Volume of region $\{ (x_1,\ldots,x_n)\in \mathbf{R}^n_{>0} \mid \prod_{i=1}^n\max(1,x_i)\leqslant d \}$

Indeed the two representations are equivalent. Substituting $x_i = e^{y_i}$, we note that

$$ \prod_{i=1}^{n} \max\{1,x_i\} < d \quad\Leftrightarrow\quad \prod_{i=1}^{n} \max\{1,e^{y_i}\} < d \quad\Leftrightarrow\quad \sum_{i=1}^{n} \max\{0,y_i\} < \log d. $$

Also, since $\mathrm{d}x_i = e^{y_i} \, \mathrm{d} y_i$, we get

\begin{align*} \operatorname{Vol}(E_n) &= \int_{\mathbb{R}_{>0}^n} \mathbf{1}_{E_n} \, \mathrm{d}x_1\cdots\mathrm{d}x_n \\ &= \int_{\mathbb{R}^n} \mathbf{1}_{\{ \sum_{i=1}^{n} \max\{0,y_i\} < \log d\}} e^{y_1+\dots+y_n} \, \mathrm{d}y_1\cdots\mathrm{d}y_n \end{align*}

From now on, we will assume $d > 1$ for an obvious reason. Write $[n] = \{1,\dots,n\}$ and set

$$ F(I) = \int_{A_I} \mathbf{1}_{\{ \sum_{i=1}^{n} \max\{0,y_i\} < \log d\}} e^{y_1+\dots+y_n} \, \mathrm{d}y_1\cdots\mathrm{d}y_n, $$

where $A_I = \bigcap_{i \in I} \{y_i \geq 0\} \cap \bigcap_{i \notin I} \{y_i < 0\} $ is the set of points $(y_1, \dots, y_n)$ such that $y_i \geq 0$ if and only if $i \in I$. Writing $k = \left| I \right|$ for the size of the set $I$, it is easy to find that the value of $F(I)$ depends only on $k$. In particular, if $k \geq 1$, then

\begin{align*} F(I) = F([k]) &= \left( \int_{-\infty}^{0} e^{y} \, \mathrm{d}y \right)^{n-k} \left( \int_{[0,\infty)^k} \mathbf{1}_{\{ \sum_{i=1}^{k} y_i \leq \log d \}} e^{y_1 + \dots + y_k} \, \mathrm{d}y_1 \dots \mathrm{d}y_k. \right) \\ &= \int_{[0,\infty)^k} \mathbf{1}_{\{ \sum_{i=1}^{k} y_i \leq \log d \}} e^{y_1 + \dots + y_k} \, \mathrm{d}y_1 \dots \mathrm{d}y_k. \end{align*}

Substituting $(z_1, \dots, z_k) = (y_1, y_1+y_2, \dots, y_1+\dots+y_k)$, the region $[0, \infty)^k$ in $y$-domain transforms to $\{(z_1, \dots, z_k) : 0 < z_1 < z_2 < \dots < z_k \}$, and so,

\begin{align*} F([k]) &= \int_{0}^{\log d}\int_{0}^{z_{k}} \dots \int_{0}^{z_2} e^{z_k} \, \mathrm{d}z_1 \dots \mathrm{d}z_k \\ &= \int_{0}^{\log d} \frac{z_k^{k-1}}{(k-1)!} e^{z_k} \, \mathrm{d}z_k. \end{align*}

If in addition $k \geq 2$, then integration by parts shows that

\begin{align*} F([k]) &= \frac{d (\log d)^{k-1}}{(k-1)!} - F([k-1]). \end{align*}

Writing $[0] = \varnothing$ for convenience and noting that $F(\varnothing) = 1$, we find that this formula also extends to $ k = 1$. Therefore

\begin{align*} \operatorname{Vol}(E_n) &= \sum_{I \subseteq [n]} F(I) \\ &= \sum_{k=0}^{n} \binom{n}{k} F([k]) \\ &= \sum_{k=0}^{n} \left[ \binom{n-1}{k} + \binom{n-1}{k-1} \right] F([k]) \\ &= \sum_{k=0}^{n-1} \binom{n-1}{k} F([k]) + \sum_{k=0}^{n-1} \binom{n-1}{k} F([k+1]) \\ &= \sum_{k=0}^{n-1} \binom{n-1}{k} \frac{d (\log d)^{k}}{k!}. \end{align*}