Fourier transform in $\mathbb{L}^2(\mathbb{R})\setminus\mathbb{L}^1(\mathbb{R})$

Question

In a book I've found the following definition for the Fourier transform in $\mathbb{L}^2(\mathbb{R})\\$:

$g(k)$=$\frac{1}{\sqrt{2\pi}}\frac{d}{dk}\int_{-\infty}^{+\infty}\frac{e^{-ikx}-1}{-ix}f(x)dx\\$ Now I don't understand why this definition is equivalent to the usual one that is: $g(k)=\lim_{N \to \infty} \frac{1}{\sqrt{2\pi}}\int_{-N}^{+N}e^{-ikx}f(x)dx\\$ where the limit is in the sense of the norm $||\cdot||_2$. Where can I find a proof of the equivalence of the two definitions?

Answer 1

This is not a proof, but a bit of unification of the two definitions.

Using the definition of derivative we have $$ \frac{d}{dk} \int_{-\infty}^{\infty} \frac{e^{-ikx}-1}{-ix} \, f(x) \, dx = \lim_{\delta\to 0} \frac{1}{\delta} \left( \int_{-\infty}^{\infty} \frac{e^{-i(k+\delta)x}-1}{-ix} \, f(x) \, dx - \int_{-\infty}^{\infty} \frac{e^{-ikx}-1}{-ix} \, f(x) \, dx \right) \\= \lim_{\delta\to 0} \int_{-\infty}^{\infty} \frac{e^{-i(k+\delta)x}-e^{-ikx}}{-i\delta x} \, f(x) \, dx = \lim_{\delta\to 0} \int_{-\infty}^{\infty} \frac{e^{-i\delta x}-1}{-i\delta x} \,e^{-ikx} \, f(x) \, dx \\= \lim_{N\to\infty} \int_{-\infty}^{\infty} \frac{e^{-ix/N}-1}{-ix/N} \,e^{-ikx} \, f(x) \, dx . $$

The other definition is $$ \lim_{N\to\infty} \int_{-N}^{N} e^{-ikx} \, f(x) \, dx = \lim_{N\to\infty} \int_{-\infty}^{\infty} \chi_{[-N,N]}(x) \, e^{-ikx} \, f(x) \, dx . $$

Thus, both definitions are of the form $$ \lim_{N\to\infty} \int_{-\infty}^{\infty} h_N(x) \, e^{-ikx} \, f(x) \, dx , $$ where $h_N \in L^2(\mathbb{R})$ and $h_N(x) \to 1$ pointwise (and on every compact set).

Answer 2

\begin{align} \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{e^{-ikx}-1}{-ix}f(x)dx &=\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\int_0^k e^{-isx}ds f(x)dx \\ &=\lim_{R\rightarrow\infty}\int_0^k\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-isx}f(x)dx ds \\ &=\int_0^k \hat{f}(s)ds \end{align}