Question Number 116096 by mathmax by abdo last updated on 30/Sep/20
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{i}}\mathrm{dx}\:\:\:\:\:\left(\mathrm{i}=\sqrt{−\mathrm{1}}\right) \\ $$
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))))](https://www.tinkutara.com/question/Q116156.png)
$${let}\:{f}\left({z}\right)=\frac{{ln}\left({z}\right)}{{z}^{\mathrm{2}} −{i}} \\ $$$$\left.{ln}\left({z}\right)={ln}\mid{z}\mid+{iarg}\left({z}\right),{arg}\left({z}\right)\in\right]−\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}}\left[\right. \\ $$$${C}_{{R}} \:=\underset{{a}\leqslant{R}} {\cup}\left\{{ae}^{{i}\theta} \in\left[\mathrm{0},\pi\right]\right\} \\ $$$${pol}\:{of}\:\frac{{ln}\left({z}\right)}{{z}^{\mathrm{2}} −{i}}\:{are}\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}},−\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f},\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\right),\:{whenR}\geqslant\mathrm{1} \\ $$$$\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\int_{−{R}} ^{\mathrm{0}} {f}\left({z}\right){dz}+\int_{\mathrm{0}} ^{{R}} {f}\left({z}\right){dz}+\int_{{Re}^{{i}\theta} } {f}\left({z}\right){dz} \\ $$$$\int_{−{R}} ^{\mathrm{0}} {f}\left({z}\right){dz}=\int_{\mathrm{0}} ^{{R}} {f}\left(−{z}\right){dz}=\int_{\mathrm{0}} ^{{R}} \frac{{ln}\left(−{z}\right)}{{z}^{\mathrm{2}} −{i}}{dz} \\ $$$$=\int_{\mathrm{0}} ^{{R}} \frac{{ln}\left({z}\right)+{i}\pi}{{z}^{\mathrm{2}} −{i}}{dz} \\ $$$$\int_{{Re}^{{i}\theta} } {f}\left({z}\right){dz}=\int_{\mathrm{0}} ^{\pi} {f}\left({Re}^{{i}\theta} \right).{iRe}^{{i}\theta} {d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{ln}\left({R}\right)+{i}\theta}{{R}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} −{i}}.{iRe}^{{i}\theta} {d}\theta \\ $$$$\mid\frac{{ln}\left({R}\right)+{i}\theta}{{R}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} −{i}}.{iRe}^{{i}\theta} \mid\leqslant\frac{{R}\sqrt{{ln}^{\mathrm{2}} \left({R}\right)+\pi^{\mathrm{2}} }}{\mid{R}^{\mathrm{2}} −\mathrm{1}\mid}\rightarrow\mathrm{0}\:{when}\:{R}\rightarrow\mathrm{0} \\ $$$$\Rightarrow\underset{\boldsymbol{{R}}\rightarrow\infty} {\mathrm{lim}}\int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f},\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({z}\right)}{{z}^{\mathrm{2}} −{i}}{dz}+\int_{\mathrm{0}} ^{\infty} \frac{{i}\pi{dz}}{{z}^{\mathrm{2}} −{i}}=\mathrm{2}{i}\pi.\frac{{ln}\left({e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}{\mathrm{2}.{e}^{{i}\frac{\pi}{\mathrm{4}}} } \\ $$$$=\mathrm{2}{i}\pi.\frac{{i}\pi}{\mathrm{4}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+{i}\right)}=−\frac{\pi^{\mathrm{2}} \left(\mathrm{1}−{i}\right)}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{i}\pi}{{z}^{\mathrm{2}} −{i}}=\int_{\mathrm{0}} ^{\infty} \frac{{i}}{\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} }\pi.\left(\frac{\mathrm{1}}{{z}−{e}^{{i}\frac{\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{z}+{e}^{{i}\frac{\pi}{\mathrm{4}}} }\right){dz} \\ $$$$={i}\frac{\pi}{\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} }\underset{{R}\rightarrow\infty} {\mathrm{lim}}\left[{ln}\left(\frac{{R}−{e}^{{i}\frac{\pi}{\mathrm{4}}} }{{R}+{e}^{{i}\frac{\pi}{\mathrm{4}}} }\right)−{ln}\left(−\mathrm{1}\right)\right) \\ $$$$=\frac{{i}\pi}{\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} }.−{i}\pi=\frac{\pi^{\mathrm{2}} }{\mathrm{2}{e}^{{i}\frac{\pi}{\mathrm{4}}} } \\ $$$$\Leftrightarrow\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({z}\right)}{{z}^{\mathrm{2}} −{i}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} −\frac{\pi^{\mathrm{2}} }{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({z}\right)}{{z}^{\mathrm{2}} −{i}}{dz}=\frac{−\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{8}}\left(\frac{\mathrm{1}−{i}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by 1549442205PVT last updated on 01/Oct/20
$$\mathrm{Do}\:\mathrm{You}\:\mathrm{need}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{z}\right) \\ $$$$=\frac{\mathrm{ln}\left(\mathrm{z}\right)}{\mathrm{z}^{\mathrm{2}} −\mathrm{i}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{analytical}\:\mathrm{function}? \\ $$