Proof for the inverse Fouriertransformation

Question

I am currently refreshing my knowledge about the fourier series and fourier transformations and had to figure out, that I never saw how the fourier transformation is derived from the fourier series. Sadly, I do not even know a book covering this derivation. Therefor I was reading stuff online and found a pdf from the University of Hamburg, Germany where the following conjecture is given:

\begin{align} f(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{\infty}^{\infty}f(\tau)e^{i\omega(t-\tau)}d\tau d\omega \end{align}

This conjecture is then given as a theorem iff $f$ is continuous on every finite intervall and the integral:

\begin{align} ||f||_{L1} := \int_{\infty}^{\infty}|f(t)|dt \end{align}

exists. Last but not least it is said that if $f$ is not continuous at $x_0$ the double integral gives the mean value of the left and right sided limit. I would really appreciate if one of you could give me a hint for literature covering the step Fourier Series -> Fourier Transformation, or give me some advice on the topic itself.

Thank you very much!

PS: The pdf I refer to can be found at: https://www.math.uni-hamburg.de/teaching/export/tuhh/cm/kf/08/vorl12.pdf Sadly it is written in german and contains some mistakes as well

PPS: My current approach is to assume, that: \begin{align} \int_{\infty}^{\infty}e^{i\omega(t-\tau)}d\omega \end{align} behaves like a $\delta$-distribution, in the form, that it is $0$ for $t\neq \tau$ but $\infty$ for $t = \tau$. If this assumption should turn out to be true the proof would follow quite easily. Sadly I can not proof this assumption either.

Answer 1

After a few days I came up with an answer:

In the PPS of the question one can read, that my approach is to show, that \begin{align} \int_{\infty}^{\infty}e^{-i\omega(t-\tau)}dt \end{align} somewhat behaves like a delta distribution. I now know, that this expression is rather sloppy. Furthermore the integral from $]-\infty, \infty[$ is not defined. So lets introduce:

\begin{align} f_v(x) = \int_{v}^{v}e^{-i\omega x}dx\quad, where\ x = t-\tau \end{align} We can now work with distributions of $f_v$:

\begin{align} T_v[\phi] &:= \frac{1}{2\pi}\int_{\infty}^{\infty}f_v(x)\phi(x)dx\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}dx\int_{-v}^{v}dw\,e^{-i\omega x}\phi(x)\\ \end{align}

Considering the limit we get:

\begin{align} \lim_{v\rightarrow\infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}dx\int_{-v}^{v}dw\,e^{-i\omega x}\phi(x) &= \lim_{v\rightarrow\infty}\frac{1}{2\pi}\int_{\infty}^{\infty}dx\,\phi(x)\int_{-v}^{v}dw\,e^{-i\omega x}\\ &= \lim_{v\rightarrow\infty}\int_{\infty}^{\infty}dx\,\phi(x)\left[\frac{-1}{ix}(e^{-ivx} - e^{ivx})\right]\\ &= \lim_{v\rightarrow\infty}\int_{\infty}^{\infty}dx\,\phi(x)\frac{sin(vx)}{\pi x} \end{align}

It is known, that the last statement converges (as a distribution) to the delta-distribution. This has to be the case as the function $f(x) = \frac{sin(vx)}{\pi x}$ is Lebesgue integrable and integrated over $[-\infty, \infty]$ the integral is $1$.