The Delta Function

By Brad Rodgers on Monday, October 22, 2001 - 08:43 pm:

If I integrate 1/x from -1 to 1, I get I complex number. I know why the mathematics yield this, but why in theory does this occur? What can be done to correct it?

Brad

By Dan Goodman on Monday, October 22, 2001 - 09:22 pm:

You can't integrate

1 / x

from -1 to 1, because integrating from 0 to 1 gives you infinity, as does integrating from -1 to 0. The formula for

\int_{a}^{b} f' (x) dx = f (b) - f (a)

is only true if there is no singularity between

a

and

b

(i.e.

a < x < b

implies

f (x)

is finite). As an example, try integrating

1 / x^{2}

from -1 to 1 using that formula and see what happens...

By Brad Rodgers on Monday, October 22, 2001 - 09:29 pm:

How then are functions like the delta function integrated rigorously?

By Dan Goodman on Monday, October 22, 2001 - 09:44 pm:

Whew! They're really complicated. Actually, I don't think they're even covered on most undergraduate mathematics courses (they're mentioned at Cambridge, but not rigourously covered). The thing is that the "delta function" is not a function, it is what's called a "distribution".

One way of thinking about it is that $δ (x) = 0$ if $x \neq 0$ , and $\int_{a}^{b} f (x) δ (x) dx = f (0)$ if $a < 0 < b$ or 0 otherwise. If you restrict yourself to these two properties you don't usually make any mistakes.

Another way of thinking about it is to use $δ_{n} (x) = k_{n} \exp (- (n x)^{2})$ where the constant $k_{n}$ is chosen to make $\int_{- \infty}^{\infty} δ_{n} (x) dx = 1$ . Whenever you see an integral involving $δ (x)$ just interpret it as meaning the limit as $n \to \infty$ of the expression with $δ_{n}$ instead of $δ$ .