Separation of arguments in the exponent Is there a way to represent $\exp({\rm i}xy)$ so that arguments $x$ and $y$ are separated?
An example would be 
$$
e^{{\rm i}xy} = \int_a^b dt~f(x,t)g(y,t),
$$
but what would $f$, $g$, $a$ and $b$ be in this case, if this is possible?
One simple version is Taylor expansion $\exp({\rm i}xy) = \sum_n i^n/n! f(x)^n g(y)^n$, where $f$ and $g$ are then identity funcitons.
 A: There are many ways of achieving this, here are some examples. First of all if we allow distributions for $f,g$ then $f(x,t) = \delta(x-t)$ (where $\delta$ is the Dirac delta function) togeather with $g(y,t) = e^{iyt}$ is a simple example since
$$e^{ixy} = \int_{-\infty}^{\infty}\delta(x-t) e^{ity}\,{\rm d}t$$
follows directly from the definition of $\delta$. We can also do without distributions. Motivated by the heat-kernel representation of the $\delta$ function $\delta(x) = \lim_{\epsilon\to 0} \frac{1}{\sqrt{2\pi \epsilon}}e^{-\frac{x^2}{2\epsilon}}$ we make the ansatz $f(x,t) =  \frac{1}{\sqrt{2\pi \epsilon}}e^{-\frac{x^2}{2\epsilon}}$ and $g(y,t) = e^{ity}$ and find
$$\int_{-\infty}^{\infty}f(x,t)g(y,t)\,{\rm d}t = e^{ixy} \cdot e^{-\frac{y^2\epsilon}{2}}$$
so for any $\epsilon \not= 0$ the pair $f(x,t) =  \frac{1}{\sqrt{2\pi \epsilon}}e^{-\frac{x^2}{2\epsilon}}$ and $g(y,t) = e^{iyt + \frac{y^2\epsilon}{2}}$ has the desired properties. 
If you want $a,b \not= \pm \infty$ then just make a change of variables in the integral for which $(-\infty,\infty)$ is mapped to any finite interval.
