Integral over the real line of a function with a second-order pole $\int_{-\infty}^\infty \frac{e^{-(A\omega+iB)^2+C}}{\omega^2} d\omega$

Question

I am trying to solve an integral of the form \begin{equation} \int_{-\infty}^\infty \frac{e^{-(A\omega+iB)^2+C}}{\omega^2} d\omega, \end{equation} where $A,B,C\in \mathbb{R}$, $A>0$.

Attempt 1: Contour integral with a dented path

On $S_\varepsilon=\left\{\varepsilon e^{i \theta}: \pi \leq \theta \leq 2 \pi\right\}$, \begin{equation} \begin{aligned} \int_{S_\varepsilon} \frac{e^{-(Az+iB)^2+C}}{z^2} dz &= e^{B^2+C}\int_\pi^{2\pi} \frac{e^{-A^2z^2-2iABz}}{z^2}dz \\ &= e^{B^2+C}\int_{S_\varepsilon} \frac{1}{z^2} \sum_{n=0}^\infty \frac{(-A^2z^2-2iABz)^n}{n!} dz \\ &= e^{B^2+C}\int_{S_\varepsilon} \frac{1}{z^2} \left( 1-A^2 z^2 - 2iABz + \frac{A^4z^4}{2} + \frac{4iA^3 B z^3}{2} - \frac{4 A^2 B^2 z^2}{2} + \cdots \right) dz \\ &= e^{B^2+C}\int_\pi^{2 \pi} \frac{i \varepsilon e^{i \theta}}{\varepsilon^2 e^{2 i \theta}}\left( 1-A^2 \varepsilon^2 e^{2 i \theta} - 2i AB \varepsilon e^{i \theta} + \frac{A^4 \varepsilon^4 e^{4 i \theta}}{2} + \cdots \right) d\theta \\ &= e^{B^2+C}\int_\pi^{2 \pi} \frac{i}{\varepsilon e^{i \theta}} + 2 AB + o(\varepsilon) d\theta \\ &= e^{B^2+C} \left( -\frac{2}{\varepsilon} + 2 \pi A B + \pi o(\varepsilon) \right) \quad \text{divergent as } \varepsilon \rightarrow 0 \end{aligned} \end{equation}

Attempt 2: By Cauchy's integral formula for $n=1$ and $C=\{Re^{i\theta}: 0 \leq \theta \leq 2\pi\}$, \begin{equation} \int_C \frac{f(\omega)}{\omega^2} d \omega = 2 \pi i f^{\prime}(0). \end{equation} Since $e^{i (\theta - \pi)}=-e^{i \theta}$, \begin{equation} \begin{aligned} \int_C \frac{e^{-(Az+iB)^2+C}}{z^2} dz &= \int_0^\pi \frac{e^{-(A \varepsilon e^{i \theta}+iB)^2+C}}{\varepsilon e^{i \theta}} d\theta + \int_\pi^{2 \pi} \frac{e^{-(A \varepsilon e^{i \theta}+iB)^2+C}}{\varepsilon e^{i \theta}} d\theta \\ &= - \int_{\pi}^{2 \pi} \frac{e^{-(A \varepsilon e^{i \theta}-iB)^2+C}}{\varepsilon e^{i \theta}} d\theta + \int_\pi^{2 \pi} \frac{e^{-(A \varepsilon e^{i \theta}+iB)^2+C}}{\varepsilon e^{i \theta}} d\theta \\ &= \cdots \\ &= 2 \pi i (-2 i AB) e^{B^2+C} \\ &= 4 \pi AB e^{B^2+C} \end{aligned} \end{equation}

I cannot seem to find any relation between the two integrals.

Attempt 3: Convolution Theorem \begin{equation} \mathfrak{F}\left\{ (f*g)(x) \right\} = \hat{f}(\omega) \hat{g}(\omega). \end{equation} Let $\hat{f}(\omega)=\frac{C}{\omega^2}$ for some constant $C$, then $\mathfrak{F}^{-1}\{\hat{f}(\omega)\}= |x|^2$. I believe the convolution theorem only applies when $f$ and $g$ are integrable. However, $|x|^2 \notin L_1$.

Is there anything that I am missing? Or is there any other way to solve this type of integrals? Any help would be appreciated!

Answer 1

Assume the following integral somehow makes sense as a tempered distribution:

$$ f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{-a \omega^2}}{\omega^2} e^{ix\omega} \, \mathrm{d}\omega $$

Then

$$ f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{-a \omega^2} - 1}{\omega^2} e^{ix\omega} \, \mathrm{d}\omega + \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{\omega^2} e^{ix\omega} \, \mathrm{d}\omega. $$

The first term is integrable and is simplifid as follows:

\begin{align*} \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{-a \omega^2} - 1}{\omega^2} e^{ix\omega} \, \mathrm{d}\omega &= -\frac{1}{2\pi} \int_{-\infty}^{\infty} \biggl( \int_{0}^{a} e^{-\omega^2 s} \, \mathrm{d}s \biggr) e^{ix\omega} \, \mathrm{d}\omega \\ &= -\int_{0}^{a} \biggl( \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-\omega^2 s} e^{ix\omega} \, \mathrm{d}\omega \biggr) \, \mathrm{d}s \\ &= -\int_{0}^{a} \frac{1}{2\sqrt{\pi s}} e^{-x^2/4s} \, \mathrm{d}s \\ &= -\frac{1}{2\sqrt{\pi}} \int_{\frac{1}{a}}^{\infty} t^{-3/2} e^{-\frac{1}{4}x^2 t} \, \mathrm{d}t \end{align*}

For the second term, consider $0 < \alpha < 1$ and note that the following equality holds:

$$ \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{|\omega|^\alpha} e^{ix\omega} \, \mathrm{d}\omega = \frac{\Gamma(1-\alpha)\sin(\pi\alpha/2)}{\pi |x|^{1-\alpha}}= \frac{1}{2\Gamma(\alpha)\cos(\pi \alpha/2) |x|^{1-\alpha}}, $$

Here, the right-hand side defines a meromorphic function of $\alpha$ on $\mathbb{C}$ for all $x \neq 0$. This allows us to define what is the natural substitute for the function $\frac{1}{w^2}$ when it is to be understood as a tempered distribution. Consequently, letting $\alpha \to 2$ to the "equality" above, we get

$$ \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{w^2} e^{ix\omega} \, \mathrm{d}\omega = -\frac{|x|}{2}. $$

More precisely, this "equality" has to be understood that "$\frac{1}{w^2}$" as a tempered distribution should be defined as the Fourier transform of the tempered distribution $-\frac{|x|}{2}$ (which makes sense, since $-\frac{|x|}{2}$ is a locally integrable function of polynomial growth).