Fourier Transform of Gaussian with imaginary pole

Question

I'm struggling at the moment with the following integral, which is essentially an inverse Fourier transform of a Gaussian with a single pole on the imaginary axis:

$$I(Q) = \frac{1}{2}\int_{-\infty}^{\infty} dx \frac{e^{ix a} e^{-\beta x^2}}{Q^2+x^2} $$

A solution to this integral is given in Ryzhik & Gradshteyn (7th ed, pg 504, 3.954) as

$$I = \frac{\pi}{4Q}e^{\beta Q^2}\left[2\cosh{Qa} - e^{-Qa}\text{erf}\left(Q\sqrt{\beta}-\frac{a}{2\sqrt{\beta}}\right)- e^{Qa}\text{erf}\left(Q\sqrt{\beta}+\frac{a}{2\sqrt{\beta}}\right)\right]$$

I've been trying to figure this solution out, as I will eventually need to take a similar integral with more complicated poles, so this is a good starting integral to solve. However, I don't intuitively see how to get to this result in R&G. I've tried rectangular contours but can't seem to see the line of thought to take me to this result. Any help would be very much appreciated! Thanks :)

Answer 1

(A method without residues to check the given answer.) By change of variables $$ I(Q) = 2\int_{-\infty}^{\infty} dx \frac{e^{ix a} e^{-\beta x^2}}{Q^2+x^2} = \frac2Q \int_{-\infty}^\infty dx \frac{e^{ix Qa} e^{-\beta Q^2 x^2}}{1+x^2} $$ so WLOG $Q=1$. Put $f(a,\beta)= I(1)$. Next, differentiate in $\beta$ and complete the square / use a standard integral: $$ f-\partial_\beta f = 2 \int_{-\infty}^\infty dx e^{ix a} e^{-\beta x^2} = 2\sqrt{\pi /\beta} e^{-a^2/4\beta} $$ We can simplify with an integrating factor: $f - \partial_\beta f = -e^{\beta}\partial_\beta (e^{-\beta} f)$ so $$ \partial_\beta (e^{-\beta} f) = -2\sqrt{\pi /\beta} e^{-a^2/4\beta - \beta} $$ Now we can differentiate $e^{-\beta} F$ wrt $\beta$, where $$ F(a,\beta) = \frac{\pi}{4}e^{\beta }\left[2\cosh{a} - e^{-a}\text{erf}\left(\sqrt{\beta}-\frac{a}{2\sqrt{\beta}}\right)- e^{a}\text{erf}\left(\sqrt{\beta}+\frac{a}{2\sqrt{\beta}}\right)\right] $$ and see if it matches. This is easy since $\text{erf}(z)=\frac{2}{\sqrt\pi}\int_0^ze^{-t^2} dt$, so that $\frac{d}{dx}\text{erf}(x) = \frac{2 e^{-x^2}}{\sqrtπ}$. To avoid mistakes I checked with sympy. Sympy says yes, but a factor of 4 is missing (as in Steven Clark's comments)

import sympy
from sympy import pi, exp, cosh, erf, sqrt

Q, a, beta = sympy.symbols("Q a beta")
I = (  # noqa: E741
    (pi / (4 * Q))
    * exp(beta * Q**2)
    * (
        2 * cosh(Q * a)  # type: ignore
        - exp(-Q * a) * erf(Q * sqrt(beta) - a / (2 * sqrt(beta)))  # type: ignore
        - exp(Q * a) * erf(Q * sqrt(beta) + a / (2 * sqrt(beta)))  # type: ignore
    )
)
f = I.subs(Q, 1)
(f * exp(-beta)).diff(beta).simplify()

This means that $$f(a,\beta) = 4F(a,\beta) + C(a).$$ One can show that $C(a)=0$ by noting that $f(a,0)=2 \int_{-\infty}^\infty dx \frac{e^{ix a}}{1+x^2}= 2 \pi e^{-|a|}$.

Answer 2

UPDATE: I just realized that the result in Ryzhik & Gradshteyn is in terms of the error function and not the complementary error function ($\operatorname{erfc}(z) = 1 - \operatorname{erf}(z)$). That's why the result I got looked different.

