find ∫_0 ^∞ ((lnx)/(x^2 −i))dx (i=(√(−1)))


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Question Number 116096 by mathmax by abdo last updated on 30/Sep/20

$find \int_{0}^{\infty} \frac{lnx}{x^{2} - i} dx (i = \sqrt{- 1})$
	Answered by mindispower last updated on 01/Oct/20
	$let f(z)=((ln(z))/(z^2 −i)) ln(z)=ln∣z∣+iarg(z),arg(z)∈]−(π/2),((3π)/2)[ C_R =∪_(a≤R) {ae^(iθ) ∈[0,π]} pol of ((ln(z))/(z^2 −i)) are ((1+i)/( (√2))),−((1+i)/( (√2))) ∫_C_R f(z)dz=2iπRes(f,((1+i)/( (√2)))), whenR≥1 ∫_C_R f(z)dz=∫_(−R) ^0 f(z)dz+∫_0 ^R f(z)dz+∫_(Re^(iθ) ) f(z)dz ∫_(−R) ^0 f(z)dz=∫_0 ^R f(−z)dz=∫_0 ^R ((ln(−z))/(z^2 −i))dz =∫_0 ^R ((ln(z)+iπ)/(z^2 −i))dz ∫_(Re^(iθ) ) f(z)dz=∫_0 ^π f(Re^(iθ) ).iRe^(iθ) dθ =∫_0 ^π ((ln(R)+iθ)/(R^2 e^(2iθ) −i)).iRe^(iθ) dθ ∣((ln(R)+iθ)/(R^2 e^(2iθ) −i)).iRe^(iθ) ∣≤((R(√(ln^2 (R)+π^2 )))/(∣R^2 −1∣))→0 when R→0 ⇒lim_(R→∞) ∫_C_R f(z)dz=2iπRes(f,((1+i)/( (√2)))) =2∫_0 ^∞ ((ln(z))/(z^2 −i))dz+∫_0 ^∞ ((iπdz)/(z^2 −i))=2iπ.((ln(e^(i(π/4)) ))/(2.e^(i(π/4)) )) =2iπ.((iπ)/4).(1/( (√2)(1+i)))=−((π^2 (1−i))/(4(√2))) ∫_0 ^∞ ((iπ)/(z^2 −i))=∫_0 ^∞ (i/(2e^(i(π/4)) ))π.((1/(z−e^(i(π/4)) ))−(1/(z+e^(i(π/4)) )))dz =i(π/(2e^(i(π/4)) ))lim_(R→∞) [ln(((R−e^(i(π/4)) )/(R+e^(i(π/4)) )))−ln(−1)) =((iπ)/(2e^(i(π/4)) )).−iπ=(π^2 /(2e^(i(π/4)) )) ⇔2∫_0 ^∞ ((ln(z))/(z^2 −i))=−(π^2 /4)e^(−i(π/4)) −(π^2 /2)e^(−i(π/4)) ⇒∫_0 ^∞ ((ln(z))/(z^2 −i))dz=((−3π^2 )/8)(((1−i)/( (√2))))$
	$l e t f (z) = \frac{l n (z)}{z^{2} - i}$ $l n (z) = l n ∣ z ∣ + i a r g (z), a r g (z) \in] - \frac{π}{2}, \frac{3 π}{2} [$ $C_{R} = \underset{a ⩽ R}{\cup} {{a e}^{i θ} \in [0, π]}$ $p o l o f \frac{l n (z)}{z^{2} - i} a r e \frac{1 + i}{\sqrt{2}}, - \frac{1 + i}{\sqrt{2}}$ $\int_{C_{R}} f (z) d z = 2 i π R e s (f, \frac{1 + i}{\sqrt{2}}), w h e n R ⩾ 1$ $\int_{C_{R}} f (z) d z = \int_{- R}^{0} f (z) d z + \int_{0}^{R} f (z) d z + \int_{{R e}^{i θ}} f (z) d z$ $\int_{- R}^{0} f (z) d z = \int_{0}^{R} f (- z) d z = \int_{0}^{R} \frac{l n (- z)}{z^{2} - i} d z$ $= \int_{0}^{R} \frac{l n (z) + i π}{z^{2} - i} d z$ $\int_{{R e}^{i θ}} f (z) d z = \int_{0}^{π} f ({R e}^{i θ}) . {i R e}^{i θ} d θ$ $= \int_{0}^{π} \frac{l n (R) + i θ}{R^{2} e^{2 i θ} - i} . {i R e}^{i θ} d θ$ $∣ \frac{l n (R) + i θ}{R^{2} e^{2 i θ} - i} . {i R e}^{i θ} ∣⩽ \frac{R \sqrt{{l n}^{2} (R) + π^{2}}}{∣ R^{2} - 1 ∣} \to 0 w h e n R \to 0$ $\Rightarrow \lim_{R \to \infty} \int_{C_{R}} f (z) d z = 2 i π R e s (f, \frac{1 + i}{\sqrt{2}})$ $= 2 \int_{0}^{\infty} \frac{l n (z)}{z^{2} - i} d z + \int_{0}^{\infty} \frac{i π d z}{z^{2} - i} = 2 i π . \frac{l n (e^{i \frac{π}{4}})}{2 . e^{i \frac{π}{4}}}$ $= 2 i π . \frac{i π}{4} . \frac{1}{\sqrt{2} (1 + i)} = - \frac{π^{2} (1 - i)}{4 \sqrt{2}}$ $\int_{0}^{\infty} \frac{i π}{z^{2} - i} = \int_{0}^{\infty} \frac{i}{2 e^{i \frac{π}{4}}} π . (\frac{1}{z - e^{i \frac{π}{4}}} - \frac{1}{z + e^{i \frac{π}{4}}}) d z$ $= i \frac{π}{2 e^{i \frac{π}{4}}} \lim_{R \to \infty} [l n (\frac{R - e^{i \frac{π}{4}}}{R + e^{i \frac{π}{4}}}) - l n (- 1))$ $= \frac{i π}{2 e^{i \frac{π}{4}}} . - i π = \frac{π^{2}}{2 e^{i \frac{π}{4}}}$ $\Leftrightarrow 2 \int_{0}^{\infty} \frac{l n (z)}{z^{2} - i} = - \frac{π^{2}}{4} e^{- i \frac{π}{4}} - \frac{π^{2}}{2} e^{- i \frac{π}{4}}$ $\Rightarrow \int_{0}^{\infty} \frac{l n (z)}{z^{2} - i} d z = \frac{- 3 π^{2}}{8} (\frac{1 - i}{\sqrt{2}})$
	Commented by 1549442205PVT last updated on 01/Oct/20

	$Do You need the condition for f (z)$ $= \frac{\ln (z)}{z^{2} - i} is an analytical function ?$
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