I'm trying to compute this convolution:

$\frac{2 \alpha}{\alpha ^2 + 4 \pi ^2 x^2} * \frac{2 \beta}{\beta ^2 + 4 \pi ^2 x^2}$

I know that the Fourier transform of a convolution of two functions is the product of Fourier transforms of the two functions.

So I need to compute $\int_{\mathbb{R}} \frac{2 \alpha}{\alpha ^2 + 4 \pi ^2 x^2} e^{-itx}dx$.

I thought it would be a good idea to use the fact that $e^{-itx} = \cos(tx) + i \sin (tx)$ and use the fac that $\sin$ is odd.

We substitute $-tx = z, \ dx = \frac{1}{t}dz$

Then the integral is:

$\int_{\mathbb{R}} \frac{2 \alpha}{\alpha ^2 + 4 \pi ^2 x^2} \cos (tx)dx$

This integral is equal to $$2 \frac{\alpha}{t} Re \lim _{R \rightarrow \infty} \int_{[-R, R]} \frac{e^{iz}}{\alpha ^2 + 4 \frac{\pi ^2}{t^2}z^2 }dz$$

$\alpha ^2 + 4 \frac{\pi ^2}{t^2} z^2 = 0 \ \ z_0 = i \frac{\alpha t}{2 \pi}$

So the integral is $$\frac{2 \alpha}{t} \cdot 2 \pi i \cdot res_{z_0} \frac{e^{iz}}{\alpha ^2 + 4 \frac{\pi ^2}{t^2}z^2 } =\frac{2 \alpha}{t} \cdot 2 \pi i \cdot \frac{e^{-\alpha t /(2 \pi)}}{2 \cdot \frac{4 \pi^2}{t^2}\cdot i \cdot \frac{\alpha t}{2 \pi}} = \alpha e^{\frac{- \alpha t}{2 \pi}}$$

Is that correct?

For a function $f(x)=\frac{1}{\pi}\frac{a}{a^2+x^2}$ (also kwnown as Lorentzian function) the Fourier transform is: $$\int_{-\infty}^{\infty}\frac{1}{\pi}\frac{a}{a^2+x^2}e^{-itx}dx=\frac{a}{\pi}\int_{-\infty}^{\infty}\frac{1}{(a+ix)(a-ix)}e^{-itx}dx$$ Now we will use Cauchy's integral formula. Although we only need to solve our integral for the real axis, we will have to use the formula once for each pole, because, if we decide to close the contour, only on the upper part of the imaginary axis, the integral on the arc will "explode" for $t >0$. For $t <0$ we will close the contour, counter clockwise, on the upper plane, and use Cauchy's formula $$\frac{a}{\pi}\int_{-\infty}^{\infty}\frac{1}{(a+ix)(a-ix)}e^{-itx}dx=\frac{a}{\pi}\oint \frac{\frac{e^{-itz}}{(a+ix)}}{(a-ix)}dz=2\pi i a\frac{e^{-i(iat)}}{2ia\pi}= e^{a t}.$$ For $t >0$ we will close the contour, clockwise, on the lower plane, not forgetting to add $(-)$ to the integral: $$\frac{a}{\pi}\int_{-\infty}^{\infty}\frac{1}{(a+ix)(a-ix)}e^{itx}dx=\frac{a}{\pi}\oint \frac{\frac{e^{itz}}{(a-ix)}}{(a+ix)}dz=-2\pi i a\frac{e^{i(-iat)}}{-2ia\pi}= e^{-a t}.$$ And so, we finally get: $$\mathcal{F}\left\{\frac{1}{\pi}\frac{a}{a^2+x^2}\right\}=F(t;a)= e^{-a |t|}.$$
So you can use this result for $f(x)=\frac{2\alpha}{\alpha^2+4\pi^2x^2}*\frac{2\beta}{\beta^2+4\pi^2x^2}$ $$\mathcal{F} \left\{f(x)\right\}=\mathcal{F} \left\{\frac{1}{\pi}\frac{a}{a^2+x^2}*\frac{1}{\pi}\tfrac{b}{b^2+x^2}\right\}=F(t;a)F(t;b)=e^{-(a+b) |t|}=e^{-\frac{\alpha+\beta}{2\pi} |t|}$$ with $a=\frac{\alpha}{2\pi}$ and $b=\frac{\beta}{2\pi}$.
• How can I calculate the inverse transform of this function if I have absolute value of $t$? – Bilbo Dec 19 '14 at 7:09
• Observe that $$\mathcal{F}\left\{\frac{1}{\pi}\frac{a}{a^2+x^2}\right\}=F(t;a)= e^{-a |t|}.$$ So you have $$\mathcal{F}^{-1}\left\{e^{-c |t|}\right\}=\frac{1}{\pi}\frac{c}{c^2+x^2} .$$ where $c=\frac{\alpha+\beta}{2\pi}=\frac{\gamma}{2\pi}$ with $\gamma=\alpha+\beta$. So you have the nice relation $$\frac{1}{\pi}\frac{a}{a^2+x^2}*\frac{1}{\pi}\frac{b}{b^2+x^2}= \frac{1}{\pi}\frac{c}{c^2+x^2}$$ or $$\frac{2\alpha}{\alpha^2+4\pi^2x^2}*\frac{2\beta}{\beta^2+4\pi^2x^2}= \frac{2\gamma}{\gamma^2+4\pi^2x^2}$$ that is convolution of Lorentzian functions is Lorentzian. – alexjo Dec 19 '14 at 7:39