find ∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))dx


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Question Number 108821 by mathmax by abdo last updated on 19/Aug/20

$find \int_{0}^{\infty} \frac{lnx}{{(x^{2} + 1)}^{2}} dx$
	Answered by mathmax by abdo last updated on 20/Aug/20
	$for this type of integral i use the formulae ∫_0 ^∞ q(x)ln(x)dx =−(1/2)Re(Σ_i Res(q(z)ln^2 z ,a_i ) with q(z)=(1/((z^2 +1)^2 )) let w(z) =((ln^2 z)/((z^2 +1)^2 )) poles of w? w(z) =((ln^2 z)/((z−i)^2 (z+i)^2 )) so the poles are +^− i (doubles) Res(w,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 w(z)}^((1)) =lim_(z→i) {((ln^2 z)/((z+i)^2 ))}^((1)) =lim_(z→i) ((2((lnz)/z)(z+i)^2 −2(z+i)ln^2 z)/((z+i)^4 )) =lim_(z→i) ((2((lnz)/z)(z+i)−2ln^2 z)/((z+i)^3 )) =((((4i)/i)ln(i)−2ln^2 (i))/((2i)^3 )) =((4(((iπ)/2))−2(((iπ)/2))^2 )/(−8i)) =((2iπ+(π^2 /2))/(−8i)) =−(π/4)+(π^2 /(16))i Res(w,−i) =lim_(z→−i) {((ln^2 z)/((z−i)^2 ))}^((1)) =lim_(z→−i) ((2((lnz)/z)(z−i)^2 −2(z−i)ln^2 z)/((z−i)^4 )) =lim_(z→−i) ((2((lnz)/z)(z−i)−2ln^2 z)/((z−i)^3 )) =((((−4i)/(−i))ln(−i)−2ln^2 (−i))/((−2i)^3 )) =((4(−((iπ)/2))−2(−((iπ)/2))^2 )/(8i)) =((−2iπ +(π^2 /2))/(8i)) =−(π/4) −(π^2 /(16))i ⇒ Σ Re(...) =−(π/2) ⇒∫_0 ^∞ ((lnx)/((x^2 +1)^2 ))dx =−(1/2)(−(π/2)) =(π/4)$
	$for this type of integral i use the formulae$ $\int_{0}^{\infty} q (x) \ln (x) dx = - \frac{1}{2} Re (\sum_{i} Res (q (z) \ln^{2} z, a_{i})$ $with q (z) = \frac{1}{{(z^{2} + 1)}^{2}} let w (z) = \frac{\ln^{2} z}{{(z^{2} + 1)}^{2}} poles of w ?$ $w (z) = \frac{\ln^{2} z}{{(z - i)}^{2} {(z + i)}^{2}} so the poles are \bar{+} i (doubles)$ $Res (w, i) = \lim_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} w (z)}^{(1)}$ $= \lim_{z \to i} {\frac{\ln^{2} z}{{(z + i)}^{2}}}^{(1)} = \lim_{z \to i} \frac{2 \frac{lnz}{z} {(z + i)}^{2} - 2 (z + i) \ln^{2} z}{{(z + i)}^{4}}$ $= \lim_{z \to i} \frac{2 \frac{lnz}{z} (z + i) - {2 \ln}^{2} z}{{(z + i)}^{3}} = \frac{\frac{4 i}{i} \ln (i) - {2 \ln}^{2} (i)}{{(2 i)}^{3}}$ $= \frac{4 (\frac{i π}{2}) - 2 {(\frac{i π}{2})}^{2}}{- 8 i} = \frac{2 i π + \frac{π^{2}}{2}}{- 8 i} = - \frac{π}{4} + \frac{π^{2}}{16} i$ $Res (w, - i) = \lim_{z \to - i} {\frac{\ln^{2} z}{{(z - i)}^{2}}}^{(1)}$ $= \lim_{z \to - i} \frac{2 \frac{lnz}{z} {(z - i)}^{2} - 2 (z - i) \ln^{2} z}{{(z - i)}^{4}}$ $= \lim_{z \to - i} \frac{2 \frac{lnz}{z} (z - i) - {2 \ln}^{2} z}{{(z - i)}^{3}} = \frac{\frac{- 4 i}{- i} \ln (- i) - {2 \ln}^{2} (- i)}{{(- 2 i)}^{3}}$ $= \frac{4 (- \frac{i π}{2}) - 2 {(- \frac{i π}{2})}^{2}}{8 i} = \frac{- 2 i π + \frac{π^{2}}{2}}{8 i} = - \frac{π}{4} - \frac{π^{2}}{16} i \Rightarrow$ $Σ Re (. . .) = - \frac{π}{2} \Rightarrow \int_{0}^{\infty} \frac{lnx}{{(x^{2} + 1)}^{2}} dx = - \frac{1}{2} (- \frac{π}{2}) = \frac{π}{4}$
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