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.

Any advice is welcome!


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 $x\delta'(x)$.

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    $\begingroup$ 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. $\endgroup$ – Ian Sep 27 '16 at 22:46
  • $\begingroup$ Hm, that's a good point. Then again, I work with "not nice" functions so often, that I just play it safe. $\endgroup$ – Sean Lake Sep 27 '16 at 22:53
  • $\begingroup$ 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 $\endgroup$ – la flaca Sep 27 '16 at 22:55
  • $\begingroup$ You should be able to do the integral again with $-k$ and show that you get the same result as with $k$. $\endgroup$ – Sean Lake Sep 27 '16 at 23:02
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    $\begingroup$ @Eliana If you pay attention to putting the limits of integration back in the same order as you started with, you can proceed directly. $\endgroup$ – Ian Sep 27 '16 at 23:17

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