Integral $\int_{\sigma}{dz\over{(z^2-1)^2}}$ with $\sigma:[0,2\pi] \to C, \sigma (t)=1+e^{it}$

Question

It appeared trying to calculate $\int_{\sigma}{\cos(\pi z)dz\over{(z^2-1)^2}}$ with $\sigma:[0,2\pi] \to C, \sigma (t)=1+e^{it}$

Any idea on how to solve it? Can't use residues.

Answer 1

Using partial fraction expansion, we can write

$$\oint_\sigma \frac{1}{(z^2-1)^2}\,dz=\frac14 \left(\oint_\sigma \frac{1}{z+1}\,dz+\oint_\sigma \frac{1}{(z+1)^2}\,dz-\oint_\sigma \frac{1}{z-1}\,dz+\oint_\sigma \frac{1}{(z-1)^2}\,dz\right) \tag1$$

Since the pole at $z=-1$ is not enclosed by $\sigma$, Cauchy's Integral Theorem guarantee that the first and second integrals on the right-hand side of $(1)$ are zero.

We can express the third integral on the right-hand side of $(1)$ as

$$\begin{align} -\oint_\sigma \frac{1}{z-1}\,dz&=-\int_0^{2\pi}\frac{1}{e^{it}}\,ie^{it}\, dt\\\\ &=-i2\pi \end{align}$$

while we can express the last integral as

$$\begin{align} \oint_\sigma \frac{1}{(z-1)^2}\,dz&=-\int_0^{2\pi}\frac{1}{e^{i2t}}\,ie^{it}\, dt\\\\ &=0 \end{align}$$

Putting it all together, we have

$$\oint_\sigma \frac{1}{(z^2-1)^2}\,dz=-i\pi/2$$

---

NOTE:

Inasmuch as $\cos(\pi z)=\sum_{n=0}^\infty \frac{(-1)^{n-1}\pi^{2n}}{(2n)!}(z-1)^{2n}$, only the first term of the series, $-1$, is implicated in the contour integral $\oint_\sigma \frac{\cos (\pi z)}{(z^2-1)^2}\,dz$. That is to say, Cauchy's Integral Theorem guarantee that the integration of all of the other terms vanish. Therefore, we have

$$\oint_\sigma \frac{\cos (\pi z)}{(z^2-1)^2}\,dz=i\pi /2$$