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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?

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3 Answers 3

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 $|e^{iz}| = e^{-\text{Im}(z)} \le e^{-y}$ on $\Gamma_2$, and $\Gamma_2$ has length $O(R)$, the integral over $\Gamma_2$ goes to $0$ as $R \to \infty$. Using Cauchy's Theorem, the integral over $\Gamma_1$ also goes to $0$, which says $\displaystyle \int_L e^{iz} \dfrac{f(z)}{g(z)}\ dz = 0$.

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Consider closing the contour along a semicircle in the upper half-plane. By assumption there are no poles inside the contour, so the integral over the closed contour must vanish. But the integral over the semicircle vanishes, so the original integral must also vanish.

Addendum: Shift the integral over the semicircle, let $\zeta = z-iy$, so $\zeta = R e^{i\theta}$, where $\theta\in[0,\pi]$. This introduces an unimportant factor of $e^{-y}$. We can now apply Jordan's lemma, which tells us that the absolute value of the integral over the semicircle is bounded by $$\pi e^{-y} \max_{\theta\in[0,\pi]} \left| \frac{f(Re^{i\theta}+iy)}{g(Re^{i\theta}+iy)}\right|.$$ By assumption, $|f/g| \sim 1/R^2$ at worst. Therefore, the integral over the semicircle vanishes in the limit.

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Why does the integral over the semicircle vanish? I don't think this is literally true, although it should tend to zero as the radius grows. –  Justin Campbell Jul 4 '12 at 1:39
    
@JustinCampbell: Hi Justin, see my edit. In the limit $\lim_{R\to\infty}$ it really does vanish! –  user26872 Jul 4 '12 at 2:57
up vote 0 down vote accepted

My apologies to the people who have already answered, but I found a way which does not involve any difficult estimates: take a rectangular contour above the poles and let the vertical edges go off to infinity. By Cauchy's theorem the integral around this loop is $0$, and since $$\bigg| e^{iz} \frac{f(z)}{g(z)} \bigg| \leq \frac{C}{x}$$ for large $|z|$ (here $C \geq 0$ is a constant and $z = x+iy$) the integral along the vertical edges tends to zero as $|x| \to \infty$. Thus the integral along any two horizontal lines above the poles is equal. But as such a horizontal line $L_y = \{ x + iy \ | \ x \in \mathbb{R} \}$ moves upward (meaning $y \to +\infty$) the integral must tend to $0$, since $$\bigg| e^{iz} \frac{f(z)}{g(z)} \bigg| \leq e^{-y}\frac{D}{x^2+1}$$ for large $|z|$ and $\int_{-\infty}^{\infty} \frac{1}{x^2+1} \ \mathrm{d}x$ is a constant independent of $y$. Thus $\int_{L_y} e^{iz} \frac{f(z)}{g(z)} \ \mathrm{d}z = 0$ for any $y$ above the poles.

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