# Strange representation of Heaviside and Delta function

1. Jun 4, 2008

2. Jun 4, 2008

### CompuChip

Probably the idea is, that we make it a contour integration. Then the only pole is at z = 0 and the result of an integration would be e.g.
$$\int \delta(x) f(x) \, dx = \int \frac{-1}{2i\pi} \frac{f(z)}{z} \, dz = (2i\pi) \operatorname{Res}_{z = 0} \frac{f(z)}{2 \pi i} = f(0)$$
if f(x) does not have any poles in the upper half plane.
but I actually doubt how valid this is (even if it works, one would need requirements on f(x) for $x \to \pm i\infty$ to close the countour; and I wonder what happens if f(x) itself has poles).

So I think that is the idea, but I wonder if it works and under what conditions.