calculate ∫_0 ^∞ ((ln(3+x^2 ))/((x^2 +1)^2 ))dx


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Question Number 72011 by mathmax by abdo last updated on 23/Oct/19

$c a l c u l a t e \int_{0}^{\infty} \frac{l n (3 + x^{2})}{{(x^{2} + 1)}^{2}} d x$
Commented by mathmax by abdo last updated on 25/Oct/19
$let A =∫_0 ^∞ ((ln(3+x^2 ))/((x^2 +1)^2 ))dx ⇒2A =∫_(−∞) ^(+∞) ((ln(3+x^2 ))/((x^2 +1)^2 ))dx let f(z)=((ln(3+z^2 ))/((z^2 +1)^2 )) ⇒f(z) =((ln(3+z^2 ))/((z−i)^2 (z+i)^2 )) so the poles of f are i and −i (doubles) residus theorem give ∫_(−∞) ^(+∞) f(z)dz =2iπRes(f,i) Res(f,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 f(z)}^((1)) =lim_(z→i) {((ln(3+z^2 ))/((z+i)^2 ))}^((1)) =lim_(z→i) ((((2z)/(3+z^2 ))×(z+i)^2 −2(z+i)ln(3+z^2 ))/((z+i)^4 )) =lim_(z→i) ((((2z(z+i))/(3+z^2 ))−2ln(3+z^2 ))/((z+i)^3 )) =((((2i(2i))/2)−2ln(2))/((2i)^3 ))=((−2−2ln(2))/(−8i)) =((1+ln(2))/(4i)) ⇒ ∫_(−∞) ^(+∞) f(z)dz =2iπ×((1+ln(2))/(4i)) =(π/2)(1+ln(2)) =2A ⇒ A =(π/4)(1+ln(2))$
$l e t A = \int_{0}^{\infty} \frac{l n (3 + x^{2})}{{(x^{2} + 1)}^{2}} d x \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{l n (3 + x^{2})}{{(x^{2} + 1)}^{2}} d x l e t$ $f (z) = \frac{l n (3 + z^{2})}{{(z^{2} + 1)}^{2}} \Rightarrow f (z) = \frac{l n (3 + z^{2})}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f f a r e i$ $a n d - i (d o u b l e s) r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} f (z) d z = 2 i π R e s (f, i)$ $R e s (f, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} f (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{l n (3 + z^{2})}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{\frac{2 z}{3 + z^{2}} \times {(z + i)}^{2} - 2 (z + i) l n (3 + z^{2})}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{\frac{2 z (z + i)}{3 + z^{2}} - 2 l n (3 + z^{2})}{{(z + i)}^{3}} = \frac{\frac{2 i (2 i)}{2} - 2 l n (2)}{{(2 i)}^{3}} = \frac{- 2 - 2 l n (2)}{- 8 i}$ $= \frac{1 + l n (2)}{4 i} \Rightarrow \int_{- \infty}^{+ \infty} f (z) d z = 2 i π \times \frac{1 + l n (2)}{4 i} = \frac{π}{2} (1 + l n (2))$ $= 2 A \Rightarrow A = \frac{π}{4} (1 + l n (2))$
	Answered by mind is power last updated on 23/Oct/19

