Exercise - Fourier transform in two variables? I need your help for the following problem:
Compute the fourier transform of the functions
$$\chi_{[0,+\infty[}e^{-x}  \quad \text{ and } \quad \frac{e^{(-\frac{x^2}{2})}}{1+iy}$$
The second function does it belong to $L^1(\mathbb{R}^2)$ and/or to $L^2(\mathbb{R}^2)$.
I have a problem for the second one, in fact I think that due to Fubini it is in $L^2(\mathbb{R}^2)$ but not in $L^1(\mathbb{R}^2)$, since $\frac{1}{1+iy}$ is not integrable.
I hope this is right so far. Then when I try to calculate the fourier transform of the second, I would like to integrate separately with respect to x and then y because $\frac{1}{1+iy}$ can be obtained by the fourier inverse from the first function and $e^{(-\frac{x^2}{2})}$ is the well-known gaussian. But this is then not possible since the second function is not integrable. So how do I have to compute the Fourier transform of the second function?
Thanks in advance!
 A: 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
$$ \hat{f}(k_x,k_y) = \sqrt{2 \pi} e^{-\frac{k_x^2}{2}} e^{-k_y} \chi_{[0,+\infty]}(k_y) $$
A: Having done the first transform, you should get:
$$
    \mathcal{F}_\omega\left(\chi_{[0,\infty]}(x) \mathrm{e}^{-x}\right) = \int_0^\infty \mathrm{e}^{i \omega x - x} \mathrm{d} x = \frac{1}{1-i \omega}
$$
Thus:
$$\begin{eqnarray}
  \chi_{[0,\infty]}(x) \mathrm{e}^{-x} &=&  \mathcal{F}^{-1}_x\left(\frac{1}{1-i \omega}\right) = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{1-i \omega} \mathrm{e}^{-i \omega x} \mathrm{d}\omega \stackrel{\omega \to - \omega}{=} \frac{1}{2\pi} \int_{-\infty}^\infty \frac{1}{1+i \omega} \mathrm{e}^{i \omega x} \mathrm{d}\omega \\ 
   &=& \frac{1}{2\pi} \mathcal{F}_x\left(\frac{1}{1+i \omega}\right) \tag{1}
\end{eqnarray}
$$
Using the equation above:
$$\begin{eqnarray}
   \mathcal{F}_{\omega_1, \omega_2}\left( \frac{\exp\left(-\frac{x^2}{2}\right)}{1+i y} \right) &=& \mathcal{F}_{\omega_1} \left(\exp\left(-\frac{x^2}{2}\right)\right) \mathcal{F}_{\omega_1} \left( \frac{1}{1+i y}\right) \\& =& \left( \sqrt{2\pi} \mathrm{e}^{-\frac{\omega_1^2}{2}} \right) \left(2 \pi \chi_{[0,\infty]}(\omega_2) \mathrm{e}^{-\omega_2}\right) \\ &=& \left(2\pi\right)^{3/2} \exp\left(-\frac{\omega_1^2}{2} - \omega_2\right) \chi_{[0,\infty]}(\omega_2)
\end{eqnarray}
$$
