Residues and Contour Integration

Parametrise $z=z_0+\rho e^{i\theta}$, so $(z-z_0)^n\,dz=\rho^{n+1}e^{i(n+1)\theta}i\,d\theta$. For $n\neq -1$ the integral over $\theta\in[0,2\pi]$ of $e^{i(n+1)\theta}$ is zero by periodicity, and for $n=-1$ the integrand is $i\,d\theta$, integrating to $2\pi i$. The Laurent series converges uniformly on every compact subset of the annulus, and the circle $\abs{z-z_0}=\rho$ is one such subset, so it may be integrated term by term, and only the $n=-1$ term contributes, giving $\int_C f=2\pi i\,a_{-1} =2\pi i\,\Res(f,z_0)$.

Around each singularity $z_j$ draw a small positively oriented circle $C_j$ inside $\gamma$, disjoint from the others. The region between $\gamma$ and the circles contains no singularity, so $f$ is holomorphic there, and the Cauchy theorem for the multiply connected region, cut into simply connected pieces by slits, gives $\int_\gamma f=\sum_j\int_{C_j}f$, the slit edges cancelling in pairs. By Lemma 2 each circle integral is $2\pi i\,\Res(f,z_j)$, and summing gives Equation (1).

At a simple pole $f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots$, so $(z-z_0)f(z)=a_{-1}+a_0(z-z_0)+\cdots \to a_{-1}$ as $z\to z_0$. For the quotient, $h(z_0)=0$ and $h'(z_0)\neq 0$ give $\frac{z-z_0}{h(z)}\to\frac 1{h'(z_0)}$ as $z\to z_0$, so $(z-z_0)f(z)=g(z)\frac{z-z_0}{h(z)}\to\frac{g(z_0)}{h'(z_0)}$, which is the residue by the first part.

Take $t\ge 0$ first. The integrand extends to $f(z)=\dfrac{e^{itz}}{1+z^2}$, holomorphic except at the simple poles $z=\pm i$. For $R>1$, close the real interval $[-R,R]$ with the semicircular arc $\Gamma_R$ of radius $R$ in the upper half plane, a contour enclosing the single pole $z=i$. By the residue theorem,

the residue computed from the quotient form of Proposition 4 with $g=e^{itz}$ and $h=1+z^2$, so $h'=2z$ and $\Res(f,i)=g(i)/h'(i)=e^{-t}/(2i)$, equivalently $e^{itz}/(z+i)$ evaluated at $z=i$. On the arc, $z=Re^{i\theta}$ with $\theta\in[0,\pi]$ gives $\abs{e^{itz}}=e^{-tR\sin\theta}\le 1$ since $t\ge 0$ and $\sin\theta\ge 0$, while $\abs{1+z^2}\ge R^2-1$, so the arc integral is bounded by $\dfrac{\pi R}{R^2-1}\to 0$ as $R\to\infty$. Letting $R\to\infty$ in Equation (2) leaves $\int_{-\infty}^\infty \frac{e^{itx}}{1+x^2}\,dx=\pi e^{-t}$. The case $t<0$ needs no separate contour. Writing $I(t)=\int_{-\infty}^\infty\frac{e^{itx}}{1+x^2}\,dx$, the substitution $x\mapsto -x$ leaves the even density $1/(1+x^2)$ fixed and sends $t\mapsto -t$, so $I(-t)=I(t)$, and $I$ is even in $t$. Hence $I(t)=I(\abs t)=\pi e^{-\abs t}$ for every real $t$.