# Why does this simple integral in electromagnetism involving the delta function seem to both vanish and not vanish depending on how you evaluate it?

This relates to David Tong's notes on electromagnetism, specifically page 18.

We have as our charge density a delta function:

$$\rho(\vec x)=Q\delta^{(3)}(\vec x).$$

We want to solve:

$$\nabla^2\phi=-\frac{Q}{\epsilon_0}\delta^{(3)}(\vec x) \tag{2.14}.$$

To do this we postulate that:

$$\phi=\frac{\alpha}{r},$$

solves $$\nabla^2\phi=0$$, and after some vector calculus and some algebra we find that it does. However, next we use Gauss' law (with $$S$$ a sphere centred on the delta function) and find that:

$$\int_S\nabla\phi\cdot d\vec S=Q/\epsilon_0,$$

which, after working through the calculation and inserting our equation for $$\phi$$ leads to:

$$\phi=\frac{Q}{4\pi r\epsilon_0} \tag{2.16}.$$

I see how we do this, but, if I apply the divergence theorem to the above integral, I find:

$$\int_V \nabla^2\phi \text{ }dV=0,$$

since, as already established, $$\nabla^2\phi=0$$? What is going on here?

Function

$$\phi = \frac{\alpha}{r}$$

obeys the Laplace equation

$$\Delta \phi = 0$$

everywhere except at at $$\vec{x} = \vec{0}$$. At $$\vec{0}$$, $$\phi$$ has a singularity and so does $$\Delta \phi$$.

So when we integrate $$\Delta \phi$$ we cannot just put in $$\Delta \phi = \vec{0}$$. That would ignore the singularity.

The Gauss divergence theorem does not work if we restrict our considerations to cases where $$\Delta \phi$$ is a function.

But it works again if we define $$\Delta\phi$$ to be a distribution. The correct expression to put in is $$\Delta \phi = -\frac{Q}{\epsilon_0}\delta^{(3)}(\vec{x})$$ as assumed in 2.14.

• Solved, thank you very much! Commented Jan 3, 2021 at 14:07