Deriving the characteristic function of the Cauchy distribution by residue theorem

Question

A random variable X is distributed by Cauchy Distribution when the density function of X is f(x) = $\frac{1}{\pi}\frac{1}{(1+x^2)}$. The characteristic function of X is defined by $$\phi(x)=\int_{-\infty}^\infty{f(x)e^{itx}dx},\;t\in {R}$$ Thus, $$\phi(x)=\int_{-\infty}^\infty\frac{1}{\pi}\frac{1}{(1+x^2)}{e^{itx}dx}$$

By the Residue theorem, it should be possible to prove that $$\phi(x) = e^{-\vert t\vert}$$

I found this mentioned in another question and have been trying to produce this proof. I am new to complex analysis, and after calculating residues I cannot derive the integral into $\phi(x) = e^{-\vert t\vert}$. My process so far is as below:

Let $f(z) = \frac{e^{itz}}{(1+z^2)}$. $f(z)$ has two simple poles, ${z_0}=-i$ and ${z_1}=i$. Evaluate $f(z)$ on two semicircle-shaped contours, one in the positive imaginary direction and another in the negative imaginary direction.

Using the formula for limit form of residues, I find that $Res(f(z),-i) = \frac{e^t}{-2\pi i}$ and $Res(f(z),i) = \frac{e^{-t}}{2\pi i}$. The Residue theorem states that $\int_{\gamma}{f(z)}dz$ = ${2\pi i}{\sum_{r=0}^n}Res(f,z_r)$ where $n$ is the total number of poles.

Inputting the residues from above into the Residue theorem yields

$$\int_{-\infty}^\infty\frac{e^{itz}}{(1+z^2)}dz={2\pi i}{\frac{e^t}{-2\pi i}}+{2\pi i}{\frac{e^{-t}}{2\pi i}}$$

How does this equal to $e^{-\vert t\vert}$? Have I simply miscalculated the residues? I am genuinely interested in the solution and any help is appreciated.

Answer 1

$$\phi(x)=\int_{-\infty}^\infty\frac{1}{\pi}\frac{1}{(1+x^2)}{e^{itx}dx}$$

When you choose the contour, it depends on the sign of $t$, if $t>0$, you choose the upper half circle as the contour, because when you evaluate the integral on the upper half circle, you let $z=Re^{i\theta}$, where $\theta$ goes from $0$ to $\pi$ along the upper half circle counter-clockwisely. For the factor:

$$e^{itz}=e^{itR\cos\theta-tR\sin\theta},~~~~~t>0,~\sin\theta>0\tag{1}$$

as $R\to\infty$, the exponential factor in (1) will decay, hence the integral along the upper half circle vanishes. (We also use this feature when prove Jordan's lemma.)

In this case, only $z=i$ is inside the upper circle contour, hence you only evaluate the residue at $z=i$, so you get

$$\int_{-\infty}^\infty\frac{e^{itz}}{1+z^2}dz=2\pi i\cdot\frac{e^{-t}}{2\pi i}=e^{-t}$$

Similarly, if $t<0$, to guarantee the integral vanish on the half circle, we need to choose the lower half circle. Let $z=Re^{i\theta}$, where $\theta$ goes from $0$ to $-\pi$ along the lower half circle clockwisely, and $$e^{itz}=e^{itR\cos\theta-tR\sin\theta},~~~~~t<0,~\sin\theta<0\tag{2}$$

Only the $z=-i$ is inside this contour, so you only compute the residue at $z=-i$. Hence, you get

$$\int_{-\infty}^\infty\frac{e^{itz}}{1+z^2}dz=\color{red}{-2\pi i}\cdot\frac{e^{t}}{-2\pi i}=e^{t}$$

Note the extra $-$ sign in the red-colored part, it is due to you go clockwisely when you do the integration along the lower half circle contour.