Delta dirac function times zero

[jashua]

Let δ(x)=∞ at x = 0, and zero elsewhere. Then

δ(x)(1-exp(x)) = ?

It seems the above expression is zero. But isn't it zero times infinity at x = 0?

 

[CompuChip]

As you have defined it, δ(x) is not really a function.
In physics, we usually call it that, but if you want to treat it rigorously you should consider it as a distribution. The most important part about that, is that δ(x) only really makes sense under an integral sign, in expressions like
\int \delta(x) f(x) \, dx = f(0)
(if x = 0 is inside the integration interval).

 

[jashua]

OK, then let us consider the following example in a probability book:

Let

f(x)=(1-e^{-\lambda x})u(x),

where

u(x)= 1, if x \geq 1, and zero elsewhere.

Then, it is found (in the book) that

\frac{d}{dx} f(x) = \lambda e^{-\lambda x} u(x)

But I think that the derivative should be:

\frac{d}{dx} f(x) = \lambda e^{-\lambda x} u(x) + (1-e^{-\lambda x})\delta (x)

So, should the second term in the above expression be zero? Then how?

 

[CompuChip]

It's easiest to split up the function first.
For x \ge 1, f(x) = (1 - e^{-\lambda x}) and the derivative is what you got in the book.
For x < 1, f(x) = 0 so f'(x) = 0.

Note that in general f(x) is not differentiable - or even continuous - at x = 1. If you want to use the product rule, you have to consider f(x) as a distribution, which is a whole area of mathematics in itself. (Oh, and you'd probably get \delta(x - 1) rather than just \delta(x)... but that's beside the point).

 

[jashua]

Thank you very much for your answer!

Let me make a correction in the book example:

u(x)=1, if x≥0 (not x≥1, as in my previous post), and zero elsewhere.

In this case f(x) becomes continuous and differentiable. As far as I understand, even in this case I should consider f(x) as a distribution in order to use the product rule. Am I right?