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?


2 Answers 2


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: $$c=a+b, \quad k = \frac {a+b}{ab}\sqrt{\frac\pi 2}$$

  • $\begingroup$ Newbie question, I was not able to calculate the Fourier transform as you did. I could not solve the integral $$\frac{1}{2\pi}\int{\frac {1}{x^2+a^2}e^{-i\omega x}}dx $$ Whats the idea here? $\endgroup$
    – AlphaBeta
    Jun 11, 2016 at 12:39
  • $\begingroup$ Easiest is to use a table of Fourier transforms. You can find one with the relevant transform for instance here: en.wikipedia.org/wiki/Fourier_transform $\endgroup$ Jun 11, 2016 at 12:54
  • 1
    $\begingroup$ Note that it is straight forward to find the inverse transform of $e^{-a|\omega|}$, proving that the transform is correct. $\endgroup$ Jun 11, 2016 at 13:06

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}


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