4

Is this proof formal enough? I plan on being a theoretical physicist one day, so I want to get into the good habit of being mathematically strict.

My proof:

$u=x$; $du=dx$

$v = \delta (x)$; $dv = -\delta (x)$

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

$$x\delta (x) = 0$$

We now integrate both sides in order to properly use the Kronecker delta function.

$$\int x \delta (x) dx = \int 0dx$$

It is known that the $\int_{-\infty}^{+\infty} f(x)\delta (x)=f(0)$. Thus,

$$0=0$$

whatwhatwhat
  • 1,657
  • Mathematically, the Dirac delta is a distribution (or what physicists call a generalized function), which is meaningful when you pair this with test functions: if $\varphi$ is any smooth compactly supported function, then as Zachary Selk computed, we get $$ \langle x \delta'(x), \varphi(x) \rangle = \int x \delta'(x) \varphi(x) , \mathrm{d}x = -\varphi(0). $$ And we know exactly what distribution gives this result: $-\delta(x)$! – Sangchul Lee Sep 03 '16 at 23:53
  • 2
    more explicitly, by definition of the distributional derivative $\langle T',\varphi \rangle = -\langle T,\varphi' \rangle$ and the multiplication of a distribution $T$ by a function $f \in C^\infty$ : $ \ \ \langle f T,\varphi \rangle = \langle T,f\varphi \rangle$ we have $\langle x \delta', \varphi \rangle =\langle \delta', x \varphi \rangle =-\langle \delta, (x\varphi)' \rangle = -\langle \delta, \varphi+x \varphi' \rangle =-\langle \delta, \varphi \rangle$ @SangchulLee – reuns Sep 04 '16 at 00:17
  • @user1952009, That's right. In order to be precise, we should work on paring itself and using the definition of derivative of distribution. Thank you for pointing out this. – Sangchul Lee Sep 04 '16 at 00:32
  • 1
    @SangchulLee $Generalized\ Functions$ were introduced, in 1935, by the russian Sergei L. Sobolev who WAS a mathematician. – Felix Marin Sep 04 '16 at 04:08
  • @FelixMarin Thank you for correcting that! It's interesting to see that the seemingly intuitive term generalized functions is less popular in math literature than more ambiguous term distribution, which has at least two more independent meanings. – Sangchul Lee Sep 04 '16 at 05:28
  • If the delta function is thought of as the limit of some regular function (e.g., a Gaussian with very small width), then $\delta'(x)$ should be an odd function (see Page 61 of R. Shankar's book: Principles of Quantum Mechanics). So naively we may have $\lim_{x\to 0}x\delta'(x)=0$. How to understand this point? – Enigma Sep 20 '23 at 05:55

1 Answers1

3

Let $f$ be any smooth compactly supported function. Then:

$$\int x\delta'(x)f(x)\ dx=\int\delta'(x) xf(x)\ dx=-\int\delta(x)(xf'(x)+f(x))\ dx=-0f'(0)-f(0)=-f(0)$$

  • In step 3, it looks like you took the derivative of $xf(x)$, took the anti-derivative of $\delta ' (x)$, and then added a negative sign. How is that allowed within the integrand??? – whatwhatwhat Sep 03 '16 at 23:52
  • 1
    @whatwhatwhat Integration by parts gives that. The intermediate term vanishes because $f$ is compactly supported. – Sangchul Lee Sep 03 '16 at 23:52
  • Why did $f(x)$ need to be introduced? – whatwhatwhat Sep 03 '16 at 23:59
  • 2
    @whatwhatwhat What's your definition of $\delta$? –  Sep 03 '16 at 23:59
  • The one-dimensional Kronecker Delta: $\delta (x) = 0$ when $x \neq 0$ and $\delta (x) = 1$ when $x = 0$ – whatwhatwhat Sep 04 '16 at 00:01
  • 1
    @whatwhatwhat, the Kronecker delta is meaningfully only under summation: $ \sum_x f(x) \delta(x) = f(0) $. Since integrals are intuitively sums of infinitesimals, say $\sum_i f(x_i) \Delta x_i$, the corresponding version of $\delta$ must be adjusted so that $ \sum_i f(x_i)\delta(x_i) \Delta x_i = f(0)$. In other words, the Dirac delta must compensate the infinitesimal factor $\Delta x$. The issue is that, there is no legitimate function that do this job. So we have to consider a wider class of objects that generalizes functions, and distributions are exactly designed for this purpose. – Sangchul Lee Sep 04 '16 at 00:09
  • @whatwhatwhat no Kronecker delta here, it is the Dirac delta distribution – reuns Sep 04 '16 at 00:23
  • Screenshot from Griffiths, Introduction to Electrodynamics. This is what is confusing me. In the integrand, the author only has $f(x)\delta (x) dx$. Intuitively, I said that the $f(x)$ in the integrand must be my $x$. Yet in the answer above both $x$ and $f(x)$ are present. – whatwhatwhat Sep 04 '16 at 00:27
  • @whatwhatwhat If you can put any nice function $f(x)$, why can't you replace this by another nice function $xf(x)$ or something like $x^{2016} + 42f(999x + 1)$? – Sangchul Lee Sep 04 '16 at 00:34
  • 1
    nice discussion .. Dirac, Kronecker .. resembles old times philosophycal discussions: would be the case to fix the postulates first ? – G Cab Sep 04 '16 at 01:09
  • @SangchulLee true, but then that seems rather superfluous, doesn't it? What was wrong with $x$ by itself? – whatwhatwhat Sep 04 '16 at 01:22
  • 2
    @whatwhatwhat, Although one conventionally use function notation for the Dirac delta, it is not a function and thus its value at a point is meaningless. So, in general, information on distribution cannot be extracted pointwise by feeding a point as argument. Instead, we can use test functions to extract information. This also means that essentially the only way we manipulate and identify a distribution is to feed them any possible test functions (by the way which we call 'pairing') and see what happens. Zachary Selk's answer demonstrates how this can be done. – Sangchul Lee Sep 04 '16 at 02:13