I'm evaluating $\frac{1}{2\pi i}\oint_C \frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-\cos t$. Here is what I've done:
$$\frac{1}{2\pi i}\oint_Cf(z)dz=a_{-1}$$ $$a_{-1}=\lim_{z \to a}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\{(z-a)^mf(z)\}$$
The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole. $$\lim_{z \to 0}\frac{1}{0!}\frac{d^0}{dz^0}\{(z-0)^1\frac{e^{zt}}{z(z^2+1)}\}=\lim_{z \to 0}\frac{e^{zt}}{z^2+1}=1$$
$$\lim_{z \to i}\frac{1}{0!}\frac{d^0}{dz^0}\{(z-i)^1\frac{e^z}{z(z^2+1)}\}=\lim_{z \to i}\frac{e^{zt}+(z-i)e^{zt}t}{2z}=\frac{e^{it}}{2i}$$
$$\lim_{z \to -i}\frac{1}{0!}\frac{d^0}{dz^0}\{(z+i)^1\frac{e^z}{z(z^2+1)}\}=\lim_{z \to -i}\frac{e^{zt}+(z+i)e^{zt}t}{2z}=\frac{e^{-it}}{-2i}$$
If I were correct, the answer would be $1+\frac{e^{it}}{2i}-\frac{e^{-it}}{2i}$.
I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:
$\lim_{z\to i}\frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -\frac{1}{2}e^{it}$ so you missed a minus sign