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 òab 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/x2 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 d(x)=0 if x ¹ 0, and
òab f(x)d(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 dn(x)=kn exp(-(n x)2)
where the constant kn is chosen to make
ò-¥¥dn(x)dx=1. Whenever you see an integral
involving d(x) just interpret it as meaning the limit as n®¥
of the expression with dn instead of d.