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)?
 A: There is also another way to calculate this integral. You know that $i$ and $-i$ are to singularities of ur function. Then you can consider the following paths.
\begin{align}
\gamma_1 & =(i-i\varepsilon)t,& & t\in [0,1] \\[6pt]
\gamma_2 & =\varepsilon e^{it}, & & t\in \left[-\frac{\pi}{2},\frac \pi 2 \right] \\[6pt]
\gamma_3 & =(i+\varepsilon)(1-t)+tiR, & & t\in [0,1] \\[6pt]
\gamma_4 & =iRe^{-it}, & & t\in \left[0,\frac \pi 2\right] \\[6pt]
\gamma_5 & =-Rt, & & t\in [-1,0]
\end{align}
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 $\varepsilon \rightarrow 0$.)
In this way, you can calculate the integral $\int_0^\infty  \! \frac{e^{itx}}{\pi (1+x^2)} \, dx $. The other integral can be calculated in the same way.
A: If you have another distribution's characteristic function(c.f.) and distribution function(d.f.) in mind, you might use it to find the c.f. of a desired distribution.
Here I will use the Laplacian d.f. and c.f. to find the c.f. of Cauchy Distribution.
We know the c.f. of Laplace Distribution($f(x) = \frac{1}{2}.e^{-|x|}$) is given by $\varphi(t) = \frac{1}{1+t^2}$.(How? Do the simple integral to find this, if already not known)
Now, we observe that the c.f. of the Laplacian r.v. is some constant(i.e. $\pi$) times the Cauchy pdf. This motivates us to use a result which buds from the Inversion Theorem which is -

If $\varphi$ is an integrable c.f., then the d.f. is continuously differentiable with continuous density given by $f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-itx}\varphi(t)dt$.

Simply plug in the c.f. and density of Laplacian Distribution in the above expression, to get,
$\frac{1}{2}.e^{-|x|} = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-itx}\frac{1}{1+t^2}dt = \frac{1}{2}E(e^{-ixY})$, where $Y$ is a Cauchy r.v.
So, we get, c.f. of Cauchy r.v. as $\varphi(-t) = e^{-|x|}\Rightarrow \varphi(t) = e^{-|x|}$.
A: 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.
