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?