# Approximation of the Fourier Transform of General Functions in a Box

I'm trying to get a general approach for the Fourier Transform of functions $f$, only in a restricted area $-\frac M2\le x \le \frac M2$, where ${\frak F}_{f(x)}(\omega)$ exists. My idea was the following:

1. Expand at $x=0$ to get $f(x)=\sum_k \frac{x^kf^{(k)}(0)}{k!}$,

2. Do the Fourier Transform elementwise $$\frac{f^{(k)}(0)}{k!} \int_{-M/2}^{M/2} x^k \exp(i2\pi x \omega) {\rm rect}(x) dx =\frac{f^{(k)}(0)}{k!}\left[-x^{n+1} E_{-n}(-i2\pi x \omega)\right]_{-M/2}^{M/2}\\ = \frac{f^{(k)}(0)}{k!}\left[\left(-\frac M2\right)^{n+1} E_{-n}(i\pi M \omega) -\left(\frac M2\right)^{n+1} E_{-n}(-i\pi M \omega)\right]\phantom{soorespace} (2_k)$$ $E_{-n}(\cdot)$ is the Exponential Integral used by W|A.

3. Summing up the $(2_k)$ would give an approximation of the Fourier Transform.

Is my reasoning correct?

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## 1 Answer

Well, for one, I'm not entirely sure where the factor of $\mathrm{rect}(x)$ came from inside the integral, seeing that you already have finite integration endpoints. But never mind that.

Your reasoning is formally correct. My major concern, however, is that this approximation has no error estimate. Is the power series representation of $f$ exact? What are we losing by throwing away the high absolute values of $x$ in the integral? By doing so, we are actually taking the FT of a potentially discontinuous function, which introduces additional high frequency content in the transform. Do you have an estimate of that?

What is the value of $M$ in relation to a scale value inherent in $f$ (e.g., $\sigma^2$ when $f$ is a gaussian)? How about the relation of $M$ to the radius of convergence of the power series representation of $f$?

Also, to be practical, why these exponential integrals? Are they inherently fast to compute or do they have lovely analytical properties that are useful to you? To be honest, these exponential integrals do not seem useful for negative indices as you want to use. (Taking a look at Numerical Recipes in C, which I admit is the 1992 version, I see that the algorithm used to compute $E_n(x)$ returns an error for $n<0$.) And even if you can compute these, how high can $n$ get before you start running into serious numerical troubles? And will you have an acceptable approximation before you reach such a value of $n$?

I don't mean to sound so discouraging. It could be a nifty idea; it's just that a lack of any means to estimate an error in the approximation produces more questions than answers.

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+1 thanks for thoughts. I'm currently trying to tackle this problem: Question on the Fourier Transform of an Integral Logarithm of Prime $\zeta$ Functions. The exponential integral also appeared in that context before it looked appealing to me. I've gotta think about your concern with the convergence radii... – draks ... Jan 27 '13 at 16:32
negative $n$ work fine, see here, at least the indefinite integral... – draks ... Jan 27 '13 at 16:44
My point had to do with the actual numerical computation of the integral, which if you were planning to do actual computation, you would have to contend with that. But maybe that is not your concern. – Ron Gordon Jan 27 '13 at 18:42