Applying the residue theorem to calculate the Fourier transform of $\frac{1}{(x-\tau)^k}$ I'm trying to do this exercise from Daniel Bump's book, Automorphic Forms and Representations.  For $f: \mathbb R \rightarrow \mathbb C$, the Fourier transform $\hat{f}$ is defined by
$$\hat{f}(v) = \int_{-\infty}^{\infty} f(u) e^{2\pi i uv} du$$

I'm having trouble with the very first part.  The residue computation after "hence" is straightforward, and the "conclude that" will come from the Poisson summation formula.
I need to compute the integral
$$\hat{f}(v) = \int_{-\infty}^{\infty} \frac{e^{2\pi i uv}}{(u - \tau)^k} du$$
I'm guessing what I ought to do is define the meromorphic function $F(z) = \frac{e^{2\pi i vz}}{(z-\tau)^k}$ and the "identity" path $\gamma_N: [-N,N] \rightarrow \mathbb C$, so that
$$\hat{f}(v) = \lim\limits_{N \to \infty}\int\limits_{\gamma_N} F(w)dw$$
For large $N, M$, I could consider the paths
$$N + it, 0 \leq t \leq M$$
$$-N + it, 0 \leq t \leq M$$
$$t + iM,  -N \leq t \leq N$$
which surround the singularity $\tau$ of $F(z)$.  This last path integral is
$$\int_{-N}^N \frac{e^{2\pi i (t+iM)v}}{(t+iM-\tau)^k}dt$$
which, when $v > 0$, tends to zero as $N$ and $M$ go to infinity at the same rate.
I haven't figured out what to do with the vertical paths.
I also haven't figured out what to do in the case $v \leq 0$.
 A: First suppose that $v>0$ and try using the contour as in this picture
Break $\gamma_R$ into the semicircular path $B_R$ and the interval $[-R,R] = I_R$. Since $\tau = x+iy$ has $y>0$, for sufficiently large $R$, $\tau$ is in the interior of $\gamma_R$. Note that on $B_R$ \begin{align*} 
\left|\frac{e^{2\pi iuv}}{(u-\tau)^k} \right| \leq \left|\frac{1}{u^k} \right| = \frac{1}{R^k} 
\end{align*}
which goes to $0$ as $R\to\infty$ and thus $\lim\limits_{R\to\infty}\int\limits_{\gamma_R}\frac{e^{2\pi iuv}}{(u-\tau)^k}du = \lim\limits_{R\to\infty}\int\limits_{I_R}\frac{e^{2\pi iuv}}{(u-\tau)^k}du = \int\limits_{-\infty}^\infty\frac{e^{2\pi iuv}}{(u-\tau)^k}du$. But by the Residue Theorem, \begin{align*}
\int\limits_{\gamma_R}\frac{e^{2\pi iuv}}{(u-\tau)^k}du &= 2\pi iRes\left(\frac{e^{2\pi iuv}}{(u-\tau)^k} \right)\vert_{u=\tau}
\end{align*}
Now if $v\leq 0$, if we let $w = -v$, we can do the same thing as above except $\gamma_R$ will be in the negative half-plane. As such, $\tau$ will not be contained in the interior of $\gamma_R$ and thus the integral will evaluate to $0$.
A: If $C_R$ is the counterclockwise contour enclosing the rectangle $0 < Im(u)< R, |Re(u)| < R$ and $k \ge 2$ and $v \ge 0$ then $\int_{-\infty}^\infty \frac{e^{2i \pi uv}}{(u-\tau)^k} du = \lim_{R \to \infty}\int_{C_R} \frac{e^{2i \pi uv}}{(u-\tau)^k} du $. 
If $Im(\tau) < 0$ then $\int_{C_R}\frac{e^{2i \pi uv}}{(u-\tau)^k} du=  0$ (the function is analytic inside the contour) 
if $Im(\tau) > 0$ and $R > |\tau|$ then $\int_{C_R}(\frac{e^{2i \pi uv}}{(u-\tau)^k}- \sum_{l=0}^{k-1} \frac{(2i \pi  v)^le^{2i \pi \tau v}}{l!} \frac{1}{(u-\tau)^{k-l}}) du = 0$ so $\int_{C_R}\frac{e^{2i \pi uv}}{(u-\tau)^k} = \int_{C_R} \frac{(2i \pi  v)^{k-1} e^{2i \pi \tau v}}{(k-1)!} \frac{1}{u-\tau} du = 2i \pi \frac{(2i \pi  v)^{k-1} e^{2i \pi \tau v}}{(k-1)!} $
For $v < 0 $ you need to consider the contour $-C_R$ enclosing the rectangle $0 > Im(u)>- R, |Re(u)| < R$
