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Question Number 32740 by caravan msup abdo. last updated on 01/Apr/18
find∫_0 ^∞ ((ln(x^2 +t^2 ))/(1+t^2 ))dt
$${find}\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$
Commented by abdo imad last updated on 04/Apr/18
let put f(x)=∫_0 ^∞ ((ln(x^2 +t^2 ))/(1+t^2 ))dt we have f^′ (x) =∫_0 ^∞ ((2x)/((x^2 +t^2 )(1+t^2 )))dt = x∫_(−∞) ^(+∞) (dt/((x^2 +t^2 )(1+t^2 ))) let untroduce the complex function ϕ(z) = (1/((z^2 +x^2 )(z^2 +1))) = (1/((z +ix)(z−ix)(z−i)(z+i))) the poles of ϕ are i,−i,ix,−ix (simples) case 1 x>0 ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(Res(ϕ,i) +Res(ϕ,ix)) but Res(ϕ,i) = (1/(2i(i+ix)(i−ix))) = (1/(−2i(1−x^2 ))) = (i/(2(1−x^2 ))) Res(ϕ,ix) = (1/(2ix(ix−i)(ix+i))) = (1/(−2ix(x^2 −1))) = (i/(2x(x^2 −1))) ∫_(−∞) ^(+∞) ϕ(z)dz = 2iπ( (i/(2(1−x^2 ))) −(i/(2x(1−x^2 )))) = ((−π)/(1−x^2 ))( 1 −(1/x))=− ((π(x−1))/(x(1−x^2 )))=((π(1−x))/(x(1−x)(1+x)))= (π/(x(1+x))) f^′ (x) = (π/(1+x)) with condition x^2 ≠1 ⇒ f(x) =πln∣1+x∣ +λ λ =f(0) =∫_0 ^∞ ((2lnt)/(1+t^2 )) dt =2 ∫_0 ^∞ ((lnt)/(1+t^2 ))dt =0(result proved) ⇒ f(x) =π ln∣1+x∣ case2 x<0 the poles are i and −ix ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ ,i) +Res(ϕ,−ix)) Res(ϕ,i) = (i/(2(1−x^2 ))) Res(ϕ,−ix) = (1/(−2ix(−ix−i)(−ix+i))) = (1/(−2ix(ix +i)(ix−i))) = (1/(2ix(x^2 −1))) = (i/(2x(1−x^2 )))
$${let}\:{put}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{we}\:{have} \\ $$$${f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:=\:{x}\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$${let}\:{untroduce}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\left({z}\:+{ix}\right)\left({z}−{ix}\right)\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:{i},−{i},{ix},−{ix}\:\:\left({simples}\right) \\ $$$${case}\:\mathrm{1}\:\:{x}>\mathrm{0} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left({Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{ix}\right)\right)\:{but} \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\:\frac{\mathrm{1}}{\mathrm{2}{i}\left({i}+{ix}\right)\left({i}−{ix}\right)}\:=\:\frac{\mathrm{1}}{−\mathrm{2}{i}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:=\:\frac{{i}}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,{ix}\right)\:=\:\:\frac{\mathrm{1}}{\mathrm{2}{ix}\left({ix}−{i}\right)\left({ix}+{i}\right)}\:=\:\frac{\mathrm{1}}{−\mathrm{2}{ix}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:=\:\frac{{i}}{\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\left(\:\:\frac{{i}}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:\:−\frac{{i}}{\mathrm{2}{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\right) \\ $$$$=\:\frac{−\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\left(\:\mathrm{1}\:−\frac{\mathrm{1}}{{x}}\right)=−\:\frac{\pi\left({x}−\mathrm{1}\right)}{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}=\frac{\pi\left(\mathrm{1}−{x}\right)}{{x}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}=\:\frac{\pi}{{x}\left(\mathrm{1}+{x}\right)} \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\pi}{\mathrm{1}+{x}}\:{with}\:{condition}\:{x}^{\mathrm{2}} \neq\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\pi{ln}\mid\mathrm{1}+{x}\mid\:+\lambda \\ $$$$\lambda\:={f}\left(\mathrm{0}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\mathrm{0}\left({result}\:{proved}\right) \\ $$$$\Rightarrow\:{f}\left({x}\right)\:=\pi\:{ln}\mid\mathrm{1}+{x}\mid \\ $$$${case}\mathrm{2}\:{x}<\mathrm{0}\:\:{the}\:{poles}\:{are}\:{i}\:{and}\:−{ix} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi\:,{i}\right)\:+{Res}\left(\varphi,−{ix}\right)\right) \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\frac{{i}}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,−{ix}\right)\:=\:\frac{\mathrm{1}}{−\mathrm{2}{ix}\left(−{ix}−{i}\right)\left(−{ix}+{i}\right)} \\ $$$$=\:\:\frac{\mathrm{1}}{−\mathrm{2}{ix}\left({ix}\:+{i}\right)\left({ix}−{i}\right)}\:=\:\:\frac{\mathrm{1}}{\mathrm{2}{ix}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:=\:\frac{{i}}{\mathrm{2}{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$
Commented by abdo imad last updated on 04/Apr/18
∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( (i/(2(1−x^2 ))) + (i/(2x(1−x^2 )))) =−(π/(1−x^2 ))( 1+ (1/x)) = ((−π(1+x))/(x(1−x^2 ))) = ((−π)/(x(1−x))) =(π/(x(x−1))) f^′ (x) = (π/(x−1)) ⇒ f(x)=πln∣x−1∣ +λ but λ=f(0)=0 ⇒ f(x)=π ln∣1−x∣ .
$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\frac{{i}}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:\:+\:\:\frac{{i}}{\mathrm{2}{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\right) \\ $$$$=−\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\left(\:\mathrm{1}+\:\frac{\mathrm{1}}{{x}}\right)\:=\:\frac{−\pi\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:=\:\frac{−\pi}{{x}\left(\mathrm{1}−{x}\right)}\:=\frac{\pi}{{x}\left({x}−\mathrm{1}\right)} \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\pi}{{x}−\mathrm{1}}\:\:\Rightarrow\:{f}\left({x}\right)=\pi{ln}\mid{x}−\mathrm{1}\mid\:+\lambda\:{but}\:\lambda={f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({x}\right)=\pi\:{ln}\mid\mathrm{1}−{x}\mid\:. \\ $$

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