Proof for the inverse Fouriertransformation 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.
 A: 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 $1$.