	$let f (t) = \int_{0}^{+ \infty} \frac{\ln (t + x^{2})}{{(x^{2} + 1)}^{2}} dx, t ⩾ 3$ $f^{'} (t) = \int_{0}^{+ \infty} \frac{dx}{(t + x^{2}) {(1 + x^{2})}^{2}}$ $f^{'} (t) = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{(t + x^{2}) {(1 + x^{2})}^{2}}$ $let Γ = C_{R} \cup [- R, R] our contor C_{R} = {Re}^{i θ}, 0 ⩽ θ ⩽ π$ $\int_{Γ} \frac{dx}{(t + x^{2}) {(1 + x^{2})}^{2}} = 2 i π (Res (f, i \sqrt{t}) + res (f, i))$ $Res (f, i \sqrt{t}) = \frac{1}{{(1 - t)}^{2} . 2 i \sqrt{t}}$ $Res (f, i) = \frac{d}{dx} \frac{{(x - i)}^{2}}{(t + x^{2}) {(x + i)}^{2} {(x - i)}^{2}} ∣_{x = i}$ $= \frac{d}{dx} (\frac{1}{(t + x^{2}) {(x + i)}^{2}}) = \frac{- 2 x {(x + i)}^{2} - 2 (x + i) (t + x^{2})}{{(t + x^{2})}^{2} {(x + i)}^{4}} ∣ x = i$ $= \frac{- 2 i {(2 i)}^{2} - 4 i (t - 1)}{{(t - 1)}^{2} (16)} = \frac{12 i - 4 it}{16 {(t - 1)}^{2}} . = \frac{- i (t - 3)}{4 {(t - 1)}^{2}}$ $f^{'} (t) = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{(t + x^{2}) {(1 + x^{2})}^{2}} = i π . (\frac{1}{{(1 - t)}^{2} 2 i \sqrt{t}} - \frac{i (t - 3)}{4 {(t - 1)}^{2}})$ $= \frac{π}{{(1 - t)}^{2} . 2 \sqrt{t}} + \frac{π (t - 1 - 2)}{4 {(t - 1)}^{2}} dt$ $f (t) = \int \frac{π dt}{{(1 - t)}^{2} . 2 \sqrt{t}} + \int \frac{π}{4 (t - 1)} dt - \frac{π}{2} \int \frac{dt}{{(t - 1)}^{2}}$ $for first u = \sqrt{t} \Rightarrow du = \frac{1}{2 \sqrt{t}} dt$ $f (t) = \frac{π}{4} \ln (∣ t - 1 ∣) + \frac{π}{2 (t - 1)} + \int \frac{π du}{{(1 - u)}^{2} {(1 + u)}^{2}}$ $\frac{a}{1 - u} + \frac{b}{1 + u} + \frac{π}{4 {(1 + u)}^{2}} + \frac{π}{4 {(1 - u)}^{2}}$ $b - a = π$ $b + a = \frac{π}{2}$ $b = \frac{3 π}{4}, a = - \frac{π}{4}$ $f (t) = \frac{π}{4} \ln (∣ t - 1 ∣) + \frac{π}{2 (t - 1)} + \int \frac{- π}{4 (1 - u)} dt + \int \frac{3 π}{4 (1 + u)} + \frac{π}{4} \int \frac{1}{{(1 + u)}^{2}} + \frac{1}{{(1 - u)}^{2}} du$ $after integration and put u = \sqrt{t}$ $f (t) = \frac{π}{4} \ln (∣ t - 1 ∣) + \frac{π}{2 (t - 1)} + \frac{π}{4} \ln (∣ \sqrt{t} - 1 ∣) + \frac{3 π}{4} \ln (1 + \sqrt{t}) - \frac{π}{4 (1 + \sqrt{t})} + \frac{π}{4 (1 - \sqrt{t)}} + c$ $f (0) = \int_{0}^{+ \infty} \frac{\ln (x^{2})}{{(1 + x^{2})}^{2}}$ $residus theorem$ $\int_{- R}^{- ϵ} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz + \int_{C_{ϵ}} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz + \int_{ϵ}^{R} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz + \int_{0}^{π} . \frac{\ln (R^{2} e^{i 2 θ)})}{{(1 + R^{2} e^{2 i θ})}^{2}} . {Rie}^{i θ} d θ$ $\int_{- R}^{- ϵ} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz = \int_{ϵ}^{R} \frac{2 \ln (z) + i π}{{(1 + z^{2})}^{2}} dz$ $\int_{C ϵ} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz = \int_{π}^{0} \frac{\ln (ϵ^{2} e^{2 i θ})}{{(1 + ϵ^{2} e^{2 i θ})}^{2}} . ϵ {ie}^{i θ} d θ$ $∣ \ln ({re}^{i θ}) ∣=∣ \ln (r) + i θ ∣= \sqrt{{(\ln (r))}^{2} + θ^{2}}$ $∣ 1 + ϵ e^{i θ} ∣⩾∣ 1 - ϵ ∣\Rightarrow$ $∣ \int_{C ϵ} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz ∣= \int_{0}^{π} ∣ \frac{\ln (ϵ^{2} e^{2 i θ})}{{(1 + ϵ^{2} e^{2 i θ})}^{2}} . ϵ {ie}^{i θ} d θ ∣⩽ \frac{ϵ \sqrt{{(\ln (r))}^{2} + θ^{2}}}{{(1 - ϵ^{2})}^{2}} . π \underset{ϵ \to 0}{\to} 0$ $meme idee marche pour R \to \infty$ $\Rightarrow \int_{Γ} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz = 2 \int_{0}^{+ \infty} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} + \int \frac{2 i π}{{(1 + z^{2})}^{2}} dz = 2 i π Res (f, i)$ $Res (f, i) = \frac{d}{dz} . \frac{\ln (z^{2})}{{(z + i)}^{2}} ∣_{z = i}$ $= \frac{\frac{2}{z} . {(z + i)}^{2} - 2 (z + i) \ln (z^{2})}{{(z + i)}^{4}} = \frac{- 2 i {(2 i)}^{2} - 4 iln (- 1)}{{(2 i)}^{4}} = \frac{8 i + 4 π}{16}$ $\Rightarrow 2 \int_{0}^{+ \infty} \frac{\ln (z^{2})}{{(1 + z^{2})}^{2}} dz = 2 i π . (\frac{8 i}{16}) = - π$ $\Rightarrow \int_{0}^{\infty} \frac{\ln (z^{2})}{(1 + z^{2})} dz = \frac{- π}{2}$ $\Rightarrow f (0) = - \frac{π}{2}$ $f (t) = \frac{π}{4} \ln (∣ t - 1 ∣) + \frac{π}{2 (t - 1)} + \frac{π}{4} \ln (∣ \sqrt{t} - 1 ∣) + \frac{3 π}{4} \ln (1 + \sqrt{t}) - \frac{π}{4 (1 + \sqrt{t})} + \frac{π}{4 (1 - \sqrt{t)}} + c$ $\Rightarrow f (0) = - \frac{π}{2} + \frac{π}{4} + \frac{π}{4} + c = - \frac{π}{2} \Rightarrow c = - \frac{π}{2}$ $\int_{0}^{+ \infty} \frac{\ln (3 + x^{2})}{{(1 + x^{2})}^{2}} dx = f (3)$
	Commented by mathmax by abdo last updated on 25/Oct/19

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 25/Oct/19

	$y^{'} re welcom$
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