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.
Best Answer
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.