= Self-convolution of a bilateral exponential
{title2=$f(x)=e^{-a|x|},\quad(f*f)(x)=(|x|+1/a)e^{-a|x|}$}
For $a>0$ and $x\geq0$, split the <convolution> integral at $y=0$ and $y=x$. The three contributions are $e^{-ax}/(2a)$, $xe^{-ax}$ and $e^{-ax}/(2a)$. Evenness gives the displayed answer for all real $x$.
With <Fourier transform> convention $\widetilde f(k)=\int f(x)e^{-ikx}dx$, one has $\widetilde f(k)=2a/(a^2+k^2)$. The <convolution theorem> and <Fourier inversion theorem> therefore yield
$$
\boxed{\int_{\mathbb R}\frac{e^{ikx}}{(a^2+k^2)^2}dk=\frac{\pi}{2a^3}(1+a|x|)e^{-a|x|}.}
$$
The value at $x=0$ is $\pi/(2a^3)$. Multiplying $f$ by $a/2$ instead gives a normalized <Laplace distribution>; its self-convolution is the density $a(1+a|x|)e^{-a|x|}/4$.
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