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Complex Integration using residue theorem

  1. Dec 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Untitled.png

    ##C_\rho## is a semicircle of radius ##\rho## in the upper-half plane.
    What is
    $$\lim_{\rho\rightarrow 0} \int_{C_{\rho}} \frac{e^{iaz}-e^{ibz}}{z^2} \,dz$$


    2. Relevant equations
    If ##C## is a closed loop and ##z_1, z_2 ... z_n## are the singular points inside ##C##,
    $$\int_C f(z) \,dz =2\pi i \sum_{k=1}^n Res_{z=z_k}f(z)$$

    3. The attempt at a solution
    According to the solution,
    $$--------------$$
    $$\begin{align}
    f(z)&= \frac{e^{iaz}-e^{ibz}}{z^2}\\
    &= \frac{1}{z^2}\left[\left(1+\frac{iaz}{1!}+\frac{(iaz)^2}{2!}+\frac{(iaz)^3}{3!}+...\right)-\left(1+\frac{ibz}{1!}+\frac{(ibz)^2}{2!}+\frac{(ibz)^3}{3!}+...\right)\right]\\
    &=\frac{i(a-b)}{z}+...~~~(0<|z|<\infty)
    \end{align}$$
    So ##z=0## is a simple pole of ##f(z)##, with residue ##B_0=i(a-b)##. Thus
    $$\begin{align} \lim_{\rho\rightarrow 0} \int_{C_{\rho}}f(z) \,dz &=-B_0\pi i\\
    &=-i(a-b)\pi i\\
    &=\pi(a-b)\end{align}$$
    $$--------------$$
    I do not understand equation (4).
    I guess Cauchy's Residue Theorem is applied here. According to that theorem, integral over a closed contour is equal to ##2\pi i## times the residue. I guess, it was assumed that for semicircular contour, the integral would be ##\pi i## times the residue, may be because of integrals over an upper semi-circle and over a lower semi-circle being equal.
    But I do not understand why the integrals over an upper semi-circle and over a lower semi-circle are equal.
    ##f(z)## is not symmetric about real-axis, because
    $$\begin{align}
    f(z)=&f(re^{i\theta})\\
    &=\frac{\exp(ia\cdot re^{i\theta})-\exp(ib\cdot re^{i\theta})}{r^2 e^{i2\theta}}\\
    &=\frac{\exp(-ar\sin{\theta+iar\cos{\theta}})-\exp(-br\sin{\theta+ibr\cos{\theta}})}{r^2e^{i2\theta}}\end{align}$$
    Now, if ##\theta## is substituted by ##-\theta##, the expression is not the same.
     
  2. jcsd
  3. Dec 7, 2016 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    To obtain eq. (4), just evaluate the integral ##\int_{C_{\rho}} (1/z) dz## explicitly; there is no need to use Theorems.
     
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