Let ⌊·⌋ denote the greatest integer function. If $$\displaystyle\int_{0}^{e^3} \Bigl\lfloor\frac{1}{e^{x-1}}\Bigr\rfloor\,dx = \alpha - \ln 2,$$ then $$\alpha^3$$ is equal to:

Step 1: Simplify the integrand

The given integral involves the greatest integer function of $$1/e^{x-1}$$ . We can simplify the expression inside the greatest integer function by rewriting $$1/e^{x-1}$$ as $$e^{-(x-1)}$$ , which is equal to $$e^{1-x}$$ . This makes the expression easier to work with.

Step 2: Determine the range of $$e^{1-x}$$

To evaluate the greatest integer function, we need to understand the range of values that $$e^{1-x}$$ takes over the integration interval $$[0, e^3]$$ . As $$x$$ goes from $$0$$ to $$e^3$$ , $$1-x$$ goes from $$1$$ to $$1-e^3$$ . Consequently, $$e^{1-x}$$ goes from $$e^1$$ down to $$e^{1-e^3}$$ . Since $$e \approx 2.718$$ , $$e^3 \approx 20.08$$ , so $$1-e^3 \approx -19.08$$ . Thus, $$e^{1-e^3}$$ is a very small positive number, approximately $$e^{-19.08}$$ .

Step 3: Identify integer values of $$\lfloor e^{1-x} \rfloor$$

The greatest integer function $$\lfloor e^{1-x} \rfloor$$ will take integer values. Since $$e^{1-x}$$ ranges from $$e^{1-e^3}$$ (a value between 0 and 1) up to $$e$$ (approximately 2.718), the possible integer values for $$\lfloor e^{1-x} \rfloor$$ are $$0, 1, 2$$ . We need to find the intervals for $$x$$ corresponding to these integer values. For $$\lfloor e^{1-x} \rfloor = 2$$ , we have $$2 \le e^{1-x} < 3$$ , which implies $$\ln 2 \le 1-x < \ln 3$$ , leading to $$1-\ln 3 < x \le 1-\ln 2$$ . For $$\lfloor e^{1-x} \rfloor = 1$$ , we have $$1 \le e^{1-x} < 2$$ , which implies $$0 \le 1-x < \ln 2$$ , leading to $$1-\ln 2 < x \le 1$$ . For $$\lfloor e^{1-x} \rfloor = 0$$ , we have $$e^{1-x} < 1$$ , which implies $$1-x < 0$$ , leading to $$x > 1$$ . Note that $$1-\ln 2 \approx 1-0.693 = 0.307$$ and $$1-\ln 3 \approx 1-1.098 = -0.098$$ .

Step 4: Split the integral and evaluate

Based on the intervals determined in the previous step, we can split the integral into three parts. The integral from $$0$$ to $$1-\ln 2$$ has $$\lfloor e^{1-x} \rfloor = 2$$ . The integral from $$1-\ln 2$$ to $$1$$ has $$\lfloor e^{1-x} \rfloor = 1$$ . The integral from $$1$$ to $$e^3$$ has $$\lfloor e^{1-x} \rfloor = 0$$ . The third integral evaluates to $$0$$ .

Step 5: Calculate the definite integrals

Now we evaluate each definite integral. The first integral is $$2x$$ evaluated from $$0$$ to $$1-\ln 2$$ , which gives $$2(1-\ln 2)$$ . The second integral is $$x$$ evaluated from $$1-\ln 2$$ to $$1$$ , which gives $$1 - (1-\ln 2) = \ln 2$$ .

Step 6: Sum the results and find $$\alpha$$

Summing the results of the definite integrals, we get $$(2 - 2\ln 2) + \ln 2 = 2 - \ln 2$$ . We are given that the integral equals $$\alpha - \ln 2$$ . By comparing, we find that $$\alpha = 2$$ .

Step 7: Calculate $$\alpha^3$$

Finally, we need to calculate $$\alpha^3$$ . Since $$\alpha = 2$$ , $$\alpha^3 = 2^3 = 8$$ .

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