From $e^{-iET} = e^{-\frac{1}{2}\mathrm{Tr}\log(\partial^2+m^2)},$ we have
$\begin{align} iET & = \frac{1}{2}\int d^4x \int \frac{d^4k}{(2\pi)^4}\int\frac{d^4q}{(2\pi)^4} e^{i(k-q)x} \langle k \vert \log(\partial^2 + m^2) \vert q \rangle \ &=\frac{1}{2}\int d^4x \int \frac{d^4k}{(2\pi)^4}\int\frac{d^4q}{(2\pi)^4} e^{i(k-q)x}\log(-k^2 + m^2) \langle k\vert q \rangle \ &=\frac{1}{2}\int d^4x \int \frac{d^4k}{(2\pi)^4}\int\frac{d^4q}{(2\pi)^4}e^{i(k-q)x}\log(-k^2 + m^2)\int d^4x'\,\langle k \vert x'\rangle\langle x'\vert q \rangle \ &=\frac{1}{2}\int d^4x \int \frac{d^4k}{(2\pi)^4}\int\frac{d^4q}{(2\pi)^4}e^{i(k-q)x}\log(-k^2 + m^2)\int d^4x'\,e^{i(q-k)x'} \ &=\frac{1}{2}\int d^4x \int \frac{d^4k}{(2\pi)^4}\int\frac{d^4q}{(2\pi)^4}e^{i(k-q)x}\log(-k^2 + m^2)(2\pi)^4\delta(q-k) \ &=\frac{1}{2}\int d^4x \int \frac{d^4k}{(2\pi)^4}\log(-k^2 + m^2)\int d^4q\,e^{i(k-q)x}\delta(k-q) \ &=\frac{1}{2}\int d^4x\int\frac{d^4k}{(2\pi)^4}\log(-k^2+m^2) \end{align}$
Now we have to do three things:
* Identify$^1$ $\int d^4x$ with $VT$;
* Change $\log(-k^2+m^2)$ to $\log(k^2-m^2)$ using$^2$ $\log(x) = \log(|x|) + i\pi$, for $x<0$;
* Make the substitution$^3$ $m^2 \rightarrow m^2-i\varepsilon$ so that we don't have to fight an infinity at $k^2 = m^2$.
We get
$\begin{align} iET &= \frac{1}{2}\left(\int d^4x\right)\int\frac{d^4k}{(2\pi)^4}\log(-k^2+m^2) \ &=\frac{1}{2}VT\int\frac{d^4k}{(2\pi)^4}\log(-k^2+m^2) \tag{1} \label{1} \ &=\frac{1}{2}VT\int\frac{d^4k}{(2\pi)^4}\log(k^2-m^2+i\varepsilon) + A \end{align}$
Where we have absorbed$^4$ $(i\pi/2)VT\int(d^4k/(2\pi)^4)$ into $A$ (Which comes from $C$, as remarked by <PERSON>).
$^1$Not the top standard of rigor, but <PERSON> does a similar step earlier in his book (page 28).
$^2$See this math.SE question for a proof of the identity.
$^3$As $\varepsilon$ is going to be set to zero later, there is no need to worry about the substitution. 592