Derivative of the Dirac delta function

Question

So, I was reading about the Dirac delta function and how its differentiation works. So, pretty much all texts and online sources I saw, define it using the integral:

$$ \int_{-\infty}^{+\infty}x\dfrac{d\delta(x)}{dx} dx= \left. x\delta(x)\right|_{-\infty}^{\infty}-\int_{-\infty}^{+\infty}\delta(x)dx$$

and here all of them say the first term is zero, as $δ(x)$ is zero at infinities. But, won't $x$ be equal to infinity at that point? Won't this mean that the first term is not zero? My point stands, even if there was a function $f(x)$ instead of $x$ in the L.H.S. of the above equation.

I originally had this doubt while solving problem 1.46 in Introduction to electrodynamics by D. J. Griffiths. Any help on this matter would be much appreciated.

Answer 1

So the Dirac delta is a distribution, that is a linear form acting on smooth compactly supported functions (the test functions). That is, a distribution $f$ is defined by the values $\langle f,\varphi \rangle$ for any test function $\varphi$. In the case when $f$ is a function, then the above bracket is just an integral, but it can be other things for more general distributions. Hence distributions generalize functions. To make this generalization coherent, as a generalization of the integration by parts formula, one defines for every distribution $f$ its derivative by the formula $$ \langle f',\varphi \rangle = -\langle f,\varphi'\rangle $$

Now that you have the definitions, let's go to your question. The Dirac delta is defined as a distribution by $$ \langle \delta_0,\varphi\rangle = \varphi(0). $$ The compact support is actually not needed for $\varphi$ because the value here just depends on the value of $\varphi$ at $0$. In particular taking $\varphi =1$ we get that $\langle \delta_0,1\rangle =1$ Hence by definition, the derivative of the Dirac delta verifies $$ \langle \delta'_0,\varphi\rangle = -\langle \delta_0,\varphi'\rangle = -\varphi'(0). $$ In particular, one can take $\varphi(x) = x$, giving $$ \langle \delta'_0,x\rangle = -1 = - \langle \delta_0,1\rangle. $$ This is the rigorous form of your formula, which could more informally be written $$ \int_{\Bbb R} x\,\delta'_0(x)\,\mathrm d x = -\int_{\Bbb R} \delta_0(x)\,\mathrm d x. $$

Answer 2

What LL says is exactly right, so this is just to add a one or two points which might help make the formal definitions fit better with the informal notions you were using in posing your question.

As LL says, a distribution is a real-valued linear map with domain $\mathcal T$, the space of smooth functions supported on an interval $[-R,R]$ for some $R>0$. If $f$ is any locally integrable function (i.e. integrable on any finite interval) and $\phi\in \mathcal T$, then $f.\phi$ is is integrable and supported in $[-R,R]$ where we assume $\phi$ is supported in $[-R,R]$. Hence we may associate to $f$ a distribution $\delta_f$ where

$$ \begin{split} \delta_f(\phi) &= \int_{-\infty}^{\infty} f(t)\phi(t) dt,\\ &= \int_{-R}^R f(t)\phi(t)dt <\infty. \end{split} $$

Now suppose that

$$ s_n = \left\{ \begin{array}{cc}n(1+nt) & \text{ if } -1/n \leq t \leq 0 \\ n(1-nt), & \text{ if } 0<t \leq 1/n \end{array} \right. $$

so that $s_n$ is nonzero only on $(-1/n,1/n)$ and in that interval is a triangular spike with height $n$ and base $2/n$, so that $\int_{-\infty}^{\infty} s_n(t)dt = 1$. Each $\phi_n$ thus defines a distribution $\delta_n$ by taking $f = s_n$ in the above. Now if $\phi \in \mathcal T$, $$ \lim_{n\to \infty} \delta_n(\phi) = \lim_{n \to \infty} \int_{-\infty} ^\infty s_n(t).\phi(t)dt = \phi(0), $$ since $\phi$ is continuous at $t=0$. Thus in some suitable sense $s_n\to \delta_0$, the Dirac delta function, as $n \to \infty$.

But now notice that for any $n$, $s_n(t)\phi(t)$ vanishes for $t \notin (-1,1)$, and in fact

$$ \begin{split} \int_{-\infty}^{\infty} s_n'(t).\phi(t)dt &= [s_n(t).\phi(t)]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} s_n(t).\phi'(t)dt \\ &\to 0 - \phi'(0) = -\phi'(0) \end{split} $$

so that $s_n'$ tends to $-\delta_0\circ \frac{d}{dx}$, which is exactly what your informal ``integration by parts'' heuristic led you to.