Fourier transform, equality and convolution

Question

I have another problem. Let $f(x) = e^{-a|x|}$ for $a>0$. Let $\widehat{f}(t) = \int_{\mathbb{R}} f(x) e^{-itx} dx = \dfrac{2a}{a^2+t^2} $ its Fourier transform.

I have to show that $\dfrac{a}{\pi(a^2+x^2)} * \dfrac{b}{\pi(b^2+x^2)} = \dfrac{a+b}{\pi((a+b)^2+x^2)}$, $b>0$.

With the convolution, I wrote $f_a(t) = \dfrac{a}{\pi(a^2+t^2)}$ when $t \in \mathbb{R}$ and $f_b(x-t) = \dfrac{b}{\pi(b^2+(x-t)^2)}$

The convolution is $f_a*f_b(x) = \int_{\mathbb{R}} f_a(t) f_b(x-t) dt$. We can compute this with a very long and tedious integration, and find the wished result. But is there any way to compute this with the Fourier transform ?

Answer 1

Indeed there is a way, denoting: $$\phi_a(x)=e^{-a|x|}$$ we have using your notations: $$\widehat{\phi_a}=f_a$$ But one of the properties of the Fourier transform is that: $$\widehat{f \times g}=f * g$$ So: $$\widehat{\phi_a \phi_b}=\widehat{\phi_a} *\widehat{\phi_b}=f_a * f_b$$

all it remains is to compute $\widehat{\phi_a \phi_b}$.

But as $e^{-a|x|} e^{-b|x|}=e^{-(a+b) x}$ you have: $$\phi_a \times \phi_b=\phi_{a+b}$$ thus: $$\widehat{\phi_a \phi_b}=\widehat{\phi_{a+b}}=f_{a+b}$$ and finally: $$f_a * f_b=f_{a+b}$$

Answer 2

Use the important relationship between convolution and Fourier transform.

$$f_a * f_b = \mathcal{F}^{-1}(\mathcal{F}(f_a) \cdot \mathcal{F}(f_b)).$$

Assuming the Fourier inversion theorem holds for these functions, all the necessary ingredients are already in your post.