# Evaluating a simple integral with the Cauchy residue theorem and a semicircular contour

I am taking a course next week that requires some basic integral techniques from complex analysis and I've been trying to quickly teach it to myself. I was given this sample problem to test my understanding, and I'm obviously missing something.

Suppose we have the function: $$f(\omega) = \frac{e^{i\omega t}}{\omega - \epsilon + i\gamma}$$ where $$\epsilon,\gamma\in\mathbb{R}$$ and $$\gamma>0$$. We want to integrate this function over the real line: $$\int_{-\infty}^\infty f(\omega)d\omega$$ Here's how I would attempt it. To compute the integral, construct a countour $$C$$ consisting of a straight line from $$-R$$ to $$R$$ on the real line, and then close it a half-circle arc in the upper half of the complex plane. We want to consider $$R\to\infty$$, in which case all poles in the upper half of the complex plane will be enclosed. The integrand approaches $$0$$ as $$|\omega|\to\infty$$, so in this limit, the semicircular arc has no contribution, and so the contour integral actually just approaches the above real integral we want to compute. The only pole of the function is at $$\epsilon-i\gamma$$, which is in the lower half of the complex plane, so it's not enclosed in the contour, and so by the Cauchy-Goursat theorem we have overall: $$\int_{-\infty}^\infty f(\omega) d\omega=\oint_C f(z)dz=0$$ Here's the source of my confusion: suppose I instead closed the contour by a semicircular arc in the lower half of the complex plane, giving a new contour $$C'$$. I'd now enclose the pole of the function, so I'd instead obtain from the residue theorem that: $$\int_{-\infty}^\infty f(\omega) d\omega=\oint_{C'} f(z)dz=2\pi i\text{Res}(f,\epsilon-i\gamma)$$ What have I misunderstood to get this discrepancy?

• What is $\;t\;$ ? Should it even be there? – DonAntonio Nov 17 '18 at 16:57

You should know that, in the upper half plane, the integral vanishes over the circular arc as $$R \to \infty$$ when $$t \gt 0$$. And analogously, the integral vanishes over the arc in the lower half plane when $$t \lt 0$$. Thus, your function will be of the form $$g(t) \theta(-t)$$, where $$g$$ is the result after taking the residue and $$\theta$$ is the Heaviside step function.
Let $$\; C_R:=\left\{\;z=Re^{i\tau}\;|\;\tau\in [0,\pi]\;\right\}\;$$ be the upper canonical semicircle of radius $$\;R\in\Bbb R^+\;$$ , and let $$\;-C_R\;$$ be corresponding lower canonical semicircle. Now
$$\left|\frac{e^{iwt}}{\omega-(\epsilon-i\gamma)}\right|=\frac{e^{-t R\sin\tau}}{|\omega-\epsilon+i\gamma|}\ge\frac{e^{-tR\sin\tau}}{R+|\epsilon+i\gamma|}$$
if $$\;t>0\;$$ and $$\;\tau\in [\pi,2\pi]\;$$ then the above is unbounded on the lower circle, and the other way around if $$\;t<0\;$$ . Thus, you cannot deduce the complex integral, in the limit, is the improper real one.