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Can the Fourier transform of a whole function $f:\mathbb R\mapsto\mathbb C$ be defined as integration over $\mathbb C$ instead of $\mathbb R$ as well, such that $$\tilde f(k) = \frac{\mathcal N}{2\pi} \int_{\mathbb C}f(x) e^{-ikx}\,dx,$$
and (if so) what is the correct value of the normalization $\mathcal N$ for consistency with the $\mathbb R$-integration?

Given a whole function $f:\mathbb R\mapsto\mathbb C$, its Fourier transform

$$ \tilde f(k) = \frac1{2\pi}\int_{-\infty}^\infty f(x)e^{-ikx}\,dx$$

can be determined by other integration paths as well by using Cauchy's Residue Theorem, for example by shifting $x$ by an imaginary constant $ic$. Assuming a sufficiently fast decaying function for $\Re(x)\to\pm\infty$ (and using the fact that a whole function has no residues), this results in simply adding that constant to the integration boundaries, i.e.

$$ \tilde f(k) = \frac1{2\pi}\int_{-\infty+ic}^{\infty+ic} f(x)e^{-ikx}\,dx,$$

which can be expressed by substitution as well:

$$ \tilde f(k) = \frac1{2\pi}\int_{-\infty}^{\infty} f(x-ic)e^{-ik(x-ic)}\,dx.$$

One can then average over different values of $c$ to obtain

$$ \tilde f(k) = \frac1{2\pi\cdot 2T}\int_{-T}^{T}\int_{-\infty}^{\infty} f(x-ic)e^{-ik(x-ic)}\,dx\,dc.$$

Now my question boils down to

1) Since infinity is involved, is this equivalent to

$$ \tilde f(k) = \frac1{4\pi T}\int_{\mathbb R\times[-iT,iT]} f(x_1+x_2)e^{-ik(x_1+x_2)}\,d^2x,$$

2) Can the limit $T\to\infty$ be taken such that $\mathbb R\times[-iT,iT]\to\mathbb C$

and 3) What would the correct normalization be?

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While $\mathbb R$ is one-dimensional, $\mathbb C$ is two-dimensional; so the integral over $\mathbb C$ -- even if it makes sense -- cannot be a line integral. –  Srivatsan Jan 3 '12 at 11:37
@Srivatsan sure, but if you take an integrational average over a constant function, the average is still that constant. What I mean is going from the sum (which I assume is valid) to a second integral over $i\mathbb R$ (which, as you mention, adds a seconds integration dimension). This should somehow be covered by the factor $\mathcal N$, which I guess will involve something like $2\pi$ if existing –  Tobias Kienzler Jan 3 '12 at 11:40
It's an interesting idea, but the integral should diverge, since every strip adds the same contribution, so you'd need to do something like a limit of the integral up to $\pm\mathrm iT$, divided by $2T$. (Of course this "limit" would be the limit of a constant.) –  joriki Jan 3 '12 at 11:51
Also, note that, even if you could make sense of this divergent integral, it wouldn't satisfy Fubini's theorem, so you can't switch from the iterated integral to the product integral, so you're really just left with an integral along $y$ of the constant integrals you get along $x$. (That's for the infinite plane; the integrals between $\pm\mathrm iT$ should probably be OK so you might gain something by switching the order of integration there.) –  joriki Jan 3 '12 at 11:58
Yes, that about sums it up. The integral over $\mathbb C$ would diverge. –  joriki Jan 3 '12 at 12:44

1 Answer 1

up vote 2 down vote accepted

No, the Fourier transform of an entire function $f:\mathbb R\mapsto\mathbb C$ cannot be defined as integration over $\mathbb C$.

The comments already contain many hints why such a definition would be problematic. However, even the definition by integration over $\mathbb R$ is problematic for an entire function, as can be seen by the following part of the Paley–Wiener theorem:

An entire function $F$ on $\mathbb C^n$ is the Fourier–Laplace transform of distribution $v$ of compact support if and only if for all $z \in \mathbb C^n$, $$ |F(z)| \leq C (1 + |z|)^N e^{R| \mathfrak{Im} z|} $$ for some constants $C$, $N$, $R$.

We learn from this that the Fourier transform of a tempered entire function is in general only a distribution, which implies that the Fourier integral will not be well defined for a general entire function.

If we look at ("tempered") functions holomorphic in an "appropriate" half-plane of $\mathbb C$ instead of entire functions, the (inverse) Laplace transform becomes the "appropriate" modification of the Fourier transform.

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