Take the 2-minute tour ×  Suppose $f,g \in \mathbb{C}[z]$ are
polynomials with $\text{deg } g - \text{deg } f \geq 2$. Is it
true that $\int_{-\infty}^{\infty} e^{iz}f(z)/g(z) dx = 0$,
where $z = x + iy$ and $y > \text{Im}(\alpha)$ for any root
$\alpha$ of $g$? If so, how does one prove this? By the residue
theorem and an easy estimate it follows that this integral does
not depend on $y$ as long as it is above all the poles, so
perhaps we could exploit this fact?  share|improve this question
add comment  3 Answers  A better way of saying this:
$\displaystyle \int_L e^{iz} \dfrac{f(z)}{g(z)}\ dz = 0$ where
$L$ is the line $\{x+Iy: x \in {\mathbb R}\}$ and
$\text{Im}(\alpha) < y$ for all roots $\alpha$ of $g$ .  It is
indeed true. Consider a closed contour $\Gamma$ consisting of
the segment $\Gamma_1$ of $C$ from $x=-R$ to $x=R$ and the arc
$\Gamma_2$ of the circle of radius $\sqrt{R^2 + y^2}$ above this
segment. Since $|f(z)/g(z)| = O(|z|^{-2})$ as $|z| \to \infty$
while $|e^{iz}| = e^{-\text{Im}(z)} \le e^{-y