I think this question boils down to two basic questions: 1) How to prove that the Fourier transform of $f(x) = \frac{1}{1+i x}$ exists, and 2) How to show that it is equal to $e^{-k} \chi_{[0,+\infty]}{(k)}$, where $\chi_{[0,+\infty]}(k)$ is the Heaviside function (i.e., $0$ for $k<0$ and $1$ when $k>0$).
Before I begin, I will define the FT of $f(x)$ by
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) e^{i k x} $$
1) The existence of the FT of $f(x)$ is justified by the Plancherel Theorem, which states that functions that are square integrable over the real line have FT's. In this case, you observe correctly that $f(x)$ is such a function.
2) You wish to compute
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{e^{i k x}}{1+i x} $$
The best way to proceed in my opinion is to apply the Residue Theorem. That is, consider the following integral in the complex plane instead:
$$\oint_{C_R} dz \: \frac{e^{i k z}}{1+i z} $$
where $C_R$ is a contour consisting of the interval $[-R,R]$ on the real axis, and the semicircle of radius $R$ in the upper half-plane (i.e., $\Im{z}>0$). This integral is equal to $i 2 \pi$ times the sum of the residues of the poles within $C_R$. In this case, there is a pole of $\frac{e^{i k z}}{1+i z}$ at $z=i$. The residue of that pole is
$$\mathrm{Res}_{z=i} \frac{e^{i k z}}{1+i z} = \lim_{z \rightarrow i} (z-i) \frac{e^{i k z}}{1+i z} = -i \, e^{-k}$$
because $\frac{e^{i k z}}{1+i z}$ is analytic outside of $z=i$. (That is, it doesn't matter from what direction in the complex plane the limit is taken.)
The integral, on the other hand, may be split into two pieces: one along the real axis, and one along the semicircle in the upper half-plane:
$$\oint_{C_R} dz \: \frac{e^{i k z}}{1+i z} = \int_{-R}^R dx \: \frac{e^{i k x}}{1+i x} + i R \int_{0}^{\pi} d \phi \: e^{i \phi} \frac{\exp{(i k R e^{i \phi})}}{1+i R e^{i \phi}} $$
In the limit as $R \rightarrow \infty$, the second integral vanishes by Jordan's Lemma when $k > 0$. Therefore, we have (so far):
$$\begin{align} \int_{-\infty}^{\infty} dx \: \frac{e^{i k x}}{1+i x} = e^{-k} & (k>0) \\ \end{align}$$
When $k<0$, the second integral diverges and we cannot use this contour. Rather, we use a similar contour in the lower half-plane. The analysis is the same, except that there are no poles inside this contour; therefore, the integral we seek is zero when $k<0$. Therefore
$$\hat{f}(k) = \begin{cases} e^{-k} & k>0 \\ 0 & k<0 \\ \end{cases} = e^{-k} \chi_{[0,+\infty]}(k) $$
EDIT
The problem calls for the FT of a function in two dimensions
$$ \hat{f}(k_x,k_y) = \int_{-\infty}^{\infty} dx \: \int_{-\infty}^{\infty} dy \: f(x,y) e^{i (k_x x+k_y y)} $$
where
$$ f(x,y) = \frac{e^{-\frac{x^2}{2}}}{1+i y} $$
Because $f$ is separable, i.e., $f(x,y) = g(x) h(y)$, $\hat{f}(k_x,k_y) = \hat{g}(k_x) \hat{h}(k_y)$. We computed $\hat{h}(k_y)$ above. To compute $\hat{g}(k_x)$:
$$ \hat{g}(k_x) = \int_{-\infty}^{\infty} dx \: e^{-\frac{x^2}{2}} e^{i k_x x} $$
Complete the square in the exponent to find:
$$ \hat{g}(k_x) = \int_{-\infty}^{\infty} dx \: e^{-\frac{(x-i k_x)^2}{2}} e^{-\frac{k_x^2}{2}} $$
Note that the integral is independent (except for the "constant" factor) of $k_x$. We may then use $\int_{-\infty}^{\infty} dx \: e^{-a x^2} = \sqrt{\frac{\pi}{a}}$ when $\Re{a} \ge 0$. The FT you seek is then
$$ \hat{f}(k_x,k_y) = \sqrt{2 \pi} e^{-\frac{k_x^2}{2}} e^{-k_y} \chi_{[0,+\infty]}(k_y) $$
I write the Fourier transform as
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} $$
Consider, rather, the integral
$$ \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i x}-e^{-i x}}{x} e^{i k x} $$
$$ = \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} - \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} $$
Consider the following integral corresponding to the first integral:
$$\oint_C dz \: \frac{e^{i (1+k) z}}{z} $$
where $C$ is the contour defined in the illustration below:
This integral is zero because there are no poles contained within the contour. Write the integral over the various pieces of the contour:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} + \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} + \int_{-R}^{-r} dx \: \frac{e^{i (1+k) x}}{x} + \int_{r}^{R} dx \: \frac{e^{i (1+k) x}}{x} $$
Consider the first part of this integral about $C_R$, the large semicircle of radius $R$:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} = i \int_0^{\pi} d \theta e^{i (1+k) R (\cos{\theta} + i \sin{\theta})} $$
$$ = i \int_0^{\pi} d \theta e^{i (1+k) R \cos{\theta}} e^{-(1+k) R \sin{\theta}} $$
By Jordan's lemma, this integral vanishes as $R \rightarrow \infty$ when $1+k > 0$. On the other hand,
$$ \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} = i \int_{\pi}^0 d \phi \: e^{i (1+k) r e^{i \phi}} $$
This integral takes the value $-i \pi$ as $r \rightarrow 0$. We may then say that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = i \pi & 1+k > 0\\ \end{align}$$
When $1+k < 0$, Jordan's lemma does not apply, and we need to use another contour. A contour for which Jordan's lemma does apply is one flipped about the $\Re{z}=x$ axis. By using similar steps as above, it is straightforward to show that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = -i \pi & 1+k < 0\\ \end{align}$$
Using a similar analysis as above, we find that
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} = \begin{cases} -i \pi & 1-k < 0 \\ i \pi & 1-k >0 \\ \end{cases} $$
We may now say that
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} = \begin{cases} \pi & |k| < 1 \\ 0 & |k| > 1 \\ \end{cases} $$
To translate to your definition of the FT, divide the RHS by $\sqrt{2 \pi}$.
Best Answer
For this particular case it's relatively simple, since we can split the 2D Gaussian into a product of two Gaussians:$$ \mathcal{F}\{\exp(-2 \pi^2 \sigma^2 (x^2+y^2))\} (u,v) = \mathcal{F}\{\exp(-2 \pi^2 \sigma^2 x^2)\} (u)\cdot \mathcal{F}\{\exp(-2 \pi^2 \sigma^2 y^2)\} (v)\\=\frac{1}{\sqrt{2\pi\sigma^2}} \exp(-\frac{u^2}{2 \sigma^2})\frac{1}{\sqrt{2\pi\sigma^2}} \exp(-\frac{v^2}{2 \sigma^2})\\=\frac{1}{2\pi\sigma^2} \exp(-\frac{u^2+v^2}{2 \sigma^2}) $$ I suggest you look at this page, in particular the section about separability of the Fourier Transform.