Bounding complex contour integrals

Question

Background: We wish to compute $$\int_0^\pi \left| i\epsilon e^{i\theta}\cdot\frac{e^{i\alpha}e^{i\alpha \epsilon e^{i\theta}}}{(1+\epsilon e^{i\theta})^4-1}\right|\,d\theta$$This arises from taking a small contour of radius $\epsilon$ around the point $z=1$ where you had some complex function with $z^4-1$ in the denominator. We have $\alpha\in\Bbb R$.

Question: How can I find the value of this as $\epsilon\to0$?

Attempt: $$=\int_0^\pi\frac{e^{-\alpha \epsilon \sin\theta}}{|\epsilon^3e^{3i\theta}+4\epsilon^2 e^{2i\theta}+6\epsilon e^{i\theta}+4|}\,d\theta$$I am not sure what to do with the denominator now. It seems as $\epsilon\to0$, we just get $\pi/4$. Is it this simple? I was expecting the whole thing to go to $0$, but it seems that is not the case?

Answer 1

The integral is equal to $\frac{\pi}4$. Taking the limit into the integral is allowed.

Just as what @JohnDoe commented, the estimation is not so useful.

Note that $$\text{Res}_{z=1}\frac{e^{iaz}}{z^4-1}=2I$$ where $I$ is your integral(without absolute values).

I am 90% confident it is not a coincidence that $$|2\pi\text{Res}_{z=k}f(z)|=|\oint f(z)dz|=\int^{2\pi}_0 |f(k+\epsilon e^{i\theta})|d\theta$$

Here, $f(z)= \frac{e^{iaz}}{z^4-1}$, You evaluated integral equals(with absolute values) $\pi/4$ while $$\text{Res}_{z=1}f(z)=\frac{e^{ia}}4$$

EDIT:

Answer to the question in your comment: Yes.

A useful lemma from this answer by @SangchulLee:

Lemma. Suppose $f(z)$ is holomorphic near $z = z_0$. Fix $\theta_0 \in (0, 2\pi)$. If $\gamma_{\epsilon}$ denotes a counter-clockwise oriented arc of angle $\theta_0$ on the circle of radius $\epsilon$ centered at $z_0$, then $$ \lim_{\epsilon\to0} \int_{\gamma_{\epsilon}} \frac{f(\zeta)}{\zeta-z_0}\;d\zeta=i\theta_0 f(z_0).$$

Proof. By the substitution $\zeta = z_0 + \epsilon e^{i\theta}$, we have $$\begin{align*} \left| \int_{\gamma_{\epsilon}} \frac{f(\zeta)}{\zeta-z_0}\;d\zeta - i\theta_0 f(z_0)\right| &= \left| i \int_{\theta_1}^{\theta_1+\theta_0} f(z_0 + \epsilon e^{i\theta})\;d\theta - i\theta_0 f(z_0)\right| \\ & \leq \int_{\theta_1}^{\theta_1+\theta_0} \left| f(z_0 + \epsilon e^{i\theta}) - f(z_0) \right| \;d\theta, \end{align*}$$ which clearly goes to zero when $\epsilon \to 0$.