# Integration over derivative of delta distribution

It is well known that $\delta'(x)\phi(x)=-\delta(x)\phi'(x)$ in a distributional sense, for some well behaved test function $\phi(x)$. However, what about the case:

$$\int_{-\infty}^\infty dz\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx \,\delta'(x+y+z)\phi(y)\psi(z)=~~???$$

Is this equal to zero? Or maybe it is equal to $\int_{-\infty}^\infty dz\int_{-\infty}^\infty dy\,(\phi'(y)\psi(z)+\phi(y)\psi'(z))$? Who knows?

• Yes, it's zero, since the function that is adjacent to the delta function is independent of x. – Paul Jul 19 '17 at 0:05
• For this case you need the co-area formula (as you would even without the derivative). – Ian Jul 19 '17 at 1:23
• The co-area formula says that to integrate over (say) $\mathbb{R}^n$ it is enough to integrate over the level sets $S_y$ of a function $f : \mathbb{R}^n \to \mathbb{R}$ (which are $(n-1)$-dimensional manifolds), and then integrate in $y$. The resulting change of variable needs a factor of $|\nabla f|$ in the denominator of the new integral. If you have a multivariable argument inside a delta function, simply take that to be your $f$; then only the level set $S_0$ will actually contribute to the integration but you will get the correct $(n-1)$-dimensional integral. – Ian Jul 19 '17 at 2:11
• Thus for example $\int_{-\infty}^\infty \int_{-\infty}^\infty g(x,y) \delta(x+y) dx dy = \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{g(x,z-x)}{\sqrt{2}} \delta(z) dz dx = \int_{-\infty}^\infty \frac{g(x,-x)}{\sqrt{2}} dx$ Perhaps it was obvious that only $g(x,-x)$ could possibly contribute to the integration, but it was probably not obvious that there should be this $\sqrt{2}$. – Ian Jul 19 '17 at 2:13
• I actually made a mistake in my example specifically: I forgot that in this case $dS(x)=\sqrt{2} dx$ so the factor is canceled. A better example: $\int_{-\infty}^\infty \int_{-\infty}^\infty g(x,y) \delta(x^2+y^2-1) dx dy = \int_0^{2 \pi} \frac{1}{2r} g(r,\theta) r dr d \theta$ (with the obvious abuse of notation about the two different $g$'s). One could also have seen this by directly changing to polar coordinates and then noting that (by the usual definition of composition involving a delta function) $\delta(r^2-1)=\frac{1}{2}(\delta(r-1)-\delta(r-1))$. – Ian Jul 19 '17 at 14:14

In this case using the co-area formula define the plane $P_c$ by $x+y+z=c$ and its surface measure by $dS_c$, then
To finish the problem you should choose a parametrization of the plane, compute the integral with $c$ as a parameter and then differentiate. I think that since there is no other dependence on $x$ that you will actually get zero (by parametrizing the plane by $(y,z)$ taking $x=c-y-z$).