let $\gamma$ be the parametrization along the square $Q_2$ (clockwise) with middle ppint 0 and side length 4. Calculate:
$$\int_\gamma \frac{e^{-z}}{z-(\pi i/2)}dz$$
I know that I could use the integral formula of Cauchy, but I'd like to do it differently.
By using the homotopy-invariance, we can change the path to:
$$\gamma: [0,1]\to\mathbb C, \quad \gamma(t)=e^{2\pi i t} + \frac{\pi}{2}i\implies \dot\gamma(t)=2\pi i \cdot e^{2\pi i t}$$
So we get:
$$\int_\gamma \frac{e^{-z}}{z-(\pi i/2)}dz=\int_0^1 \frac{e^{-\gamma(t)}}{\gamma(t)-(\pi i / 2)}dt=\int_0^1\frac{e^{-[e^{2\pi i t} + \frac{\pi}{2}i]}}{e^{2\pi i t}}\cdot 2\pi i \cdot e^{2\pi i t}dt=2\pi i\int_0^1e^{-[e^{2\pi i t} + \frac{\pi}{2}i]}dt=2\pi i\int_0^1e^{-e^{2\pi i t}}\cdot \underbrace{e^{-\frac{\pi}{2}i}}_{\frac{1}{i}}dt=2\pi\int_0^1e^{-e^{2\pi i t}}dt=2\pi\int_0^1\exp(-\exp2\pi i t)dt$$
now, I kind of have trouble integrating $\exp(-\exp(2\pi i t))$ . What I tried is a substituion $u=-e^{2\pi i t}$ but then I get $du=-2\pi i e^{2\pi i t}dt$ which I don't like/know how to process since there's still a t in there.