Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Derivation of Fourier Transform?

How is the Fourier transform obtained by taking the limit of the Fourier series as the period goes to infinity? In particular I am unsure about how to transform the sum to an integral

share|improve this question

marked as duplicate by Qiaochu Yuan Jun 18 '11 at 6:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
I believe the Fourier series can be interpreted as a Riemann sum as the period goes to infinity. –  Qiaochu Yuan Jun 17 '11 at 13:02
add comment

1 Answer 1

up vote 3 down vote accepted

(Note, personally I don't really like this as a heuristic, but here's one possible argument that vaguely resembles what you want.)

We shall write the Fourier series in a slightly non-standard fashion, to make the "limit" more apparent, by choosing more suitable normalisations. In particular, consider a periodic function of period $L$. In particular, we will define the "Fourier series coefficients" as the following (the main difference from the usual definition is that I dropped the $1/L$ normalisation factor in front of the integral):

$$ \hat{f}(\xi) = \int_{-L/2}^{L/2} f(x) e^{-2\pi i \xi x} dx \qquad \xi = \frac{n}{L}, n\in \mathbb{Z} $$

Then the Fourier series for $f$ can be written as (equals sign in quotes because there are issues with convergence)

$$ f(x) "=" \sum_{n\in\mathbb{Z}} \frac{1}{L}\hat{f}\left(n/L\right)e^{2\pi i \frac{n}{L} x}$$

Then very formally speaking, the first expression, when you take $L\nearrow \infty$ can be seen as the expression for the Fourier transform

$$ \hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i \xi x} dx$$

while the second expression formally looks like the Riemann sum of the integral

$$ f(x) "=" \int_{-\infty}^{\infty} \hat{f}(\xi) e^{2\pi i \xi x} d\xi $$

with step size $1/L$.

In certain situations, these convergenses can be made precise. For example, Let $L > L_0$ and let $f$ be a smooth function compactly supported in the interval $[-L_0/2, L_2/2]$. Then for every $\xi$, one can find an $n$ sufficiently large so that $n/\xi > L_0$, and so $\hat{f}(\xi)$ is well-defined for all $\xi$. Furthermore, using compact support, it is easy to show that $\hat{f}(\xi)$ is continuous (and in fact smooth) in $\xi$. And the usual textbook proof also shows that $\hat{f}(\xi)$ decays faster than any polynomial. Therefore the inversion formula makes sense and $\hat{f}(\xi)$ is integrable. Using the uniform continuity of $\hat{f}(\xi)$, you also have that the Riemann sum converges to the integral, and so in this case the argument is rigorous:

If $f$ is a smooth function compactly supported in $[-\frac{L_0}{2},\frac{L_0}{2}]$, then for any $L > L_0$, the Fourier series coefficient $\hat{f}_L(n/L)$ agrees with the Fourier transform coefficient $\hat{f}(\xi = n/L)$. Furthermore, on any compact subset $K$ of the real line, the Fourier series $\sum L^{-1} \hat{f}_L(n/L) e^{2\pi i n x / L}$, converges uniformly in any $C^k$ norm to the inverse Fourier transform $\int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i \xi x} d\xi$.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.