TITLE: Proving Continuous Fourier Transform Formulas
QUESTION [0 upvotes]: Given a continuous non-periodic function, its Fourier transform is defined as:
$$f(x) = \int_{-\infty}^\infty c(k) e^{ikx} dk, \ \ \ \ \ \ \ \ \ \ \ \ \ c(k) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-ikx} dx$$
The problem is proving this is true by evaluating the integral when $c(k)$ is plugged into the equation for $f(x)$.
This ends up with a long integral:
$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{-ikx'} dx' \right) e^{ikx} dk$$
I'm not sure really how to proceed from here. I moved the $e^{ikx}$ into the inner integral, which I figured was fine since it's constant relative to $x'$.
$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{ik(x-x')} dx' \right) dk$$
I tried to kill at least one of the integrals by seeing if something evaluated to a Dirac Delta but I can't seem to get that result. I also tried integrating by parts, but that led me nowhere.

REPLY [0 votes]: Start by assuming $f$ is absolutely integrable. Then
$$
    \frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}f(x')e^{-ikx'}dx' e^{ikx}dk \\
   = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-R}^{R}e^{ik(x-x')}dk dx' \\
   = \frac{1}{\pi}\int_{-\infty}^{\infty}f(x')\frac{\sin(R(x-x'))}{x-x'}dx'.
$$
The kernel $K(x,x')=\frac{1}{\pi}\frac{\sin(R(x-x'))}{x-x'}$ has a principal value integral in $x'$ equal to $1$. So, if $f$ is differentiable at $x$, then the above may be written as
$$
    \lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{-R}^{R} \frac{f(x')-f(x)}{x-x'}\sin(R(x-x'))dx'+f(x),
$$
and the limit of the integral is $0$ by the Riemann-Lebesgue lemma. So, you can a sense in which $\frac{1}{2\pi}\int_{-R}^{R} e^{ik(x-x')}dk \rightarrow \delta_{x}(x')$ as $R\rightarrow\infty$.
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