Problem Analysis: Definite Integral with Greatest Integer Function

The problem asks us to evaluate the definite integral $\int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx$. We are given that the value of this integral is equal to $\alpha - \log_e 2$, and our goal is to find the value of $\alpha^3$. The notation $\left[ \cdot \right]$ represents the greatest integer function.

Evaluating the Integrand $ \left[e^{1-x}\right] $

The function inside the integral is $f(x) = \left[\frac{1}{e^{x-1}}\right] = \left[e^{1-x}\right]$. We need to understand how the value of this function changes within the integration limits, from $x=0$ to $x=e^3$.

• When $x=0$, the value is $e^{1-0} = e \approx 2.718$. Thus, $\left[e^{1-0}\right] = 2$.
• When $x=e^3$, the value is $e^{1-e^3}$, which is a small positive number slightly greater than 0. Thus, $\left[e^{1-e^3}\right] = 0$.
• The greatest integer function $\left[e^{1-x}\right]$ changes its value only when $e^{1-x}$ becomes an integer. Let $e^{1-x} = k$, where $k$ is an integer. Taking the natural logarithm on both sides gives $1-x = \log_e k$. Rearranging this, we get the value of $x$ where the function changes: $x = 1 - \log_e k$.

Determining Integration Intervals

As $x$ increases from 0 to $e^3$, $e^{1-x}$ decreases from $e$ to $e^{1-e^3}$. The integer values that $e^{1-x}$ can take are 2, 1, and 0.

• Interval for $\left[e^{1-x}\right] = 2$: This happens when $2 \le e^{1-x} < 3$. Taking logarithms: $\log_e 2 \le 1-x < \log_e 3$. Solving for $x$: $1 - \log_e 3 < x \le 1 - \log_e 2$. Within the integration range $[0, e^3]$, this corresponds to the interval $[0, 1 - \log_e 2]$.
• Interval for $\left[e^{1-x}\right] = 1$: This happens when $1 \le e^{1-x} < 2$. Taking logarithms: $\log_e 1 \le 1-x < \log_e 2$, which means $0 \le 1-x < \log_e 2$. Solving for $x$: $1 - \log_e 2 < x \le 1$. This corresponds to the interval $(1 - \log_e 2, 1]$.
• Interval for $\left[e^{1-x}\right] = 0$: This happens when $0 < e^{1-x} < 1$. Taking logarithms: $1-x < \log_e 1$, which means $1-x < 0$, or $x > 1$. This corresponds to the interval $(1, e^3]$.

Calculating the Definite Integral

We split the integral into parts based on the intervals identified:

$ \int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \int_{0}^{1-\log_e 2} 2 \, dx + \int_{1-\log_e 2}^{1} 1 \, dx + \int_{1}^{e^3} 0 \, dx $

Now, we evaluate each integral separately:

• $ \int_{0}^{1-\log_e 2} 2 \, dx = 2 \times \left[ x \right]_{0}^{1-\log_e 2} = 2 \times (1 - \log_e 2 - 0) = 2 - 2\log_e 2 $
• $ \int_{1-\log_e 2}^{1} 1 \, dx = 1 \times \left[ x \right]_{1-\log_e 2}^{1} = 1 \times (1 - (1 - \log_e 2)) = \log_e 2 $
• $ \int_{1}^{e^3} 0 \, dx = 0 $

Adding the results from the three intervals:

$ \text{Total Integral Value} = (2 - 2\log_e 2) + (\log_e 2) + 0 = 2 - \log_e 2 $

Finding the Value of $\alpha^3$

The problem states that the integral equals $\alpha - \log_e 2$. We equate our calculated value to this expression:

$ 2 - \log_e 2 = \alpha - \log_e 2 $

Comparing both sides, we find:

$ \alpha = 2 $

Finally, we compute $\alpha^3$:

$ \alpha^3 = 2^3 = 8 $

Therefore, the value of $\alpha^3$ is 8.