fourier analysisintegration
I need to calculate the Inverse Fourier Transform of this Gaussian function:
$\frac{1}{\sqrt{2\pi}} exp(\frac{-k^2 \sigma^2}{2})$
where $\sigma > 0$, namely I have to calculate the following integral:
$\int{\frac{1}{\sqrt{2\pi}} \exp(\frac{-k^2 \sigma^2}{2})} \exp(ikx) dk$.
How can I do that? If I am not wrong (in which case please correct me) the result should be $\frac{1}{\sqrt{2\pi}\sigma}exp(\frac{-x^2}{2\sigma^2})$…but I don't know how to get there!
$\operatorname{sinc}(x)$ is defined in two different ways; one of them is $$\operatorname{sinc}(x)=\begin{cases}\frac{\sin(\pi x)}{\pi x},&x \neq 0,\\1,&x = 0.\end{cases}$$ So, multiply your answer by $\frac{\pi}{\pi}$, simplify, use the definition above, and enjoy!
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
The Gaussian $e^{-\alpha x^{2}}$ satisfies the differential equation $$ \frac{df}{dx} = -2\alpha x f,\;\;\; f(0)=1. $$ The Fourier transform turns differentiation into multiplication, and multiplication into differentiation. So you get back the same differential equation with different constants. For example, if $f(x)=e^{-\alpha x^{2}}$, then \begin{align} \frac{d}{dk}\hat{f}(k) & =\frac{d}{dk}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-ikx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-ikx}(-ix)dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(-2\alpha xe^{-\alpha x^{2}})(\frac{i}{2\alpha}e^{-ikx})dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\frac{d}{dx}e^{-\alpha x^{2}})(\frac{i}{2\alpha}e^{-ikx})dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}\left(-\frac{d}{dx}\frac{i}{2\alpha}e^{-ikx}\right)dx \\ & = -\frac{k}{2\alpha}\hat{f}(k). \end{align} And, $$ \hat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx $$ Therefore, $$ \hat{f}(k) = \left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\right]e^{-k^{2}/4\alpha} $$ Determining the multiplier constant is normally done with a trick using polar coordinates: \begin{align} \left[\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\right]^{2} & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}e^{-\alpha y^{2}}dxdy \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{\infty}e^{-\alpha r^{2}}rdrd\theta \\ & = \int_{0}^{\infty}e^{-\alpha r^{2}}rdr \\ & = -\frac{1}{2\alpha}\int_{0}^{\infty}\frac{d}{dr}e^{-\alpha r^{2}}dr \\ & = \frac{1}{2\alpha} \end{align} Finally, $$ \hat{f}(k) = \frac{1}{\sqrt{2\alpha}}e^{-k^{2}/4\alpha}. $$