Derive Fourier transforms from Fourier expansion. How are they related?

Question

I am just trying to relate Fourier Series expansion to Fourier Transforms. If someone could show how one value on the middle of the table is derived (from expansion) as opposed to deriving their respective equations, that would be much appreciated.

Answer 1

I'm not sure what you mean when you say derive from expansion of (I assume) Fourier Series because that won't work for the entire table. Fourier Transforms, like Fourier Series, expands a function to it's trigonometric sum. However, Fourier Series only works with periodic functions - with Fourier Transforms, the period is assumed to tend to infinity. Formally, the Fourier Transform is denoted as

\begin{equation} F(k) = \int_{-\infty}^{\infty} g(x)e^{-ikx}\mathrm{d}x \end{equation}

So, working with $f(t)=1$,

$F(k) = \int_{-a}^{a} 1e^{-ikx}\mathrm{d}x \rightarrow F(k) = \int_{-a}^{a} e^{-ikx}\mathrm{d}x \rightarrow \dfrac{-e^{-ikx}}{ix} \Big|_{-a}^{a} \rightarrow = i \dfrac{-e^{-iax} - e^{iax}}{x}$

where the top part of the last equation is $=-2isin(ax)$

But because the period tends to infinity, a limit needs to be taken. So, continuing,

$ \rightarrow = i \dfrac{-2isin(ax)}{x} \rightarrow = 2\dfrac{sin(ax)}{x} \rightarrow \lim_{a \to \infty} \int_{-a}^{a} e^{-ikx}\mathrm{d}x = \lim_{a \to \infty} \int_{-a}^{a} 2\dfrac{sin(ax)}{x}\mathrm{d}x = 2{\pi}{\delta}(x)$