I'm looking for some help regarding the derivation of the Fourier Sine and Cosine transforms, and more specifically how is it that we get to the inversion formula that the Wikipedia page claims.
Now, I recently read some elementary calculations on how to formally obtain the Fourier transform (and its inversion formula), as defined by
$$
\hat{f}(\omega) = \frac{1}{2 \pi} \int _{-\infty} ^{\infty} f(\mu) e^{-i\omega \mu} \, d\mu
$$
via the corresponding Fourier series
$$
f(x) = \lim _{N \rightarrow \infty} \sum _{n=-N}^{N} e^{inx} \cdot \frac{1}{2\pi} \int _{-\pi}^{\pi} f(\mu) e^{-in \mu} \, d\mu.
$$
To the best of my understanding, the argument goes as follows:
-
We consider an extended domain for the Fourier series, given by $(-\pi M, \pi M)$, and accomodate the series to this by modifying the sum and coefficients, now given by
$$
c_n = \frac{1}{M} \cdot \frac{1}{2\pi} \int _{-\pi M}^{\pi M} f(\mu) e^{-in \mu / M} \, d\mu = \frac{1}{M} g \left( \frac{n}{M} \right)
$$
with
$$
g(\omega) = \frac{1}{2\pi} \int _{-\pi M}^{\pi M} f(\mu) e^{-i \omega \mu} \, d\mu.
$$ -
Recognize the sum
$$
\lim _{N \rightarrow \infty} \sum _{n=-N}^{N} e^{inx / M} \cdot \frac{1}{M} g \left( \frac{n}{M} \right), \quad M \rightarrow \infty
$$
as a Riemann sum that converges to the integral
$$
\int _{-\infty}^{\infty} g(\omega) e^{i\omega x} \, d\omega.
$$ -
Recall the definition of $g(\omega)$; this is the Fourier transform (ignoring convergence shenanigans):
$$
g(\omega) = \frac{1}{2\pi} \int _{-\infty}^{\infty} f(\mu ) e^{-i\omega \mu} \, d\mu.
$$
I've been trying to replicate this same argument for the case of a trigonometric Fourier series
$$
f(x) = a_0 + \sum _{n=1}^{\infty} a_n \sin (nx) + b_n \cos (nx),
$$
which seems to hint at the inversion formula from the wikipedia page
$$
f(x) = \int _{0}^{\infty} \hat{f}^c (\omega) \cos (\omega x) \, d\omega + \int _{0}^{\infty} \hat{f}^s (\omega) \sin (\omega x) \, d\omega,
$$
but I've had no luck! Not because it's impossible to rewrite the coefficients for an extended domain as before, but because of that pesky $a_0$. If the normalizing factor for $a_0$ was given by
$$
\int _{-\pi M}^{\pi M} \cos ^2 \left( \frac{nx}{M} \right) \, dx = \pi M ,\quad n=1,2,3,\dots
$$
instead of $2\pi M$, we could carry over as usual and simply sum from $n=0$ without that extra term. However, this particular case won't allow me to do things as simply. Still, even if I were missing something that would allow me to perform the sum from $n=0$ and take the limit $M \rightarrow \infty$, the quoted inversion formula should at least take into account for this extra term, and I certainly don't see that happening.
Could someone help clear this up for me? Thanks!
Best Answer
If $f$ is a function on $[0,\infty)$, you can extend it to an odd function $f_o$ on $\mathbb{R}$ and then deal look at the Fourier transform and it's inversion for $f_o$. $$ f_o(x) = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}e^{isx}\int_{-\infty}^{\infty}e^{-isy}f_o(y)dy ds \\ = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}e^{isx}\int_0^{\infty}2i\sin(-sy)f(y)dy ds \\ = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_0^{R} 2i\sin(sx)\int_0^{\infty}2i\sin(-sy)f(y)dy ds \\ = \frac{2}{\pi}\int_0^{\infty}\left(\int_0^{\infty}f(y)\sin(sy)dy\right)\sin(sx) ds $$ This is a legitimate derivation. The ones where you try to derive the continuous transform from the discrete all suffer from serious defects. Fourier was the first to offer such a derivation, and his was never made rigorous.