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

