 Self-convolution of a bilateral exponential

For $a>0$ and $x\ge 0$, split the convolution integral at $y=0$ and $y=x$. The three contributions are $e^{-ax}/(2a)$, $x{e}^{-ax}$ and $e^{-ax}/(2a)$. Evenness gives the displayed answer for all real $x$.

With Fourier transform convention $\tilde{f}(k)=\int f(x)e^{-ikx}dx$, one has $\tilde{f}(k)=2a/(a^2+k^2)$. The convolution theorem and Fourier inversion theorem therefore yieldThe 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 $a(1+a|x|)e^{-a|x|}/4$.