How to proceed in the computation of this Fourier transform. I want to solve the following problem:
Let $R(x) = P (x)/Q(x)$ be a rational function with $(\text{degree}\: Q)≥ (\text{degree}\: P )+2$ and $Q(x) \not= 0$ on the real axis. Then I want to prove that if $α_1 , . . . , α_k$ are the roots of R in the upper half-plane, then there exists polynomials $P_j(ξ)$ of degree less than the multiplicity of $α_j$ so that
$$\int_{-\infty}^{\infty}R(x) e^{-2 \pi i x ξ}dx= \sum_{j=1}^k P_j(ξ)e^{-2 \pi i \alpha_j ξ}$$
But I think there has to be something wrong with this because How Can I apply the residue theorem to the zeros of a function?, Can I suppose that the poles of $R(x)$ are simple? and How Can I compute the residue?
I have read this reference but is not too clear (page 8)

Edition:

The firs thing is to write the following $$Q(z)=(z-\alpha_{1})^{j_1}...(z-\alpha_{k})^{j_k}M(z)$$
where $H(z)$ encodes the roots of $Q$ but in the lower-half plane, then we define $n=\text{max}  \{j_1,...,j_k\}$ then we have that:
$$\frac{P(z)}{Q(z)}=\frac{1}{(z-\alpha_n)^{n}} \frac{P(z)}{H(z)}$$
and then  I choose the contour as the semicircle in the upper half plane enclosing $\alpha_n$, the problem is the derivatives right?
 A: I noticed you had a link to the text in another problem of yours. Thanks. So I checked. You left out a detail in the statement of the problem: the formula applies only in the case that $\xi < 0$. That helps with some of the confusion, and proves--at least to me--that there is a typo. And I'll explain why.
In the first part of the question, they talk about the roots of $R=P/Q$, and there is a formula given for $\xi < 0$. For $\xi < 0$, the evaluation of the integral is easily carried out by closing a contour in the upper half-plane because $|e^{-2\pi i z \xi}|=e^{-2\pi i(i\Im z)\xi}=e^{2\pi(\Im z)\xi}$ decays as $\Im z\rightarrow \infty$ because $\xi < 0$ is assumed (which, is a statement you neglected to copy.) The evaluation of that integral has nothing to do with the zeros of $R$, but has everything to do with the poles of $R$, which are the zeros of $Q$. To further promote this as the correct interpretation, you will notice that in the second part of the question, the authors write

(b) In particular, if $Q$ has no zeros in the upper half-plane, then $\int_{-\infty}^{\infty}R(x)e^{-2\pi i x\xi}dx=0$ for $\xi < 0$.

So the first part of the problem undoubtedly has to do with the zeros of $Q$. That makes sense, and (b) would be a non-sequitur otherwise. So, assume that the $\alpha_j$ are the roots of $Q$ (or assume they are the poles of $R$); then you can do the problem. And it makes sense.
Evaluation of Integral: For $\xi < 0$, the line integral on the real axis may be evaluated using a finite contour in the complex plane that consists of a segment along the real axis from $-T$ to $T$ and the upper half of a circle of radius $T$ centered at the origin. Call that positively oriented contour $C_{T}$. Then
$$
      \int_{-\infty}^{\infty}R(x)e^{-2\pi i x\xi}dx
    = \oint_{C_{T}} \frac{P(z)}{R(z)}e^{-2\pi i z\xi} dx
$$
If $\lambda_1,\lambda_2,\cdots,\lambda_n$ are the roots of $Q$ the $P/Q$ will be singular at $\lambda_j$ unless $Q$ has a root with at least as large of a multiplicity. So you only need to include the $\lambda_j$'s in the list for which $P/Q$ is singular near $\lambda_j$. Then you can write
$$
             \frac{P(z)}{Q(z)} = \frac{1}{(z-\lambda_j)^{r_j}}R_{j}(z)
$$
where $R_{j}(z)$ is holomorphic and non-zero in a neighborhood of $\lambda_j$. Then you can reduce the contour integral over $C_{T}$ to a sum of circular integrals $C_j$ that are non-overlapping and lie in the upper half-plane:
\begin{align}
       \oint_{C_{T}}R(z)e^{-2\pi i z \xi}dz &= \sum_{j=1}^{n}\oint_{C_j}\frac{1}{z-\lambda_j}R_{j}(z)e^{-2\pi i z\xi}dz \\
   &= \left.2\pi i\sum_{j=1}^{n}\frac{d^{r_j}}{dz^{r_j}}\left(R_j(z)e^{-2\pi i z\xi}\right)\right|_{z=\lambda_j}
\end{align}
That gives the form of answer stated in the problem. You'll get powers of $\xi$ for every derivative of the exponential function. That's how you get polynomials in $\xi$, and they can be different for every $\lambda_j$.
