In Griffith's Introduction to Electrostatics, International 4th ed, pg-94 these two equations are given for calculating energy of continous charge distributions:

$$ W= \frac12 \int \rho V d \tau \tag{1}$$

And after some simplifications, this other equation is given $$ W= \frac{\epsilon_o}{2} \int_{whole space} \vec{E}^2 d \tau \tag{2}$$

And, there are also some exercises to find the energy of conductors using equation (2) but after some careful thought, I realize that in practice, equation (2) may have to integrate over a boundary. This is problem because the component of electric field normal to a surface is discontinuous by an amount $\frac{\sigma}{\epsilon}$. Now, since we have a discontinuity at the boundary, how can is the expression integrable?(*)

As far as I learned, discontinuous functions such as $ \frac{1}{x}$ etc are not integrable on a set containing their discontinuity.

On some reflection, I started realize voltage of a conductor maybe suffering from the same problem because when we integrate from infinity to a point inside a conductor, there is a sudden discontinuity of the electric field at the boundary again. Hence, (1) also suffers the same problem.


1 Answer 1


A discontinuity is not a problem for the integrability of a function. The reason $\frac{1}{x}$ cannot be integrated over $x=0$ is because the integral diverges, but this is related to the fact that $|1/x| \rightarrow \infty$ as $x \rightarrow 0$, not because it is discontinuous. So to answer your question; the integral is in fact well-defined, even though the field may be discontinuous at the boundary.

  • $\begingroup$ If I got this right, generally speaking, integrability and discontinuity are disjoint mathematical properties of a function? $\endgroup$
    – Babu
    Commented May 26, 2021 at 11:12
  • $\begingroup$ I am not 100% sure what you mean by disjoint, but they are definitely different properties and one does not imply (or exclude) the other. However, a function must be continuous at a point in order to be differentiable at that point. I don't know if maybe this is what you were thinking of? $\endgroup$
    – Jakob KS
    Commented May 26, 2021 at 11:26
  • $\begingroup$ In fact a divergent integrand is compatible with a finite integral, e.g. $\int_0^1x^{-1/2}dx=2$. $\endgroup$
    – J.G.
    Commented May 26, 2021 at 11:46

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