}}{\mathrm{8i}}\:=−\frac{\pi}{\mathrm{4}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\mathrm{i}\:\Rightarrow \\ $$$$\Sigma\:\mathrm{Re}\left(…\right)\:=−\frac{\pi}{\mathrm{2}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{lnx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{4}} \\ $$

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