 Solution

For integrable $f,g$, define their convolution by $(f\ast g)(x)={\int }_{\mathbb{R}}f(y)g(x-y)dy$. Absolute integrability of the double integral permits Fubini's theorem. With $s=x-y$,The convolution theorem, together with uniqueness of the Fourier transform, therefore identifies the inverse transform of that product with $f\ast g$. With the stated convention the inverse carries $1/(2\pi )$, so there is no extra factor in the convolution itself.

Now take $f=e^{-|x|}$. For $x\ge 0$, split the convolution at $y=0$ and $y=x$:The convolution of two even functions is even, so the self-convolution of a bilateral exponential givesIn particular the centre value is one; both pieces extend continuously there.

Splitting the Fourier integral at zero givesHence $\tilde{h}(k)=4/(1+k^2{)}^2$. The Fourier inversion theorem yieldsThe integral is absolutely convergent. At $x=0$ it gives ${\int }_{\mathbb{R}}(1+k^2{)}^{-2}dk=\pi /2$, providing a normalization check.