Let
$ X = X_1 - X_2 / $
where $ X_1 $ and $ X_2 $ are iid scalar random variables.
Also, let $ Y $ be a chi-squared variable with 1 degree of freedom. So,
$ f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}} $
Additionally, if $ Y= X^2 $, then
$ E[e^{-jty}] = E[e^{-jtx^2}] = \int_{-\infty}^{\infty}e^{-jtx^2}f_X(x)dx $
Let $ u = x^2 $, then
$ dx = \frac{du}{2x} $
and we have that
$ \begin{align} E[e^{-jtx^2}] &= \int_{-\infty}^{\infty}e^{-jtx^2}f_X(x)dx \\ &= \int_{-\infty}^{\infty}e^{-jtu}\frac{f_X(x)}{2x}du \\ &= \int_{-\infty}^{\infty}e^{-jtu}\frac{f_X(\sqrt{u})}{2\sqrt{u}}du \\ &= \mathcal{F}\{\frac{f_X(\sqrt{x})}{2\sqrt{x}}\} \end{align} $
Since $ Y= X^2 $, we have that
$ \begin{align} \mathcal{F}\{\frac{f_X(\sqrt{x})}{2\sqrt{x}}\} &= \mathcal{F}\{\frac{e^{-\frac{x}{2}}}{\sqrt{2\pi x}}\} \\ \Leftrightarrow \frac{f_X(\sqrt{x})}{2\sqrt{x}} &= \frac{e^{-\frac{x}{2}}}{\sqrt{2\pi x}} \\ \Leftrightarrow f_X(\sqrt{x}) &= \frac{\sqrt{2}e^{-\frac{x}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow f_X(x) &= \frac{\sqrt{2}e^{-\frac{x^2}{2}}}{\sqrt{\pi}} \end{align} $
Recall that $ X = X_1 - X_2 $, where $ X_1 $ and $ X_2 $ are iid. Therefore,
$ \begin{align} f_1(x) * f_2(x) &= \frac{\sqrt{2}e^{\frac{-x^2}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow f_1(x) * f_1(-x) &= \frac{\sqrt{2}e^{\frac{-x^2}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow F_1(t) . F_1(-t) &= \frac{\sqrt{2}e^{\frac{-t^2}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow |F_1(t)|^2 &= \frac{\sqrt{2}e^{\frac{-t^2}{2}}}{\sqrt{\pi}} \end{align} $
where $ g(t) $ is the phase.
$ F_1(t) $ is real if we can assume that $ X_1 $ and hence $ X_2 $ are even functions. Then $ g(t) = 0 $ and $ F_1(t) $ is given by
$ F_1(t) = $
Taking the inverse Fourier transform, we get
$ f_1(x) = \frac{2e^{-x^2}}{\sqrt{\pi}} $
