Where to find a table of fair Fourier transforms? [closed]

Question

I am greatly dissatisfied with those tables of Fourier transforms, available online. I simply have no guess what method they use to derive their tables, but it seems completely off to me.

For instance, some transforms from the tables that disturb me:

• The shift rule:

$$f (x-a) \to e^{-iaw}\hat {f}(w ) $$

It is given in the tables without exceptions but obviously it should not work for instance, for Heaviside Theta function and other functions that do not tend to zero at infinity.

Update. I want to clarify that I want the following:

$$\theta(x+a)\to ~PV\frac{-i}{k} +\pi\delta(k)+ a$$

(which follows from Sokhotski–Plemelj theorem)

instead of

$$\theta(x+a)\to -\frac{i e^{i a w} }{w}+\pi \delta (w)$$

which is given in the tables ($\theta(x)$ is the Heaviside Theta).

• Fourier transform for negative even powers:

$$\frac1{x^2}\to -\pi w \operatorname{sgn}(w)$$

Since this function has divergent integral over zero, the Fourier transform obviously should be infinite at $w=0$. But the table formula gives $0$!

Where can I find fair tables for Fourier transforms? Particularly, such that if $\int_{-\infty}^\infty f(x)dx=\infty$ the expression for Fourier transform at $w=0$ should take infinite values.

Answer 1

Let me add the comment that in the article of Sebastião e Silva I quoted in a recent post, some of the questions you pose are addressed.

a) Distributions are not introduced as functionals on spaces of test functions but as derivatives in a generalised sense of continuous functions. Thus the delta distribution is the second derivative of $\frac 12 |x|$. Any locally integrable function is a distribution--take the distributional derivative of its primitive. Hence, $\log |x|$ is a distribution and so are its distributional derivatives--this gives the negative powers of $x$. In a similar manner, any rational (even meromophic) function can be regarded as a distribution. The details of all this are in the article I cited.

b) Within this framework the concept of limits of distributions at finite or infinite points, values of distributions at a point and integrals of distributions are developed. Although a general distribution doesn't have a value at a point, most ones which occur in practice do at many points and this is significant (despite some comments to the contrary on this post)

c) Most significantly for your question, Sebastião e Silva developed the concept of a parametrised integral, i.e., the integral of a family of distributions which depend on a parameter--distributionally. This is particularly important for Fourier transforms which are defined exactly as classically, not via duality. The resulting integrals converge in the above sense, even when they diverge in the usual sense,

d) The mathematician's dream then comes true--one has Fubini and can differentiate under the integral in parametrised integrals in situations where this would forbidden in the classical one. This explains why many formal computations give the correct answer and puts them firmly in a rigorous framework.

Sebastião e Silva's article is completely elementary and he works through several examples very much in the style of your question. You can also find more sophisticated material at the site I pointed out, some in French but also some in Portuguese. He was working on an encyclopedic monograph of his ideas and their applications but his untimely passing prevented its completion.

Answer 2

Although this question probably should be moved to Math Stack Exchange, some remarks are easy:

For example, (the distribution) "$1/x^2$" is certainly not locally integrable near $0$, so ... it's not that we conclude that its Fourier transform blows up at $0$, although, indeed, the literal integral that would express this Fourier transform pointwise value does blow up... to make sense of this Fourier transform (rather than declaring that it has no sense) is to analytically continue in the family of tempered distributions $1/|x|^s$ (and, similarly, $\mathrm{sgn}\,x\cdot |x|^{-1}$). This is what Riesz showed that Hadamard did, in effect, in his "finie partie" device, c. 1930.

That is, to reiterate, many Fourier transforms (already even on $L^2$, as opposed to $L^1$) cannot possibly be the literal integral, but some sort of extension (by continuity) of that integral.

In particular, if your are wanting/needing tables/ideas of Fourier transforms that blow up in certain ways for distributions such as $1/x^2$, you will be frustrated to some degree by tables and explanations, because a big part of the underlying goal is to make sense of things even when the literal integral definition of Fourier transform gives a divergent integral.

So, yes, essentially all identities that do hold when the literal integral converges do also hold when Fourier transform is extended by continuity to larger classes of functions (or/and "generalized functions", so, here, tempered distributions).

And, I add, all these things are self-consistent, if one does not put too much stock in "pointwise values"... which, unsurprisingly, are not a good way to understand generalized functions (measure theory notwithstanding).

EDIT: and, certainly, as @ChristianRemling notes, another viable notion of the distribution $1/x^2$ is as the distributional derivative of the principal-value integral against $1/x$. And it is worth noting that already this "principal value integral" is not a literal integral, but something more exotic.

And, to respond to the revision of the original post, I think it is not wise to want the interaction of Fourier transform with translation to be anything other than how it behaves with $L^1$ functions. E.g., with Heaviside's step function. The extension-by-continuity, or by duality, of Fourier transform to tempered distributions gives a certain result... which is "good" because it is consonant with many structural ideas, e.g., "continuity" in weaker topologies, and, thus, extension-by-continuity.

If, instead, we insist on literal integrals and insist on making their divergence "matter", we find ourselves (I think) in a not-so-useful situation.