Question on evaluating $\int_{C}\frac{e^{iz}}{z(z-\pi)}dz$ without the residue theorem

Question

I am trying to figure out how to evaluate the integral $\int_{C}\frac{e^{iz}}{z(z-\pi)}dz$ where $C$ is any circle centered at the origin with radius greater than $\pi$. I can see that $\frac{e^{iz}}{z(z-\pi)}$ is analytic everywhere except where $z=0$ and $z=\pi$, both of which are in the region bounded by $C$. I can also see that by using the Taylor expansion of $e^{iz}$ that $$\int_{C}\frac{e^{iz}}{z(z-\pi)}dz = \sum_{n=0}^{\infty}\frac{i^{n}}{n!}\int_{C}\frac{z^{n-1}}{z-\pi}dz$$

I'm supposed to apply Cauchy's Theorem or Cauchy's Integral Theorem to evaluate the integral along this curve but I am not sure how to do so. I do not yet have the residue theorem in my tool box.

Answer 1

Using partial fraction expansion we have, for every $R>\pi$,

$$\oint_{|z|=R} \frac{e^{iz}}{z(z-\pi)}\,dz=\frac1\pi\oint_{|z|=R} \frac{e^{iz}}{z}\,dz-\frac1\pi\oint_{|z|=R}\frac{e^{iz}}{z-\pi}\,dz$$

Now finish by using Cauchy's Integral Formula (or the residue theorem).

Answer 2

By symmetry the residues of $\frac{e^{iz}}{z(z-\pi)}$ at $z=0$ and $z=\pi$ are the same and they equal $-\frac{1}{\pi}$.
It follows that for any $R>\pi$ we have $$ \oint_{\|z\|=R}\frac{e^{iz}}{z(z-\pi)}\,dz = 2\pi i\cdot\left(-\frac{2}{\pi}\right) = -4i.$$