---

I'm going to assume that $a$, $\beta$, and $Q$ are positive.

Using the Laplace transform of cosine, along with the integral formulas $$\int_{0}^{\infty} e^{-px^{2}} \cos(rx) \, \mathrm dx = \frac{1}{2} \sqrt{\frac{\pi}{p}} \, e^{-r^{2}/(4p)}, \quad \left(p>0, r \in \mathbb{R}\right), $$

and $$\int_{0}^{\infty} e^{-px^{2}} e^{-rx} \, \mathrm dx = \frac{1}{2} \sqrt{\frac{\pi}{p}} \, e^{r^{2}/(4p)} \operatorname{erfc} \left(\frac{r}{2 \sqrt{p}} \right), \quad \left(p >0, r \in \mathbb{R} \right), $$

we have

$ \begin{align} I &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{e^{ixa} e^{-\beta x^{2}}}{Q^{2}+x^{2}} \, \mathrm dx \\ &= \int_{0}^{\infty} \frac{\cos(ax) e^{-\beta x^{2}}}{Q^{2}+x^{2}} \, \mathrm dx \\ &= \frac{1}{Q} \int_{0}^{\infty} \cos(ax) e^{-\beta x^{2}}\int_{0}^{\infty} e^{-Qt} \cos(xt) \, \mathrm dt \, \mathrm dx \\ &= \frac{1}{Q} \int_{0}^{\infty} e^{-Qt} \int_{0}^{\infty} e^{- \beta x^{2}}\cos(ax) \cos(tx) \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2Q} \int_{0}^{\infty} e^{-Qt} \int_{0}^{\infty} e^{- \beta x^{2}} \left(\cos\left((t-a)x\right) + \cos \left((t+a)x \right) \right) \, \mathrm dx \, \mathrm dt \\ &= \frac{\sqrt{\pi}}{4Q \sqrt{\beta}} \int_{0}^{\infty} e^{-Qt} \left(e^{-(t-a)^{2}/(4 \beta)} + e^{-(t+a)^{2}/(4 \beta)} \right) \, \mathrm dt \\ &= \frac{\sqrt{\pi}}{4Q \sqrt{\beta}} \, e^{-a^{2}/(4 \beta)}\int_{0}^{\infty} e^{-t^{2}/(4 \beta)} \left(e^{- t(Q+ a/(2 \beta))} + e^{-t(Q-a/(2 \beta))} \right) \, \mathrm dt \\ &= \small \frac{\sqrt{\pi}}{4Q \sqrt{\beta}} \, e^{-a^{2}/(4 \beta)} \frac{\sqrt{\pi}}{2 \sqrt{\frac{1}{4 \beta}}} \left( \exp \left(\tfrac{\left(Q+ \frac{a}{2 \beta}\right)^{2}}{\frac{1}{\beta}} \right) \operatorname{erfc} \left(\frac{Q+ \frac{a}{2 \beta}}{2 \sqrt{\frac{1}{4 \beta}}} \right) + \exp \left(\tfrac{\left(Q- \frac{a}{2 \beta}\right)^{2}}{\frac{1}{\beta}} \right) \operatorname{erfc} \left(\frac{Q- \frac{a}{2 \beta}}{2 \sqrt{\frac{1}{4 \beta}}} \right)\right) \\ &= \frac{\pi}{4Q} \ e^{ \beta Q^{2}} \left(e^{a Q} \operatorname{erfc} \left(Q \sqrt{\beta} + \frac{a}{2 \sqrt{\beta}} \right) + e^{-aQ} \operatorname{erfc} \left(Q \sqrt{\beta} - \frac{a}{2 \sqrt{\beta}}\right) \right) \\&= \frac{\pi}{4Q} \ e^{ \beta Q^{2}} \left(2 \cosh(aQ) -e^{a Q} \operatorname{erf} \left(Q \sqrt{\beta} + \frac{a}{2 \sqrt{\beta}} \right) - e^{-aQ} \operatorname{erf} \left(Q \sqrt{\beta} - \frac{a}{2 \sqrt{\beta}}\right) \right). \end{align}$