Another revision question I need a bit of help with, sorry guys!
By choosing an appropriate contour, evaluate the following integral;
$$\int_{-\infty}^\infty \frac{e^{iz}}{(z-i)^2}$$
Every time I attempt this question, I get the denominator to equal 0 somehow. I'm sure there must be a smarter way to manipulate the denominator into getting the z not equal to i, but I'm stuck how to at the moment. Thanks in advance.
For a large value of $R\in\mathbb{R}^+$, consider the countour $\gamma_R$ made by a segment $S_R$ between $-R$ and $R$ and the semicircle $\Gamma_R$ in the upper half-plane joining $R$ with $-R$ (let all the path be counter-clockwise oriented). By the residue theorem, $$\oint_{\gamma_R}\frac{e^{iz}}{(z-i)^2}\,dz = 2\pi i\cdot\text{Res}\left(\frac{e^{iz}}{(z-i)^2},z=i\right)=2\pi i\cdot \frac{i}{e}=\color{red}{-\frac{2\pi}{e}}$$ and you just need to prove that $$ \lim_{R\to +\infty}\int_{\Gamma_R}\frac{e^{iz}}{(z-i)^2}\,dz = 0.$$