Why is the first step in the Lovasz-Vempala convex volume algorithm correct?
It is also unclear to me what $\epsilon$ means in that line of the paper.
The approximation error for $Z(a_0)$ is never mentioned again;
my interpretation is that the authors are saying the error is so small it can
be ignored.

The bound I get is:
$$ \tag{*} Z(a_0)=Z(2n) \geq (1-(2/e)^n \mathrm{poly}(n)) \int_0^\infty e^{-2nt} t^n \pi_n dt.$$

So the approximation should be ok for $n &gt; 100 \log \epsilon^{-1}$ and large $n$.
I don't know if it is necessary to use a different analysis or algorithm
to get Theorem 1.1 as stated for the regime where
$\epsilon$ is exponentially small compared to $n$.

To show (*), I claim the stronger statement that
$\int_1^\infty e^{-at}t^n dt \leq e^{-a}/n$
for all $a\geq 2n$ - this can be shown by induction on $n$.
Using Stirling's approximation on $\int_0^\infty e^{-2nt}t^n dt=n!(2n)^{-(n+1)}$,
$$ \frac{\int_1^\infty e^{-2nt}t^n dt}
{\int_0^\infty e^{-2nt}t^n dt}
\leq \frac{e^{-2n}(2n)^n}{n^n e^{-n}} \mathrm{poly}(n)
= (2/e)^n \mathrm{poly}(n).
$$