Let [.] denote the greatest integer function. If $\int_\limits0^{e^3}\left[\frac{1}{e^{x-1}}\right] d x=\alpha-\log _e 2$, then $\alpha^3$ is equal to _________.

<p>To solve this, we start by evaluating the integral:</p> <p>$ I = \int_0^{e^3} \left[ \frac{1}{e^{x-1}} \right] \, dx $</p> <p>The greatest integer function $[\cdot]$ returns the largest integer less than or equal to the input value. Here's how we can approach the problem:</p> <p><p><strong>Determine the function inside the integral</strong>: </p> <p>$\frac{1}{e^{x-1}} = e^{1-x}$.</p></p> <p><p><strong>Identifying the intervals</strong>:</p></p> <p><p>When $e^{1-x} \geq 2$, which simplifies to $x \leq 1 - \ln 2$, we have $\left[e^{1-x}\right] = 2$.</p></p> <p><p>When $1 \leq e^{1-x} < 2$, simplifying gives $1 - \ln 2 < x \leq 1$, and thus $\left[e^{1-x}\right] = 1$.</p></p> <p><p>When $0 \leq e^{1-x} < 1$, which holds for $x > 1$, thus $\left[e^{1-x}\right] = 0$ from $x = 1$ to $x = e^3$.</p></p> <p><p><strong>Evaluate the integral on these intervals</strong>:</p> <p>$ \int_0^{1-\ln 2} 2 \, dx = 2(1-\ln 2) $ </p> <p>$ \int_{1-\ln 2}^1 1 \, dx = 1 - (1 - \ln 2) = \ln 2 $ </p> <p>$ \int_1^{e^3} 0 \, dx = 0 $</p></p> <p><p><strong>Combine these results</strong>:</p> <p>$ I = 2(1-\ln 2) + \ln 2 + 0 = 2 - \ln 2 $</p></p> <p>Thus, we are given that:</p> <p>$ \alpha - \ln 2 = 2 - \ln 2 $</p> <p>This implies that:</p> <p>$ \alpha = 2 $</p> <p>Therefore, $\alpha^3 = 2^3 = 8$.</p>