Fourier Transform of Sine I'm having trouble calculating the Fourier Transform of the sin function. Specifically, the function
$ G(\omega)=\int _{-\infty}^{\infty} g(t)\ e^{-i \omega t} dt $ 
For the fourier transform of 
$ g(t)=\sin (2 \pi A t)$  
the answer is:
$ \dfrac{\pi}{i}[ \delta(\omega - 2 \pi A)-\delta(\omega + 2 \pi A)] $.
My only problem is that I don't understand why the factor in the front is $\pi/i $ , and not just the $ 1/(2i) $ that falls out of the expansion of the sine function.
 A: The short answer is that the $2 \pi$ comes from the inversion formula.
Here is an informal perspective which gives as hint as to how this can be made
more rigorous:
A distribution is defined by its
action on a space of nicely behaved test functions.
For a function $f$ from a restricted class of ordinary functions, we can define a distribution $T_f$ by $T_f(\phi) = \int f \phi$. For the function
$t \mapsto 1$, we get 
$T_1(\phi) = \int \phi$.
The '$\delta$ function' is defined by the distribution
$T_\delta(\phi) = \phi(0)$. That is, it takes a test function $\phi$ and
returns its value at $0$.
The Fourier transform of a distribution is defined by
$\hat{T_f}(\phi) = T_f(\hat{\phi})$, where $\hat{\phi}$ is the ordinary  Fourier transform of $\phi$.
We see that ${\hat{T}_1}(\phi) = T_1(\hat{\phi}) = \int \hat{\phi}$.
The standard inversion formula shows that $\int \hat{\phi} = 2 \pi \phi(0)$,
which gives ${\hat{T}_1}(\phi) = 2 \pi T_\delta(\phi)$, or more succinctly,
${\hat{T}_1} = 2 \pi T_\delta$.
The same sort of analysis shows that
${\hat{T}_{t \mapsto e^{iat}}}(\phi) = 2 \pi \phi(a) = 2 \pi T_\delta(\omega \mapsto \phi(\omega+a))$. The last expression may be written informally as
$T_{ \omega \mapsto 2 \pi \delta(\omega-a) } (\phi)$, which is the desired result.
A: Remember 
$ \sin( 2 \pi A t) = (e^ { i\ 2 \pi A t} - e^ {- i\ 2 \pi A t} )/ ( 2 i) $
A: Use the concept of duality property
$$x(t) \rightleftharpoons X(\omega$$
$$X(t) \rightleftharpoons 2\pi x(-\omega)$$
$$\delta(t) \rightleftharpoons 1$$
$$ 1 \rightleftharpoons 2\pi \delta(-\omega)$$
Since  $\delta(\omega)$ is a even function 
$$ 1 \rightleftharpoons 2\pi \delta(\omega)$$
$$e^{j\omega_0t} \rightleftharpoons 2\pi \delta(\omega-\omega_0)$$
$$e^{-j\omega_0t} \rightleftharpoons 2\pi \delta(\omega+\omega_0)$$
$$sin(\omega_0t) \rightleftharpoons \frac{\pi}{j} [ \delta(\omega-\omega_0) - \delta(\omega+\omega_0)]$$
