Computing Complex Integral to Determine Analytic Continuation of $f(z) = \int_0^\infty {{e^{-zt}} \over {1 + t^2}} dt$ My question is the following:

Find the analytic continuation of the function $f(z)$ defined by
  $$ f(z) = \int_0^\infty {{e^{-zt}} \over {1 + t^2}} dt, \ \ \vert \arg(z) \vert < {1 \over 2}\pi,$$
  to the domain $\{ z \in \Bbb C \mid -{1 \over 2}\pi < \arg(z) < \pi \}$.

My attempt to do this is by calculating the above integral using Cauchy's Residue Theorem (especially as the previous question had me to just that for a different situation!). I observe that the poles are at $\pm i$. My issue is in determining which contour to use. I can't take a semicircle in either the upper or lower half planes since then the "$-\infty$ to $0$" integral won't converge, and neither will the integral over the semicircle. I have tried using a quarter-circle in the first quadrant, but that gives a similarly unpleasant integral along the imaginary axis.
Any advice would be most appreciated.
Important note:
 There is this question here,
Find the analytic continuation of the $ f(z) = \int_{0}^{\infty} \frac{exp(-zt)}{1+t^2} dt$,
which is entirely the same as my question, however that answer is not using complex variables techniques. I am interested in a complex analysis proof, not one that gives an expression in terms of an infinite sum.
Thanks! :)
Sam
 A: Some definitions:
$$\operatorname{C_i}(x)=-\int_{x}^{\infty} \frac{\cos t}{t} dt=\gamma +\log x +\sum_{n=1}^\infty \frac{(-x^2)^n}{2n(2n)!},$$
$$\operatorname{S_i}(x)=\int_0^x \frac{\sin t}{t} dt=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^{2n-1}}{(2n-1)(2n-1)!}
.$$
Replacing $x$ by $z$, we get $\operatorname{C_i}(z)$ and $\operatorname{S_i}(z)$. $\operatorname{C_i}(z)$ is analytic and single-valued in $\mathbb{C}\setminus (-\infty, 0]$ , $\operatorname{S_i}(z)$ is entire. So
$$G(z)=\operatorname{C_i}(z)\sin z -\operatorname{S_i}(z)\cos z+\frac{\pi}{2}\cos z$$
is analytic in $\mathbb{C}\setminus (-\infty, 0]$.
First we evaluate the integral$$f(a) = \int_0^\infty {{e^{-at}} \over {1 + t^2}} dt=\int_0^\infty {{ae^{-x}} \over {x^2 + a^2}} dx
$$for a real positive $a$.
Integrate $\varphi (z)=\frac{ae^{-z}}{z^2+a^2}$ along a quarter-circle in the first quadrant, then
$$\int_0^R {{ae^{-x}} \over {x^2 + a^2}} dx+\int_{C_R}\varphi (z)dz-i(PV)\int_0^R \frac{ae^{-iy}}{y^2-a^2}dy=\pi i Res(\varphi , ia)  \tag{1}$$
because $\varphi (z)$ has a simple pole at $z=ia$. We note
$$\lim_{ R\to \infty} \int_{C_R}\varphi (z)dz=0,$$
$$ Res(\varphi , ia)=-\frac{ie^{-ia}}{2},$$
$$(PV)\int_0^\infty \frac{ae^{-iy}}{y^2-a^2}dy=i\operatorname{C_i}(a)\sin a -i\operatorname{S_i}(a)\cos a -\frac{\pi}{2}\sin a.\tag{2}$$
Our technique employed to obtain (2) is as follows:
$$\begin{align}
(PV)\int_0^\infty \frac{e^{-iy}}{y-a}dy&=(PV)\int_0^\infty \frac{e^{-i(y-a)}\cdot e^{-ia}}{y-a}dy\\
&=(PV)\int_{-a}^a \frac{e^{-it}\cdot e^{-ia}}{t}dt+\int_a^\infty \frac{e^{-it}\cdot e^{-ia}}{t}dt\\
&=-2ie^{-ia}\int_0^a \frac{\sin t}{t}dt+\int_a^\infty \frac{e^{-it}\cdot e^{-ia}}{t}dt.
\end{align}
$$
Therefore from (1) we have $$f(a)=\operatorname{C_i}(a)\sin a -\operatorname{S_i}(a)\cos a+\frac{\pi}{2}\cos a.$$
Since this holds for all $a>0$, we have $$f(z)=G(z) $$for $z$ with $\operatorname{Re} z>0$.
We know  that $G(z)$ is the analytic continuation of $f(z)$ to the domain $\mathbb{C}\setminus (-\infty, 0]$ .
Addemdum 
Equality (1) should be
$$\int_0^R {{ae^{-x}} \over {x^2 + a^2}} dx+\int_{C_R}\varphi (z)dz+i(PV)\int_0^R \frac{ae^{-iy}}{y^2-a^2}dy=\pi i Res(\varphi , ia)  \tag{1}$$
Details of evaluation of $(PV)\int_0^\infty \frac{ae^{-iy}}{y^2-a^2}dy$:
$$\begin{align}
2\cdot (PV)\int_0^\infty \frac{ae^{-iy}}{y^2-a^2}dy
&=(PV)\int_0^\infty \frac{e^{-i(y-a)}\cdot e^{-ia}}{y-a}dy-\int_0^\infty \frac{e^{-i(y+a)}\cdot e^{ia}}{y+a}dy\\
&=e^{-ia}\left((PV)\int_{-a}^a \frac{e^{-it}}{t}dt+\int_a^\infty \frac{e^{-it}}{t}dt\right)-e^{ia}\int_a^\infty \frac{e^{-it}}{t}dt\\
&=-2ie^{-ia}\operatorname{S_i}(a)+(e^{-ia}-e^{ia})\int_a^\infty \frac{e^{-it}}{t}dt\\
&=-2ie^{-ia}\operatorname{S_i}(a)+2i\left(\operatorname{C_i}(a)+i\int_a^\infty \frac{\sin t}{t}dt\right)\sin a\\
&=-2ie^{-ia}\operatorname{S_i}(a)+2i\operatorname{C_i}(a)\sin a-2\left(\frac{\pi}{2}-\operatorname{S_i}(a)\right)\sin a\\
&=-2i\operatorname{S_i}(a)\cos a+2i\operatorname{C_i}(a)\sin a-\pi \sin a\\
(PV)\int_0^\infty \frac{ae^{-iy}}{y^2-a^2}dy&=-i\operatorname{S_i}(a)\cos a+i\operatorname{C_i}(a)\sin a-\frac{\pi}{2} \sin a
\end{align}$$
