Suppose the functions $e(x)$ and $h(x)$ with units $\left[\frac{V}{m}\right]$ and $\left[\frac{A}{m}\right]$ and their Fourier transforms $E(k)$ and $H(k)$ with units $\left[V\right]$ and $\left[A\right]$. The time avaverage A of $e(x)h(x)$ in function of their Fourier transforms $E(k)$ and $H(k)$ can be written as \begin{align} A &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} e(x)h(x) dx \\ &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} \int\limits_{-\infty}^{\infty} E(k)e^{ikx} dk \int\limits_{-\infty}^{\infty} H(k')e^{ik'x} dk' dx \\ &= \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(k) H(k')\left[ \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X}e^{i(k+k')x} dx \right] dk dk' \\ &= 2 \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(k) H(k')\delta(k'+k) dk dk' \\ &= 2 \int\limits_{-\infty}^{\infty} E(k) H(-k)dk \\ \end{align}
The question: the unit of the time average A is $\left[\frac{VA}{m^2}\right]$ whereas the unit of the final result is $\left[\frac{VA}{m}\right]$. How can this be and how to interpret it?
Example: if you know the Fourier transforms $E(k)$ and $H(k)$, and you want to calculate the average A, then you will use $A=2 \int\limits_{-\infty}^{\infty} E(k) H(-k)dk$ but this will give another unit as expected? So, then you have the wrong result?
Your error is in the integral converging to the Dirac delta. Normally it should be $$\lim_{X\to\infty} \int_{-X}^X e^{i(k+k')x}\,\mathrm d x = \delta_{k+k'}$$ so without dividing by $X$. This is at this step that you are loosing a $m$.