# Prove that $x\frac{d}{dx} \delta (x) = -\delta (x)$

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,

$$0=0$$

• 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: $-\delta(x)$! Sep 3 '16 at 23:53
• 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 Sep 4 '16 at 0:17
• @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. Sep 4 '16 at 0:32
• @SangchulLee $Generalized\ Functions$ were introduced, in 1935, by the russian Sergei L. Sobolev who WAS a mathematician. Sep 4 '16 at 4:08
• @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. Sep 4 '16 at 5:28

Let $f$ be any smooth compactly supported function. Then:
$$\int x\delta'(x)f(x)\ dx=\int\delta'(x) xf(x)\ dx=-\int\delta(x)(xf'(x)+f(x))\ dx=-0f'(0)-f(0)=-f(0)$$
• @whatwhatwhat Integration by parts gives that. The intermediate term vanishes because $f$ is compactly supported. Sep 3 '16 at 23:52
• @whatwhatwhat What's your definition of $\delta$?
• @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. Sep 4 '16 at 0:09