# Proving delta dirac properties

I'm currently studying electromagnetism from Reitz's Foundations on electromagnetic theory, in that book delta dirac is presented as a "function" $\delta$ satisfying:

$\delta(x)=0$ for any $x \not=0$

$\int \delta(x)dx=1$

and is also stated that $\int f(x) \delta (x)dx=f(0)$ for any functon $f$. Now I'm trying to prove the following two properties:

a) $\delta(kx)=\frac{1}{|k|}\delta (x)$ for any constant $k \not=0$

b)$x\frac{d \delta(x)}{dx}=-\delta(x)$

For the first one I tryed integrating $\delta(kx)$ and by using the substitution $u=kx$ I get:

$$\int\delta(kx)dx=\int\frac{1}{k}\delta(u)du$$ but I don't know why that implies that the integrands must be equal, and even if they were equal I don't realize why I should obtain an absolute value.

For the second one I don't understand how I can differetiate it, intuitively I can see that if I take a point $x$ approaching zero by the left, then $\lim_{n\to\infty} \frac{d\delta_n(x)}{dx}$, where $\delta_n$ is a sequence of continuous functions converging to delta dirac, gets bigger and bigger as $x$ gets closer to zero and if I do the same thing approching zero by the right that limit gets smaller and smaller. But formally I don't know how to interpret the derivative since it seems to me that the slope of delta dirac is zero everywhere except at zero, where it's not defined.

For a you use the fact that $\delta(x) = \delta(-x)$, ie that the delta function is even.
For b you integrate by parts, taking advantage of the fact that $\delta(x) = 0$ on the boundaries. The tricky part with b is that the equality only holds if there are no other functions of $x$ multiplying $x\delta'(x)$.
• I'm not sure what you mean when you say that last thing. $x \delta'(x)=-\delta(x)$ means that (in the heuristic integral notation) $\int_{-\infty}^\infty x \delta'(x) f(x) dx = -\int_{-\infty}^\infty \delta(x) f(x) dx$ if $f$ is a smooth compactly supported function. The only way for there to be a problem when you multiply by something else would be if you multiplied by something that isn't a smooth compactly supported function. – Ian Sep 27 '16 at 22:46
• Actually the part a) was as follows "show that $\delta(kx)=\frac{1}{|k|}\delta(x)$ for any$k\not=0$ and in particular $\delta(-x)=\delta(x)$". So it seems that I should prove it without using the even property of dirac delta but concluding it as a corollary – la flaca Sep 27 '16 at 22:55
• You should be able to do the integral again with $-k$ and show that you get the same result as with $k$. – Sean Lake Sep 27 '16 at 23:02