# Cauchy distribution characteristic function

I know that it's easy to calculate integral $\displaystyle\int_{-\infty}^\infty \frac{e^{itx}}{\pi(1+x^2)} \, dx$ using residue theorem. Is there any other way to calculate this integral (for someone who don't know how to use residue theorem)?

• Could by any chance post the calculation using the residues? – helplessKirk Sep 1 '14 at 22:40

Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by \begin{align} F(x)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-ix t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-ix t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-at}e^{-ix t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-ix)t}}{a-ix} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+ix)t}}{a+ix} \right|_{0}^{t=v}\\ &=\frac{1}{a-ix}+\frac{1}{a+ix}\\ &=\frac{2a}{x^2+a^2}. \end{align} Next, the inverse Fourier transform of $F(x)$ is \begin{align} f(t)=\mathcal{F}^{-1}[F(x)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(x)e^{ix t}\,dx\\ e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{x^2+a^2}e^{ix t}\,dx\\ \frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{ix t}}{x^2+a^2}\,dx. \end{align} Thus, putting $a=1$, the given integral turns out to be $$\frac1\pi\int_{-\infty}^{\infty}\frac{e^{ix t}}{x^2+1}\,dx=\large\color{blue}{e^{-|t|}}.$$ Other method using double integral technique can be seen here.
There is also another way to calculate this integral. U know that $i$ and $-i$ are to singularities of ur function. Then you can consider the following paths.
We define $\gamma_w:=\gamma_1+\cdots+\gamma_5$ and then our integral is zero (Cauchy's integral theorem). On the other hand we can calculate the several integrals separately with (let $R \rightarrow \infty$ and $\varepsilon \rightarrow 0$.)
I this way u can calculate the integral $\int_0^\infty \! \frac{e^{itx}}{\pi (1+x^2)} \, dx$. The other integral can be calculated in the same way.