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?
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
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 $\lambda_j$.