I am trying to evaluate $$\int_{\gamma(0;2)}\frac {e^{i\pi z/2}}{z^2-1}\, \mathrm d z$$ using the Cauchy integral formula
The problem is it is not holomorphic at $1$ and $-1$. My textbook suggests partial fractions, but surely the issue still exists?
On
Since $$ \frac1{z^2-1}=\frac12\left(\frac1{z-1}-\frac1{z+1}\right) $$ using partial fractions, we can compute the integral as $$ \begin{align} \int_{\gamma(0;2)}\frac{e^{i\pi z/2}}{z^2-1}\,\mathrm{d}z &=\frac12\int_{\gamma(0;2)}\frac{e^{i\pi z/2}}{z-1}\,\mathrm{d}z -\frac12\int_{\gamma(0;2)}\frac{e^{i\pi z/2}}{z+1}\,\mathrm{d}z\\ &=\pi ie^{i\pi/2}-\pi ie^{-i\pi/2}\\[9pt] &=-2\pi \end{align} $$ but we still need to use residues as in Jack's answer.
$f(z)=\frac{e^{i\pi z/2}}{z^2-1} $ is a meromorphic function with simple poles at $z=\pm 1$, having residues:
$$ \text{Res}\left(f(z),z=\pm 1\right) = \lim_{z\to \pm 1}\frac{e^{i\pi z/2}}{z\pm 1}=\frac{i}{2}, $$ so, by the residue theorem: $$ \oint_{\|z\|=2} f(z)\,dz = 2\pi i\left(\frac{i}{2}+\frac{i}{2}\right)=\color{red}{-2\pi}.$$