# Dirac delta function with complex arguments

1. May 13, 2008

### JayFsd

This is probably a silly question to some, but I've been struggling to understand how the delta function behaves when given a complex argument, that is \delta(z), z \in C. I guess the basic definition is the same that the integral over all space is 1, but I'm looking for a more detailed guide on the inner workings. Does someone know of a good reference?

TIA.

2. May 13, 2008

### HallsofIvy

Staff Emeritus
The (Dirac) delta function, which is a "Distribution" or "Generalized Function", rather than a true "function", can be defined in a number of ways. The most common definition is that
$$\int_C f(x)\delta (x)dx$$
is equal to 1 if 0 is in set C, equal to 0 if 0 is not in set C.

More generally
$$\int_C f(x)\delta (x-a) dx[/itex] is equal to 1 if a is in set C, equal to 0 if 0 is not in set C. There is no reason C cannot be a subset of the complex numbers and a a complex number. It can be thought of, roughly, as the limit of a sequence of functions, fn, such that fn= n if |z|< 1/n, 0 other wise. Of course, that sequence doesn't actually converge which is why $\delta(z)$ is not a true function! 3. Apr 8, 2009 ### yaychemistry I kind of have a follow-up question to JayFsd (a year later, ha). Suppose that [tex]a$$ is complex and you have the integral

$$\int_C f\left(x\right) \delta\left(x - a\right)dx$$

such that $$a$$ lies on the countour $$C$$, you would expect the result to be:

$$\int_C f\left(x\right) \delta\left(x - a\right)dx = f\left(a\right)$$.

However, to evaluate this integral for a specific contour, let's say, we do the substitution $$x = x\left(t\right),\, 0 <t < 1$$ and get

$$\int_0^1 f\left(x\left(t\right)\right)\delta\left(x\left(t\right) - x\left(t_a\right)\right)\frac{dx}{dt}dt$$

where $$x\left(t_a\right) = a$$

Ok, so the question I have:

According to Wikipedia, the dirac delta scales as

$$\delta\left(g\left(x\right)\right) = \sum_i\frac{\delta\left(x - x_i\right)}{|g'\left(x_i\right)}$$

where the $$x_i$$ are the roots of $$g$$.

So if I use that scaling property in the above integral, then I get

$$\int_0^1 f\left(x\left(t\right)\right)\delta\left(x\left(t\right) - x\left(t_a\right)\right)\frac{dx}{dt}dt = \frac{dx}{dt_a}/\left|\frac{dx}{dt_a}\right|f\left(x\left(t_a\right)\right)$$

which doesn't agree with the expected result. It seems as if I don't understand the derivation of the scaling property well enough -- that is with real numbers it seems fine, but when you throw complex numbers in the mix I get hopelessly lost.

Anyone care to chime in and help me?