MHB Solve Fourier Transform: f(t)=sin(2πt)/t

[lucad93]

I don't know if it is the right section to post in. I have a problem with a "simple" Fourier transform. This is the function to transform: $$f(t)=\frac{\sin\left({2\pi t}\right)}{t}$$. My first idea was to write that as $$\sin\left({2\pi t}\right)\cdot\frac{1}{t}$$ but then my fantasy crashed against a wall not finding the right transform for $$\frac{1}{t}$$. Can you help me please?

 

[Joppy]

I don't know if it is the right section to post in. I have a problem with a "simple" Fourier transform. This is the function to transform: $$f(t)=\frac{\sin\left({2\pi t}\right)}{t}$$. My first idea was to write that as $$\sin\left({2\pi t}\right)\cdot\frac{1}{t}$$ but then my fantasy crashed against a wall not finding the right transform for $$\frac{1}{t}$$. Can you help me please?

The following may be useful,

$$\frac{sin(2\pi t)}{t} = 2\pi sinc(2\pi t)$$.

and,

$$\int_{-\infty}^{\infty} \,2\pi sinc(2\pi t) = \pi$$.

Also, if $$f(t) = f(-t)$$ and $$f: \Bbb{R}\rightarrow \Bbb{R}$$ then $$\mathbb{I}\mathbb{m}({{F}_{n}}) = 0$$

EDIT: I forgot to add that this function is not periodic ($$f(t) \ne f(t+T)$$). We can deal with this by taking the limit as $$T \rightarrow \infty$$ or equivalently, the limit as $$f \rightarrow 0$$. This is because, very loosely speaking, the function is periodic with $$\infty$$.

Did you manage to solve this?

 

[Opalg]

I don't know if it is the right section to post in. I have a problem with a "simple" Fourier transform. This is the function to transform: $$f(t)=\frac{\sin\left({2\pi t}\right)}{t}$$. My first idea was to write that as $$\sin\left({2\pi t}\right)\cdot\frac{1}{t}$$ but then my fantasy crashed against a wall not finding the right transform for $$\frac{1}{t}$$. Can you help me please?

This is not an easy problem, but there is a "back door" way to sneak up on the answer.

Start with the "rectangular" function $$g(x) = \begin{cases}1&\text{if }|x|\leqslant 1, \\ 0&\text{otherwise.} \end{cases}$$ Its Fourier transform is given by $$\hat{g}(t) = \int_{\mathbb R}g(x)e^{-itx}dx = \int_{-1}^1e^{-itx}dx = \Bigl[\frac{e^{-itx}}{-it}\Bigr]_{-1}^1 = \frac{e^{-it} - e^{it}}{-it} = \frac{2\sin t}t.$$ So the Fourier transform of the rectangular function is your function (give or take a couple of constants). You can now apply the Fourier inversion theorem to deduce that the Fourier transform of your function is a rectiangular function.