Exponential derivative of delta distribution? 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!
 A: 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.
A: 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.
A: 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)
