Complex integration around a singularity

Question

I am trying to integrate the function $f(z)=$$\frac{5}{z^2}$ from -3 to 3 and I am supposed to develop a closed region that avoids the origin and use the analyticity of the function in this region to integrate the function in a way that is simpler than parametrizing along some path. Any suggestions?

Answer 1

Do line integrals as usual in real analysis, i.e. parametrization and stuff:

$$\begin{align*}\text{From $\;-3\;$ to}\;\,i:&\;\;\; -3\le x\le0\;,\;\;\;y=\frac13x+1\\{}\\ \text{From $\;i\;$ to}\;\,3:&\;\;\;\;\;\;\;0\le x\le3\;,\;\;y=-\frac13x+1\end{align*}$$

So you integral is

$$\int\limits_{-3}^0\frac5{x^2-y^2+2xyi}dxdy+\int\limits_0^3\frac5{x^2-y^2+2xyi}dxdy$$

Answer 2

You can pick any other piecewise smooth path $\alpha$ from -3 to 3 in the complex plane so that the sum of the path you've given, which I'll call $\beta$ (piecewise linear from -3 to $i$ and then $i$ to 3), and $\alpha$ is simple and does not enclose the origin. Then Cauchy's theorem says that $\int_{\alpha + \beta} \frac{5}{z^2}dz=0$.

You have $$\int_{\alpha + \beta} \frac{5}{z^2}dz = \int_{\alpha} \frac{5}{z^2}dz +\int_{\beta} \frac{5}{z^2}dz=0 \rightarrow-\int_{\alpha} \frac{5}{z^2}dz = \int_{\beta} \frac{5}{z^2}dz$$

You get to pick any $\alpha$ satisying the condition in the first sentence above to make things easier on yourself for the integration. A simple choice for $-\alpha$ is to go from -3 to $-\frac{1}{2}$ along the real line, then follow the arc of the circle centered at the origin of radius $\frac{1}{2}$ clockwise to $\frac{1}{2}$, and then follow the real line to 3.

You end up with: $$\int_{\beta} \frac{5}{z^2}dz=\int_{-3}^{1/2} \frac{5}{x^2}dx+\int_{\pi}^{0}\frac{5}{((1/2)e^{i \theta})^2}\cdot (1/2)ie^{i \theta}d\theta+\int_{1/2}^{3} \frac{5}{x^2}dx$$

The three integrals on the RHS above are all pretty simple to compute.