Derivation of Fourier's inversion theorem from Fourier series

Question

I am a bit lost in my textbook's derivation of the fourier inversion theorem

My text book shows a derivation of:

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega \ e^{i\omega t} \ \int_{-\infty}^{\infty}duf(u) \ e^{-i\omega u} \tag{1}$$

from:

$$f(t) = \sum_{r=-\infty}^{\infty}c_r \ e^{i \omega_r t} \tag{2}$$

and: $$c_r = \frac{1}{T} \int_{-T/2}^{T/2} f(t) \ e^{-i \omega_r t} \tag{3}dt$$

where $\omega_r$ are the allowed frequencies given by $\omega_r = \frac{2\pi r}{T}$

First, it defines a "quantum frequency", where $\Delta \omega = \frac{2\pi (r+1)}{T} - \frac{2\pi r}{T} = \frac{2\pi}{T}$

then $(2)$ becomes:

$$c_r = \frac{\Delta \omega}{2 \pi} \int_{-T/2}^{T/2} f(u) \ e^{-i \omega_r u}du \tag{4}$$

Note a variable change from $t$ to $u$ is done to prevent confusion. Substitute $(4)$ into $(2)$ and we obtain:

$$f(t) = \sum_{r=-\infty}^{\infty} \frac{\Delta \omega}{2\pi} \biggl(\int_{-T/2}^{T/2}f(u) e^{-i\omega_ru} \ du\biggl) \ e^{i\omega_r t} \tag{5}$$

Taking limit as $T\to \infty$ and hence $\Delta \omega \to 0$,

we obtain:

$$g(\omega_r) = \int_{-T/2}^{T/2}f(u) e^{-i\omega_ru} \ du \to \int_{-\infty}^{\infty}f(u) e^{-i\omega_ru} \ du \tag{6} $$

and

$$\sum_{r=-\infty}^{\infty} \frac{\Delta \omega}{2\pi} g(\omega_r) \ e^{i\omega_r t} \to \frac{1}{2\pi}\int_{-\infty}^{\infty}g(\omega) \ e^{i\omega t} d\omega \tag{7}$$

and hence by combining $(6)$ and $(7)$, $(4)$ becomes:

$$f(t) = \int_{-\infty}^{\infty} \frac{1}{2\pi} \biggl(\int_{-\infty}^{\infty}f(u) e^{-i\omega_ru} \ du\biggl) \ e^{i\omega_r t} \ d\omega \\ = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(u) e^{-i\omega_ru} \ e^{i\omega_r t} \ du\ d\omega\tag{8}$$

which apparently is equal to $(1)$

Doubts

So I have a few doubts about this derivation, mainly:

Firstly, in step $(7)$, why does a fixed value: $\omega_r$ become a distribution: $\omega$, if $T \to \infty$ ?

Secondly, why is step $(8)$ = $(1)$? We can see in $(1)$ that the $e^{i\omega t}$ and $e^{-i\omega u}$ terms have been brought OUTSIDE the $d\omega$ integral, even though they contain the variable $\omega$ (Note that here, $\omega_r$ in $(8)$ becomes a distribution as well, $\omega$ in $(1)$)

Any hints?

Thanks!!!

Answer 1

I think your second confusion is only due to notation: Sometimes taking the integral of a function with respect to $x$ is written as $\int dxf(x)$ instead of $\int f(x) dx$. The idea behind this notation is that $\int dx$ represents some operator which is applied to whatever comes right to it. So this would mean that (1) is equal to $\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(u)exp(-i\omega u)exp(i\omega t)du d\omega$.

For your first question: It helps me to think of $\omega_r$ as a function of $r$ (which it kinda is). Meaning that you choose some $r$ and calculate the corresponding $\omega_r$. Now as $T \rightarrow \infty$ consider what happens to the difference of $\omega_r$ and $\omega_{r+1}$. It becomes infinitely small. Thus summing over all possible $r$ and evaluating $\omega_r$ for each of those $r$'s is like integrating over the reals with respect to $\omega$.

Hope this helps.

Answer 2

If $f$ is $C^1$ and supported on $[-A,A]$ for $T >2A, t \in [-T/2,T/2]$ we have the Fourier series $$ f(t) = \frac{1}{T}\sum_n c_n(T) e^{2i \pi n t/T}, \qquad c_n(T) = \int_{-T/2}^{T/2} f(t)e^{-2i \pi nt/T}dt$$

Let $$F(u) = \int_{-\infty}^\infty f(t) e^{-2i \pi tu}dt, \qquad c_n(T) = F(n/T)$$ If also $|F(u)| < c|u|^{-2}$ we have $$\int_{-\infty}^\infty F(u)e^{2i \pi ut}du = \lim_{T \to \infty} \frac{1}{T}\sum_n F(n/T) e^{2i \pi n t/T}=\lim_{T \to \infty} \frac{1}{T}\sum_n c_n(T)e^{2i \pi n t/T}= f(t)$$ Without those three restrictions ($f \in C^1$ supported on an interval and $F$ decreasing fast enough) it is not obvious everything converges as needed.