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This question is from here. I'm asking it here as well to increase the number of people who see it and might be able to help.

The question is, what is the result of the following integral for integer $n$ and real $x$?

$$\int_{-\infty}^\infty dy\, e^{ny}e^{iyx}$$

Does it diverge and give infinity, or is it actually just equal to the following?

$$=\sum_{m=0}^\infty\frac{n^m}{m!}\int_{-\infty}^\infty dy\, y^m e^{iyx}=2\pi\sum_{m=0}^\infty\frac{1}{m!}\left(-in\frac{\partial}{\partial x}\right)^m\delta(x)=2\pi \left(e^{-in\frac{\partial}{\partial x}}\delta(x)\right)$$

More importantly, is there some book or scientific article on distributions that discusses this? Thanks for any suggestion!

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    $\begingroup$ the formal answer is $2\pi\delta(x-in)$ $\endgroup$
    – Carlo Beenakker
    Commented Oct 26, 2017 at 15:04
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    $\begingroup$ To study the properties of this integral I would suggest the Feynman trick of replacing $n$ by a real parameter $a$ and looking at the derivatives $\frac{\partial^N}{\partial a^N}$ of the integral after changing the integration domain to something like $[-R,R]$ so that there are no convergence problems. Then one can play with the limits $R\to \infty$ and $N\to \infty$. $\endgroup$
    – B K
    Commented Oct 26, 2017 at 17:42

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Note that $$e^{a\partial/\partial x}f(x)=f(x+a)$$ is the translation operator, so your exponent of the delta function gives $2\pi \delta(x-in)$, which is indeed consistent with

$$\int_{-\infty}^\infty e^{izy}dy=2\pi\delta(z)$$

for $z=x-in$.

All of this is purely formal, but there have been attempts to put it on a more secure ground by defining the delta function of a complex argument as an "ultradistribution". See for example Theories of Generalised Functions (page 121) and Distribution Theory and Transform Analysis (page 204). Caveat emptor.

To avoid a possible confusion, there is also the definition of a delta function in the complex plane as $\delta^2(x+iy)=\delta(x)\delta(y)$, which is just the product of two delta functions of real argument. Here we are concerned with a single delta function $\delta(x+iy)$ of a complex argument. The two objects are contrasted in this reference.

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    $\begingroup$ I was under the impression that the integral representation of the delta function is only defined for real arguments $x$. That is why I'm not sure $\delta(x-in)$ exists? Which limit or integral representation does it have to come from, so that it converges and we get a single spike at a point $x-in=0$ in the complex plane? Perhaps you know of some reference that discusses this? $\endgroup$
    – Kagaratsch
    Commented Oct 26, 2017 at 15:15
  • $\begingroup$ @CarloBeenakker correct me if I'm wrong, it seems you already have given negative answer to this question here mathoverflow.net/questions/118101/… ? $\endgroup$
    – Nemo
    Commented Oct 26, 2017 at 15:21
  • $\begingroup$ @Nemo Very useful link, thank you! There Carlo Beenakker writes "Dirac never considered the delta function of a complex argument, only of real numbers." and suggests to split the delta function into two, one for real and one for imaginary part of the argument. In case at hand that would demand $n=0$. However, the expression $e^{-in\frac{\partial}{\partial x}}\delta(x)$ seems to allow for more general $n$ under integration against a test function. Thus my whole question above. $\endgroup$
    – Kagaratsch
    Commented Oct 26, 2017 at 15:25
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    $\begingroup$ Thank you all for your feedback. I have added a few references that may provide some justification for this delta function of a complex argument. The previous MO question @Nemo referred to was in response to a query what Paul Dirac meant when he wrote about the delta function of a "c-number". In that context the "c-number" is still purely real ("c" then stands for "classical", as opposed to "quantum"). So that is not directly related to this question. $\endgroup$
    – Carlo Beenakker
    Commented Oct 26, 2017 at 21:14
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    $\begingroup$ To make sense to $\delta(x-a)$ for $a \in \mathbb{C}$ you need to look at analytic functionals (the equivalent of distributions but acting on some space of analytic functions instead of the Schwartz space and $C^\infty_c$). In that case the Fourier/Laplace transform formulation of $\delta(x-a)$ makes sense on the smaller space of analytic functions being the Fourier/Laplace transform of a function decreasing faster than every exponential (a space containing the Gaussians) @Kagaratsch $\endgroup$
    – reuns
    Commented Oct 27, 2017 at 0:59
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I think Schwartz and Gelfand-Shilov, et al, had already considered the possibility of having "Fourier transform" map from all distributions to the dual of the Paley-Wiener space (the latter being the image of test functions under literal Fourier transform).

So although the literal integral you write does not converge (nor do many for tempered distributions, either), it does have a sense compatible with other things.

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The integral always diverges and what you really want is probably not an integral, but the Fourier transform of an exponential function, which also does not exist in general because $y\mapsto e^{ny}$ is (except for $n=0$) not a tempered distribution. This is also reproduced in Carlo's answer as it gives an undefined expression for $n\neq 0$; $\delta$ is only defined as a distribution on $\mathbb{R}$ (at least in this context, otherwise the integral/FT would look different)

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    $\begingroup$ What about the Schwartz's Paley–Wiener theorem? en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem It seems to suggest that tempered distribution is not the only admissible kind. $\endgroup$
    – Kagaratsch
    Commented Oct 26, 2017 at 17:08
  • $\begingroup$ Does it? How so? Remember that compact support implies tempered. And the PWT for more general supports (say convex cones) still requires temperedness to make sense of the Fourier transform. $\endgroup$
    – Johannes Hahn
    Commented Oct 26, 2017 at 18:50
  • $\begingroup$ This is the same old canard that one finds so often on MO. The fact that one can define the FT for tempered distributions doesn‘t mean that one cannot define it for more general ones. In the last 50, 60 years, a plethora of pairs of suitable spaces of test functions connected by the FT have been constructed and the latter then extends to the corresponding distribution spaces by duality. (The case of tempered distributions has a special symmetry due to the fact that the two spaces of test functions coincide). In particular, the FT has been extended to ALL distributions. $\endgroup$
    – reliquia
    Commented May 31, 2021 at 17:54
  • $\begingroup$ In this case, the range of the FT has to be larger than the space of tempered distributions, of course. As has already been mentioned in this thread, it is, in fact, a space of analytic functionals, i.e., elements of the dual of a suitable space of entire functions. Another canard is that there is something mysterious about a delta function with a complex argument. Despite it‘s name, it is simply a point measure and so can be defined as a Radon measure on any topological space, in particular as a distribution on any manifold. $\endgroup$
    – reliquia
    Commented May 31, 2021 at 18:05

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