How can I convolve $\mathrm{e}^{-\mid t \mid }$ by $\mathrm{e}^{-t^2}$ I have to solve the following differential equation using Fourier Transform: $$ -f''(t) +f(t) = \mathrm{e}^{-t^2} $$
I will use the following convention for the Fourier Transform formula and the inversion formula :
$$ F(\xi)=\int_{R}{f(t)\mathrm{e}^{-2i\pi\xi t}dt}  $$
$$ f(-t) = \int_{R}{F(\xi)\mathrm{e}^{-2i\pi\xi t}d\xi} $$
This is what I have:
$$ -(2i\pi\xi)^2F(\xi) + F(\xi) = \mathscr{F}[\mathrm{e}^{-t^2}] \\
 F(\xi) = \frac{1}{1 + 4\pi^2 \xi^2} \mathscr{F}[\mathrm{e}^{-t^2}]$$
I noticed that $ \frac{1}{1 - 4\pi^2 \xi^2}$ is the Fourier transform of $\frac{1}{2}\mathrm{e}^{-\mid t \mid} $. $$ f(-t) = \frac{1}{2}\mathrm{e}^{- \mid x \mid} *\mathrm{e}^{-x^2}\\
 f(-t) = \frac{1}{2}\int_{-\infty}^{+\infty}{\mathrm{e}^{-\mid t-x \mid}\mathrm{e}^{-x^2} dx}\\
 f(-t) = \frac{1}{2}\int_{-\infty}^{+\infty}{\mathrm{e}^{-x^2 -\mid t-x \mid} dx} \\
 f(t) = \frac{1}{2} \int_{-\infty}^{+\infty}{\mathrm{e}^{-x^2 -\mid -t-x \mid}dx} \\
 f(t) = \frac{1}{2} \int_{-\infty}^{+\infty}{\mathrm{e}^{-x^2 -(t+x)} dx} \\
 f(t) = \frac{1}{2} \mathrm{e}^{-t} \int_{-\infty}^{+\infty}{ \mathrm{e}^{-x^2 -x} dx}$$
I'm stuck here because I don't know how to integrate this function. Any tips to do it ?
 A: Using the fact that $-|t-x| = x-t$ if $x≤t$ and $t-x$ when $x≥t$, you get
$$
\begin{align*}
2\,f(-t) &= e^{-t}\int_{-\infty}^t e^{x-x^2} \mathrm d x + e^t \int_t^{\infty} e^{-x-x^2} \mathrm d x
\\
&= e^{-t}\int_{-\infty}^t e^{-(x-1/2)^2+1/4} \mathrm d x + e^t \int_t^{\infty} e^{-(x+1/2)^2+1/4} \mathrm d x
\\
&= e^{1/4-t}\int_{-\infty}^{t-1/2} e^{-x^2} \mathrm d x + e^{1/4+t} \int_{t+1/2}^{\infty} e^{-x^2} \mathrm d x 
\end{align*}
$$
However, the primitive of the Gaussian cannot be expressed with functions simpler than the special function $\mathrm{erf}(x) = \frac{2}{\sqrt \pi} \int_0^x e^{-y^2}\mathrm d y$ (the error function). Using this function we get
$$
\begin{align*}
\frac{4}{\sqrt \pi} f(-t) &= e^{1/4-t}\big[\mathrm{erf}(x)\big]_{-\infty}^{t-1/2} + e^{1/4+t} \big[\mathrm{erf}(x)\big]_{t+1/2}^{\infty}
\\
&= e^{1/4-t}\,\mathrm{erf}(t-1/2) + e^{1/4-t} + e^{1/4+t} - e^{1/4+t}\, \mathrm{erf}(t+1/2)
\\
&= e^{1/4}\left( 2\cosh(t) + e^{-t}\,\mathrm{erf}(t-1/2) - e^{t}\, \mathrm{erf}(t+1/2)\right).
\end{align*}
$$
Therefore
$$
f(t) = \frac{\sqrt{\pi}\,e^{1/4}}{4} \left( 2\cosh(t) + e^{t}\,\mathrm{erf}(-t-1/2) - e^{-t}\, \mathrm{erf}(-t+1/2)\right)
$$
