Suppose the functions $E(k)$ and $H(k')$ with units $[V]$ and $[A]$ and the unit of $k$ and $k'$ being $\left[\frac{1}{m}\right]$. Using the Kronecker delta function $\delta_{i,j}$ we can write $$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} E(k)H(k') \delta_{k,-k'} dk dk' = \int\limits_{-\infty}^{\infty} E(k)H(-k) dk$$
If now the units on both sides of the equation are compared, different results are obtained. The left side has unit $ \left[V A \frac{1}{m}\frac{1}{m} = \frac{VA}{m^2}\right]$ whereas the right side has $\left[ \frac{VA}{m} \right]$. How can this be explained?
The Kronecker delta function can be defined as $$\delta_{k,-k'} = \lim\limits_{X\to\infty} \frac{1}{X}\int\limits_{-X}^{X} e^{i(k+k')x}dx $$ and is thus dimensionless.
If $k$ has units $[ \frac{1}{m}]$ then the kroncker delta for k has units $[m]$. note that for any function f, $\int_{\mathbb{R}} f(x)\delta(x)dx = f(0)$ so if $dx$ has units of length and $f$ is, say, unitless, see that $\delta(x)$ must have units of $\frac{1}{length}$. Your intuition could be that under a change of units $x \to \alpha x$, we get $\delta(\alpha x) = \frac{1}{| \alpha |} \delta(x)$