Vacuum energy of a free scalar field from path integral

My question has been asked two other times:
Spinor vacuum energy (misleading title) and
Vacuum Energy Calculation using Path Integral. I am not completely satisfied with the answers and it looks like they both have errors in their algebraic steps. Since it has been asked twice, I hope you will look at my VERY DETAILED question which contains and exceeds the clarifications sought in these other questions.

I am using Zee’s QFT book and he skipped over far too many steps in section II.5. My questions are about the missing steps. They are in bold below. (I think I will read through Zee’s section III and then switch to a non-nutshell book but in the meantime I am working through it.) Let $varphi$ be a scalar field with ground state $|0rangle$. We have, by identity, the energy of the vacuum $E_{text{vac}}$ as

$$ Z=langle 0|e^{-ihat H T} |0rangle=e^{-iE_{text{vac}}T} $$

and we want to determine exactly what $E_{text{vac}}$ is. We also let the time $Ttoinfty$ so our integrals are over all of spacetime. We write out $Z$ as the generating functional

$$ Z=int Dvarphi e^{ iint d^4xfrac{1}{2}[(partialvarphi)^2-m^2varphi^2 ] } .$$

By a standard Gaussian identity and a magical procedure for "discretizing" infinite dimensional path integrals, and for some "non-essential" stuff $C$, we obtain

$$ Z=Cleft( frac{1}{det [partial^2+m^2]} right) =Ce^{ -frac{1}{2}text{Tr}log(partial^2+m^2) } .$$

Therefore, setting the exponentials equal, the energy of the vacuum has the form

$$ iE_{text{vac}}T varphi= frac{1}{2}text{Tr}log(partial^2+m^2)varphi . $$

(Since $C$ has exponential dependence, this gives the additional energy $A$ obtained below.) Now this is where Zee skips some steps. He writes

$$ text{Tr} log(partial^2+m^2)=int !d^4x,langle x| log(partial^2+m^2)|xrangle . $$

Is this an identity for the trace? I kind of see that by the orthogonality of $|xrangle$ and $|yrangle$, we will only pick out the diagonal elements of the operator but he introduces this formula from nowhere. He proceeds to solve the integral inserting the identity twice as

$$ text{Tr} log(partial^2+m^2)= int!d^4xint!frac{d^4k}{(2pi)^4} int!frac{d^4q}{(2pi)^4} langle x|kranglelangle k| log(partial^2+m^2) |qranglelangle q| xrangle. $$

What is $q$? Is it momentum written as a second dummy variable akin to $(k,q)sim (k_1,k_2)$? As if by magic, Zee uses "we obtain" to write

$$ iE_{text{vac}}T =frac{1}{2} VTint!frac{d^4k}{(2pi)^4} log(k^2-m^2+ivarepsilon) +A $$

WHAT HAPPENED HERE? (How did he know to insert the identity two times?!?!) I see we get $VT$ from $int d^4x$, kind of. I see the $ivarepsilon$ appeared magically in the usual way. I don’t see what else happened there. Both of the above linked previous questions (Spinor vacuum energy and
Vacuum Energy Calculation using Path Integral) try to explain this, but I am not satisfied and I will begin my own computation. Assuming the trace identity, we have

begin{align}
iE_{text{vac}}T&=frac{1}{2} int!d^4xint!frac{d^4k}{(2pi)^4} int!frac{d^4q}{(2pi)^4} langle x|kranglelangle k| log(partial^2+m^2) |qranglelangle q| xrangle.
end{align}

Use $langle x| krangle=e^{ikx}$, $langle q| xrangle=e^{-iqx}$, and $-ipartial|qrangle=q|qrangle$ to obtain

begin{align}
&=frac{1}{2} int!d^4xint!frac{d^4k}{(2pi)^4} int!frac{d^4q}{(2pi)^4} e^{ix(k-q)}log(-q^2+m^2) langle k |qrangle .
end{align}

Now I use

$$delta(k-q)=int frac{d^4x}{(2pi)^4}e^{ix(k-q)}$$

to obtain

begin{align}
&=frac{1}{2} int!frac{d^4k}{(2pi)^4} int !d^4!q,delta(k-q)log(-q^2+m^2) langle k |qrangle
&=frac{1}{2} int!frac{d^4k}{(2pi)^4} log(-k^2+m^2) langle k |krangle
&=frac{1}{2} int!frac{d^4k}{(2pi)^4} log(-k^2+m^2) .
end{align}

If I proceed here, I do not get the correct answer. Even if I add $ivarepsilon$ and use an identity for the complex logarithm, there’s no way I could get $VT$. The steps are worked out most clearly in Spinor vacuum energy, but I do not like what he has done. For instance, his partial operator should have acted to the right to return $q$ but he has acted to the left to obtain $k$. Seems like he messed up a factor of $(2pi)^4$ as well. Mostly my question is about why he delayed the creation of the Dirac delta until after the insertion of a third resolution of the identity.