Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function, denoted by $[x]$, gives the largest integer less than or equal to $x$.
• Definite Integral of a Step Function: To integrate a function involving the greatest integer function, $[f(x)]$, we identify the intervals where $f(x)$ crosses integer values. The integral is then split into sub-integrals over these intervals, where $[f(x)]$ takes a constant integer value.
• Integral of a Constant: $\int_a^b c \, dx = c(b-a)$.

Step-by-Step Solution

Step 1: Analyze the Integrand and the Integration Interval

The integral is given by I = \int_\limits0^{e^3}\left[\frac{1}{e^{x-1}}\right] d x. The function inside the greatest integer bracket is $f(x) = \frac{1}{e^{x-1}} = e^{-(x-1)} = e^{1-x}$. We need to analyze the behavior of $f(x) = e^{1-x}$ over the interval $[0, e^3]$.

• At the lower limit, $x=0$: $f(0) = e^{1-0} = e^1 = e \approx 2.718$.
• At the upper limit, $x=e^3$: $f(e^3) = e^{1-e^3}$. Since $e^3$ is a large positive number, $1-e^3$ is a large negative number, so $f(e^3)$ is a very small positive number, close to 0.
• Since the exponent $1-x$ is a decreasing function of $x$, $f(x) = e^{1-x}$ is also a decreasing function. Thus, as $x$ ranges from $0$ to $e^3$, $e^{1-x}$ decreases from approximately $2.718$ to a value close to $0$. The integer values that $[e^{1-x}]$ can take are $2, 1, 0$.

Why this step is taken: Understanding the range and monotonicity of the function inside the greatest integer bracket is crucial for determining the integer values it will take and identifying the points where these values change.

Step 2: Determine the Critical Points where the Integrand Changes Value

The value of $[e^{1-x}]$ changes when $e^{1-x}$ crosses an integer. We set $e^{1-x} = k$, where $k$ is an integer. Taking the natural logarithm of both sides: $1-x = \ln k$ $x = 1 - \ln k$ We need to find the values of $x$ within the interval $[0, e^3]$ where $[e^{1-x}]$ changes its integer value.

1. For $[e^{1-x}] = 2$: This occurs when $2 \le e^{1-x} < 3$. The upper bound $e^{1-x} < 3$ is always satisfied because at $x=0$, $e^{1-x}=e \approx 2.718 < 3$. The lower bound is $e^{1-x} \ge 2$. This implies $1-x \ge \ln 2$, so $x \le 1 - \ln 2$. Thus, for $x \in [0, 1 - \ln 2]$, $[e^{1-x}] = 2$. Note that $1-\ln 2 \approx 1-0.693 = 0.307$, which is within $[0, e^3]$.

2. For $[e^{1-x}] = 1$: This occurs when $1 \le e^{1-x} < 2$. The upper bound $e^{1-x} < 2$ implies $1-x < \ln 2$, so $x > 1 - \ln 2$. The lower bound $e^{1-x} \ge 1$ implies $1-x \ge \ln 1 = 0$, so $x \le 1$. Thus, for $x \in (1 - \ln 2, 1]$, $[e^{1-x}] = 1$.

3. For $[e^{1-x}] = 0$: This occurs when $0 \le e^{1-x} < 1$. The condition $e^{1-x} < 1$ implies $1-x < \ln 1 = 0$, so $x > 1$. Since $e^{1-x}$ is decreasing and $f(e^3) > 0$, this condition holds for $x \in (1, e^3]$. Thus, for $x \in (1, e^3]$, $[e^{1-x}] = 0$.

Why this step is taken: Identifying these critical points ($x = 1-\ln 2$ and $x=1$) allows us to divide the integration interval into sub-intervals where the greatest integer function has a constant value.

Step 3: Split the Integral Based on Critical Points

We split the integral $I$ into three parts corresponding to the intervals where $[e^{1-x}]$ is constant: $I = \int_0^{1-\ln 2} \left[e^{1-x}\right] dx + \int_{1-\ln 2}^1 \left[e^{1-x}\right] dx + \int_1^{e^3} \left[e^{1-x}\right] dx$ Substituting the constant integer values determined in Step 2: $I = \int_0^{1-\ln 2} 2 \, dx + \int_{1-\ln 2}^1 1 \, dx + \int_1^{e^3} 0 \, dx$

Why this step is taken: This is the core technique for integrating step functions. By splitting the integral, we transform the problem into integrating simple constant functions over specific intervals.

Step 4: Evaluate Each Sub-integral

Now, we evaluate each integral:

1. First integral: $\int_0^{1-\ln 2} 2 \, dx = 2[x]_0^{1-\ln 2} = 2((1-\ln 2) - 0) = 2(1-\ln 2) = 2 - 2\ln 2$

2. Second integral: $\int_{1-\ln 2}^1 1 \, dx = [x]_{1-\ln 2}^1 = 1 - (1-\ln 2) = 1 - 1 + \ln 2 = \ln 2$

3. Third integral: $\int_1^{e^3} 0 \, dx = 0$

Why this step is taken: Evaluating these simple integrals directly yields the components of the total definite integral.

Step 5: Sum the Results and Find $\alpha$

Summing the results from Step 4 gives the total value of the integral $I$: $I = (2 - 2\ln 2) + (\ln 2) + 0$ $I = 2 - \ln 2$ We are given that $I = \alpha - \log_e 2$. Comparing our result with the given form: $2 - \ln 2 = \alpha - \log_e 2$ Since $\log_e 2 = \ln 2$, we have: $2 - \ln 2 = \alpha - \ln 2$ This implies $\alpha = 2$.

Why this step is taken: By equating our calculated integral value with the given expression, we can solve for the unknown constant $\alpha$.

Step 6: Calculate $\alpha^3$

We found that $\alpha = 2$. We need to calculate $\alpha^3$. $\alpha^3 = 2^3 = 8$

Why this step is taken: This is the final calculation required by the problem statement.

Common Mistakes & Tips

• Incorrectly identifying integer values: Ensure you correctly determine the range of $[f(x)]$ and the points where it changes. For $e^{1-x}$, the starting value is $e \approx 2.718$, so the first integer it crosses downwards is 2.
• Algebraic errors with logarithms: Be careful when solving equations involving logarithms and exponents, especially when isolating $x$.
• Forgetting the upper limit of integration: In this case, the integral from $1$ to $e^3$ involves $[e^{1-x}] = 0$, making its contribution zero, but it's important to acknowledge the full integration range.

Summary

The problem required evaluating a definite integral involving the greatest integer function. We first analyzed the behavior of the integrand $e^{1-x}$ over the given interval $[0, e^3]$. By identifying the points where $e^{1-x}$ crossed integer values (specifically at $x = 1-\ln 2$ and $x=1$), we split the integral into three sub-intervals. We then replaced $[e^{1-x}]$ with its constant integer value in each sub-interval and evaluated the resulting simple integrals. Summing these values gave us the total integral $I = 2 - \ln 2$. Comparing this with the given form $I = \alpha - \log_e 2$, we found $\alpha = 2$. Finally, we calculated $\alpha^3 = 2^3 = 8$.

The final answer is $\boxed{8}$.