Residue theorem - simple integral around a singularity

Question

Could anyone explain in detail how do you take the integration contours in order to calculate the following integral?

$$\int_{y_c-\delta}^{y_c+\delta} \frac{1}{(y-y_c)^2} dy$$

where $\delta$ is very close to $y_c$ and there is a singularity at $y=y_c$.

Answer 1

I will denote $y_c$ as $c$.

In general, your integral equals $$I(\delta)=\int_{\gamma}(y-c)^{-2}dy$$

where $\gamma$ is a smooth curve comnecting $c-\delta$ to $c+\delta$.

Claim $$\lim_{\delta\to 0^+} |I(\delta)|=\infty$$

Proof:

By Residue theorem,

$$I(\delta)=\int_\rho (y-c)^{-2}dy$$ where $\rho$ is an infinitely small upper semicircle centered at $c$. By parametrization $$\int_\rho (y-c)^{-2}dy=\int^\pi_0 (c+\delta e^{it} -c)^{-2} i\delta e^{it}dt$$ which equals $$\frac i\delta\int^\pi_0 e^{-it}=\frac 2\delta$$

Hence, $$\lim_{\delta\to 0^+}|I(\delta)|=\lim_{\delta\to0^+}\frac 2\delta=\infty$$