What is the Cauchy's integral formula for this?
Solution 1:
z = 0 is not inside the contour. That leaves only one pole inside the contour at $z = \pi i$
Cauchy integral formula says that is $f$ is holomorphic, for every $a$ inside the contour,
$\oint \frac {f(z)}{(z-a)^{n+1}} \ dz = \frac {2\pi i}{n!} f^{(n)}(a)$
$\frac {d}{dz} \frac {e^{iz}}{z} = \frac {e^{iz}(iz - 1)}{z^2}$
Evaluated at $z = \pi i$
$\frac {2\pi i}{1!}\frac {e^{-\pi}(-\pi - 1)}{-\pi^2}\\ \frac {2ie^{-\pi}(\pi + 1)}{\pi}$