Fourier Transform of $e^{-k^2/(a^2+k^2)}$

Question

In my work, I encountered a function of the following form,

$$ f(k) = \exp\bigg[-\frac{k^2}{a^2 + k^2}\bigg] $$

In the limit $a\rightarrow 0$, the Fourier transform of this function will be $e^{-1}\delta(x)$
and in the limit $a\rightarrow\infty$ the Fourier transform of this function will be a Gaussian (by expanding the denominator binomially).

Is it possible to calculate the Fourier transform of this function for any real values of $a$?

EDIT 1:

According to Mark's answer, we can rewrite $f(k)$ as follows, $$ f(k) = e^{-1}(e^{a^2/(a^2+k^2)} - 1) + e^{-1} \\ = e^{-1}(e^{(1+(k/a)^2)^{-1}} - 1) + e^{-1} $$ Expanding the first term binomially (for $(k/a) << 1$) will give, $f(k) \approx e^{-(k^2/a^2)}$

For the limit $(k/a) << 1$, does the Fourier Transform of $f(k)$ become a Gaussian function?

Answer 1

We can write the distribution $f(k)$ as $f(k)=g(k)+e^{-1}$, where $g(k)$ is the $L^1$ function given by

$$g(k)=e^{-1}\left(e^{a^2/(a^2+k^2)}-1\right)$$

The Fourier transform of $g$ is given by

$$\begin{align} \mathscr{F}\{g\}(x)&=e^{-1}\int_{-\infty}^\infty \left(e^{a^2/(a^2+k^2)}-1\right)e^{ikx}\,dk\\\\ &=|a|e^{-1}\int_{-\infty}^\infty \left(e^{1/(1+k^2)}-1\right)e^{ik(x|a|)}\,dk \end{align}$$

According to WolframAlpha, there is no result found in terms of standard functions. You could try writing $e^{1/(1+k^2)}-1=\sum_{n=1}^\infty \frac{1}{n!(1+k^2)^n}$ and seeing if that leads to anything useful.

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In this section, we find a recursive equation for the function $f_n(t)$ where

$$f_n(t)=\int_{-\infty}^\infty \frac{e^{ikt}}{(1+k^2)^n}\,dk\tag1$$

For $n=1$, it is easy to show that $f_1(t)=\pi e^{-|t|}$. Now, for $n>2$ we differentiate $(1)$ twice to find that

$$f_n''(t)-f_n(t)=-f_{n-1}(t)\tag 2$$

with $f_n'(0)=0$ and $f_n(0)=\int_{-\infty}^\infty \frac1{(1+k^2)^n}\,dk=\pi\frac{(2n-3)!!}{(2n-2)!!} $ (See THIS ANSWER).

The general solution to $(2)$ can be written in terms of the Green function (or Green's function) for the ODE $y''(x)-y(x)=\delta(x)$, along with the prescribed initial conditions. We are now equipped with a recursive algorithm to calculate $f_n(t)$ for any $n$.

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EDIT:

As @Metamorphy pointed out, we can express the integral $\int_{-\infty}^\infty \frac{e^{ikx|a|}}{(1+k^2)^n}\,dk$ in terms of the second kind modified Bessel function. Specifically, we have

$$\begin{align}\int_{-\infty}^\infty \frac{e^{ikx|a|}}{(1+k^2)^n}\,dk&=2\int_0^\infty \frac{\cos(k|xa|)}{(1+k^2)^n}\,dk\\\\ &=\frac{\sqrt{\pi}\,|xa|^{n-1/2}}{2^{n-3/2}(n-1)!}K_{n-1/2}(|xa|)\tag3 \end{align}$$

Using $(3)$ we can write

$$\mathscr{F}\{f\}(x)=2\pi e^{-1}\delta(x)+|a|e^{-1}\sqrt{\pi}\sum_{n=1}^\infty \frac{K_{n-1/2}(|xa|)|xa|^{n-1/2}}{2^{n-5/2}n!(n-1)!} $$

Answer 2

Since there are several choices of parameters for the Fourier transform, the answer will depend on the version of the Fourier transform adopted. Now, a residue calculation shows that

\begin{align*} \hat{f}(\xi)&=\int_{-\infty}^{\infty} \exp\left(-\frac{k^2}{k^2+a^2}\right)e^{-ik\xi} \, \mathrm{d}k \\ &= 2\pi e^{-1} \Biggl[ \delta(\xi) + e^{-|a\xi|} \sum_{n=0}^{\infty} \sum_{j=0}^{\infty} \frac{(n+2j)!}{(n+j+1)!(n+j)!j!n!} \frac{|a|^{n+1} |\xi|^n}{2^{n+2j+1}} \Biggr] \tag{1} \end{align*}

in distribution sense. I utilized contour integral to compute the integral, but this is essentially what we would obtain from @Mark Viola's approach as well. As a sanity check, we find that $\text{(1)}$ correctly produces the result

$$ \hat{f}(\xi) = \int_{-\infty}^{\infty} \exp\left(-\frac{k^2}{k^2+a^2}\right)e^{-ik\xi} \, \mathrm{d}k \to 2\pi e^{-1} \delta(\xi) $$

as $a \to 0$. To verify $\text{(1)}$, note that the integral is invariant under the substitution $a \mapsto |a|$ and $\xi \mapsto |\xi|$ by the symmetry. So, we may assume that $a > 0$ and $\zeta \geq 0$. Then we decompose the left-hand side as

\begin{align*} \hat{f}(\xi) &= e^{-1} \int_{-\infty}^{\infty} \exp\left(\frac{a^2}{k^2+a^2}\right)e^{-ik\xi} \, \mathrm{d}k \\ &= e^{-1} \Biggl[ 2\pi \delta(\xi) + \int_{-\infty}^{\infty} \biggl( \exp\left(\frac{a^2}{k^2+a^2}\right) - 1 \biggr) e^{-ik\xi} \, \mathrm{d}k \Biggr] \\ &= e^{-1} \Biggl[ 2\pi \delta(\xi) + \sum_{m=1}^{\infty} \frac{1}{m!} \int_{-\infty}^{\infty} \left( \frac{a^2}{k^2 + a^2} \right)^m e^{-ik\xi} \, \mathrm{d}k \Biggr] \\ &= e^{-1} \Biggl[ 2\pi \delta(\xi) + \sum_{m=1}^{\infty} \frac{1}{m!} (-2\pi i) \underset{k=-ia}{\mathrm{Res}} \left( \frac{a^2}{k^2 + a^2} \right)^m e^{-ik\xi} \Biggr]. \end{align*}

Here, the last line follows by employing the lower semicircular contour. Computing and simplifying the residue then leads to $\text{(1)}$.