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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!

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2 Answers

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

$$ \hat{f}(k_x,k_y) = \sqrt{2 \pi} e^{-\frac{k_x^2}{2}} e^{-k_y} \chi_{[0,+\infty]}(k_y) $$

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Thank you for your answer, but you are considering the problem in one variable but, there are two variables (x and y) –  Mathoman Jan 24 '13 at 14:28
    
@Mathoman: yes, but since the functions are separable, they may be considered independently. Do you need the FT of $e^{-x^2/2}$ done out? –  Ron Gordon Jan 24 '13 at 14:59
    
Ok, thank you! I've got it now! Everything is fine! Thanks a lot! –  Mathoman Jan 24 '13 at 15:41
    
@Mathoman: you're welcome. Keep in mind that the FT exists from the Plancherel Theorem mentioned above. The integrals may be done individually because 1) they each exist and 2) the original function is separable. –  Ron Gordon Jan 24 '13 at 15:44
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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: $$\begin{eqnarray} \mathcal{F}_{\omega_1, \omega_2}\left( \frac{\exp\left(-\frac{x^2}{2}\right)}{1+i y} \right) &=& \mathcal{F}_{\omega_1} \left(\exp\left(-\frac{x^2}{2}\right)\right) \mathcal{F}_{\omega_1} \left( \frac{1}{1+i y}\right) \\& =& \left( \sqrt{2\pi} \mathrm{e}^{-\frac{\omega_1^2}{2}} \right) \left(2 \pi \chi_{[0,\infty]}(\omega_2) \mathrm{e}^{-\omega_2}\right) \\ &=& \left(2\pi\right)^{3/2} \exp\left(-\frac{\omega_1^2}{2} - \omega_2\right) \chi_{[0,\infty]}(\omega_2) \end{eqnarray} $$

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Yes that's what I did, but I am wondering why you can write (how is it justified) the fourier transform in two variables as a product of two fourier transforms of one variable... and which fourier transform (of which space?) you take? Thanks –  Mathoman Jan 24 '13 at 14:23
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