Fourier transform of $\frac{\sin{x}}{x}$ can you help me with this question 
find $\sin$ Fourier transform of $\frac{\sin{x}}{x}$ 
 A: Note that the sinc function is not in $L^1$; therefore you cannot compute its Fourier transform in the "obvious" way by means of the residue theorem or something like that. But this function is in $L^2$, so it is indeed Fourier related to some other function  $f\in L^2$. If we can guess an $f$ such that its inverse Fourier transform is sinc we are done.
Now we know from Fourier series as well as from Fourier transform that sinc turns up when processing a rectangle pulse. Therefore we try
$$f(\omega):=\cases{c\quad &$\bigl(|\omega|< \Omega\bigr)$ \cr 0&$\bigl(|\omega|> \Omega\bigr)$ \cr}$$
with parameters $c$ and $\Omega$ yet to be determined. At any rate
$${\cal F}^{-1}f(x)=\int_{-\infty}^\infty f(\omega)e^{i\omega x}\ d\omega=c\int_{-\Omega}^\Omega e^{i\omega x}\ d\omega={2c\sin(\Omega x)\over\Omega}\ .$$
Now adjust the parameters $c$ and $\Omega$ to your needs. (Note that your definition of ${\cal F}$, resp. ${\cal F}^{-1}$, might lead to slightly different values.)
A: You've got:
$$F(w) = \int \frac{\sin x}{ x} exp(-iw x) dx$$
Differentiate wrt $w$ to get:
$$F'(w) = -i \int \sin x exp(- iwx)dx $$
You know how to find fourier transform of sine and then you should integrate your result.
A: I write the Fourier transform as
$$\hat{f}(v) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{-i 2 \pi v x} $$
Consider, rather, the integral
$$ \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i x}-e^{-i x}}{x} e^{-i 2 \pi v x} $$
$$ = \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i (1-2 \pi v) x}}{x} - \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{-i (1+2 \pi v) x}}{x} $$
Consider the following integral corresponding to the first integral:
$$\oint_C dz \: \frac{e^{i (1-2 \pi v) z}}{z} $$
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:
$$\int_{C_R} dz \: \frac{e^{i (1- 2 \pi v)z}}{z}  + \int_{C_r} dz \: \frac{e^{i (1- 2 \pi v) z}}{z}  + \int_{-R}^{-r} dx \: \frac{e^{i (1- 2 \pi v) x}}{x} + \int_{r}^{R} dx \: \frac{e^{i (1- 2 \pi v) x}}{x} $$
Consider the first part of this integral about $C_R$, the large semicircle of radius $R$:
$$\int_{C_R} dz \: \frac{e^{i (1- 2 \pi v)z}}{z} = i \int_0^{\pi} d \theta e^{i (1-2 \pi v) R (\cos{\theta} + i \sin{\theta})}  $$
$$  = i \int_0^{\pi} d \theta e^{i (1-2 \pi v) R \cos{\theta}} e^{-(1- 2 \pi v) R \sin{\theta}} $$
By Jordan's lemma, this integral vanishes as $R \rightarrow \infty$ when $1-2 \pi v > 0$.  On the other hand,
$$ \int_{C_r} dz \: \frac{e^{i (1-2 \pi v) z}}{z} = i \int_{\pi}^0 d \phi \: e^{i (1-2 \pi v) r e^{i \phi}} $$
This integral takes the value $-i \pi$ as $r \rightarrow 0$.  We may then say that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1-2 \pi v) x}}{x} = i \pi & 1-2 \pi v > 0\\ \end{align}$$
When $1-2 \pi v < 0$, 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
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1-2 \pi v) x}}{x} = -i \pi & 1-2 \pi v < 0\\ \end{align}$$
Using a similar analysis as above, we find that
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i (1+2 \pi v) x}}{x} = \begin{cases} -i \pi & 1+2 \pi v < 0 \\ i \pi & 1+2 \pi v >0 \\ \end{cases} $$
We may now say that
$$\hat{f}(v) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{-i 2 \pi v x} = \begin{cases} \pi & |v| < \frac{1}{2 \pi} \\ 0 & |v| > \frac{1}{2 \pi} \\ \end{cases} $$
A: Here is another perspective on the problem: Using $\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$ one easily calculates
$$ \mathcal{F} \sin = \frac{1}{2} i \left( \delta_{-\frac{1}{2\pi}} - \delta_{\frac{1}{2\pi}} \right), \tag{1} $$
where $\delta_x$ is the Dirac distribution concentrated at $x$. Now, we use that - intuitively speaking - the Fourier transform changes differentiation into multiplication with $x$. More precisely, we have (using the notation $\chi: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto x$)
$$ \left( \mathcal{F} \, \frac{\sin(\chi)}{\chi} \right)' = \mathcal{F} (-2 \pi i \chi) \frac{\sin(\chi)}{\chi} = (-2 \pi i) \mathcal{F} \sin $$
and substituting (1) yields
$$ \left( \mathcal{F} \, \frac{\sin(\chi)}{\chi} \right)' = \pi \left( \delta_{-\frac{1}{2\pi}} - \delta_{\frac{1}{2\pi}} \right). $$
This tells us that $\mathcal{F} \, \frac{\sin(\chi)}{\chi}$ makes an upward jump of height $\pi$ at $-\frac{1}{2\pi}$ and a downward jump of height $\pi$ at $\frac{1}{2\pi}$. From that one obtains
$$ \mathcal{F} \, \frac{\sin(\chi)}{\chi} = \pi 1_{(-\frac{1}{2\pi}, \frac{1}{2\pi})} + C $$
for some constant $C$, where $1_A$ denotes the characteristic function of $A$. Finally, it is no problem to check that $C$ must be $0$.
