We all know that for Joule losses \$P_j = V(t)I(t)\$, and the integral form is \$P_j=\iint \rho j^2 \, dV \$.
When trying to do some integrals, I found this expression in a document:
$$P_i = \frac{1}{2 \sigma} \iint_S \pmb{J}_i \cdot \pmb{J}_i^* \, dS $$
I don't understand where the 1/2 factor comes from.
I don't know what the subscript \$i\$ is referring to here, so I will ignore it for now.
Short Hand-Waving Answer
The \$\mathbf{J}\$ are amplitudes of harmonic components in this expression, meaning they multiply terms like \$cos(\omega t)\$. When you take the time average of a the \$cos^2\$, you get a factor of \$\frac{1}{2}\$.
More Formal Answer
Factors such as \$\frac{1}{2}\$ or \$\pi\$ often depend upon the conventions used. One common representation used in derivations of energy transfer is to represent harmonic quantities such as the electric field \$\mathcal{E}(x,y,z,t)\$ and the current density \$\mathcal{J}(x,y,z,t)\$ like this:
$$ \mathbf{\mathcal{J}}(x,y,z,t) = Re[\mathbf{J}(x,y,z)e^{j\omega t}] \\ \mathbf{\mathcal{E}}(x,y,z,t) = Re[\mathbf{E}(x,y,z)e^{j\omega t}] $$
\$\mathbf{J}\$ and \$\mathbf{E}\$ are complex quantities here that allow you to represent both magnitude and phase information. If you do this for \$\mathcal{E}\$ and \$\mathcal{J}\$ you find
$$ <P> = <\mathcal{E}(t)\cdot\mathcal{J}(t)> = \frac{1}{\sigma}<\mathcal{J}(t)\cdot\mathcal{J}(t)> \\= <\frac{1}{\sigma}Re[\mathbf{J}(x,y,z)e^{j\omega t}]\cdot Re[\mathbf{J}(x,y,z)e^{j\omega t}]>, $$
where the brackets \$<>\$ denote the time average, which is taken over a duration that is long compared to one period of the harmonic variation, but short enough so that the amplitude is relatively constant. I am assuming \$\sigma\$ is real. It doesn't have to be, but based on the resulting formula, I believe your reference is making that assumption.
For any complex number \$z\$, the real part can be obtained with \$Re[z] = \frac{1}{2}(z+z^*)\$. Using \$Re[\mathbf{J}(x,y,z)e^{j\omega t}]\$ = \$\frac{1}{2}[\mathbf{J}(x,y,z)e^{j\omega t} + \mathbf{J^*}(x,y,z)e^{-j\omega t}]\$, we find
$$ <P> = \frac{1}{\sigma}<\frac{1}{2}(\mathbf{J}e^{j\omega t}+\mathbf{J^*}e^{-j\omega t})\cdot\frac{1}{2}(\mathbf{J}e^{j\omega t}+\mathbf{J^*}e^{-j\omega t})>\\ =\frac{1}{\sigma}\frac{1}{4}<[\mathbf{J}\cdot\mathbf{J^*}+\mathbf{J^*}\cdot\mathbf{J} + \mathbf{J}\cdot\mathbf{J}e^{2j\omega t} + \mathbf{J^*}\cdot\mathbf{J^*}e^{-2j\omega t}]> $$
Here I am omitting the spatial dependence arguments for convenience.
The terms with exponentials have a zero time average (they represent reactive power), and we are left with
$$ \\ <P> = \frac{1}{2\sigma}\mathbf{J}\cdot\mathbf{J^*} \\ $$
Perform the spatial integral to get your equation. Note also that this is the power at one frequency only, so \$\mathbf{J}\$ really means \$\mathbf{J}(x,y,z,\omega)\$ using the representation at the beginning of this post for representing the harmonic quantities.
