Convolution and Fourier transform problem I was struggling with this question, can use some help.
given that $a\not=0$
$$f_a(x) =\frac{1}{x^2+a^2}$$
I'm trying to find k and c dependent on a and b   $$(f_a ∗ f_b) (x) = kf_c(x) $$
I know that the convolution can be written as $$ 2 \pi \hat f\hat g$$ but trying to make the calculation just leaves me in more trouble. I also tried directly calculating the convolution but i just hit an unsolvable integral. 
Any ideas?
 A: Fourier transform of $f_a$ is:
$$\hat f_a = \frac 1a\sqrt{\frac\pi 2}e^{-a|\omega|} \tag 1$$
The fourier transform of a convolution is a regular product:
$$\mathscr F\Big[ f_a\star f_b \Big] = \hat f_a\cdot\hat f_b = \frac 1a\sqrt{\frac\pi 2}e^{-a|\omega|} \cdot \frac 1b\sqrt{\frac\pi 2}e^{-b|\omega|} 
= \frac 1{ab}\frac\pi 2e^{-(a+b)|\omega|} \tag 2$$
The relevant inverse transform is:
$$\mathscr F^{-1}\Big[e^{-a|\omega|}\Big] = \sqrt {\frac 2\pi}\frac{a}{x^2+a^2} \tag 3$$
Combining (2) and (3):
$$f_a\star f_b = \frac 1{ab}\frac\pi 2 \sqrt {\frac 2\pi}\frac{(a+b)}{x^2+(a+b)^2} 
= \frac {a+b}{ab}\sqrt{\frac\pi 2} \frac{1}{x^2+(a+b)^2} \tag 4
 $$
Therefore:
$$c=a+b, \quad k = \frac {a+b}{ab}\sqrt{\frac\pi 2}$$
A: Another approach. Assume without loss of generality that $a > 0$, $b > 0$. Then
\begin{align}
    (f_a * f_b)(x) &= \int_{-\infty}^{\infty}\frac{1}{(x-t)^2+a^2}\frac{1}{t^2+b^2}dt \\
   &=\int_{-\infty}^{\infty}\frac{1}{2ia}\left[\frac{1}{t-x-ia}-\frac{1}{t-x+ia}\right]\frac{1}{2ib}\left[\frac{1}{t-ib}-\frac{1}{t+ib}\right]dt.
\end{align}
The function $F(z)=1/(z-x-ia)$ is holomorphic in the lower half plane, while $1/(z-x+ia)$ is holomorphic in the lower half plane. So, only mixed terms are non-zero, which gives
$$
    (f_a * f_b)(x) = -\frac{1}{4ab}\int_{-\infty}^{\infty}
   \frac{1}{t-x-ia}\frac{1}{t+ib}+\frac{1}{t-x+ia}\frac{1}{t-ib}dt
$$
You can close the contour in the upper half-plane. Then the integral of the first two terms becomes the Cauchy integral formula for the evaluation of $1/(z+ib)$ at $z=x+ia$, and the integral of the second two terms becomes the Cauchy integral formula for the evaluation of $1/(z-x+ia)$ at $z=ib$. That gives
\begin{align}
    (f_a *f_b)(x) & =-\frac{2\pi i}{4ab}\left[\frac{1}{(x+ia)+ib}+\frac{1}{(ib)-x+ia}\right] \\
  & = -\frac{2\pi i}{4ab}\cdot\frac{2i(a+b)}{x^2+(a+b)^2} \\
  & = \frac{\pi(a+b)}{ab}f_{a+b}.
\end{align}
