Cauchy's Integral Formula states that if $\displaystyle f(z)$ is analytic in $\displaystyle C$ and the point $\displaystyle z$ is not on C, then:

$\displaystyle f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi

$

Ok, $\displaystyle f(z)=e^{-z}$ and that's analytic in C. Don't confuse the ENTIRE integrand with the expression for $\displaystyle f(z)$ when using this formula.

So plugging in $\displaystyle z=\pi i/2$, then the integral is just:

$\displaystyle 2\pi i f(\pi i/2)$ right?

Try and get good at calculating these in Mathematica so that you can check your results. For example (where I used an equivalent circle of radius 2 via deformation of path theorem):

Code:

In[111]:=
NIntegrate[(Exp[-z]/(z - Pi*(I/2)))*2*I*
Exp[I*t] /. z -> 2*Exp[I*t],
{t, 0, 2*Pi}]
N[2*Pi*I*Exp[(-Pi)*(I/2)]]
Out[111]=
6.283185307179801 -
2.708944180085382*^-14*I
Out[112]=
6.283185307179586