`rArrint_(0)^(oo) [(2)/(e^(x))]dx` (where [*] denotes the greatest integer function) equals

Options

A. `log_(e)2`
B. ` e^(2)`
C. 0
D. `2//e`

Explanation

To solve the integral \( I = \int_{0}^{\infty} \left\lfloor \frac{2}{e^x} \right\rfloor dx \), we will break it down step by step.

### Step 1: Analyze the function \( \frac{2}{e^x} \)

As \( x \) varies from \( 0 \) to \( \infty \):
- At \( x = 0 \), \( \frac{2}{e^0} = 2 \).
- As \( x \to \infty \), \( \frac{2}{e^x} \to 0 \).

Thus, the function \( \frac{2}{e^x} \) decreases from \( 2 \) to \( 0 \) as \( x \) increases from \( 0 \) to \( \infty \).

### Step 2: Determine the range of \( \left\lfloor \frac{2}{e^x} \right\rfloor \)

- For \( 0 \leq x < \ln(2) \):
- \( \frac{2}{e^x} \geq 1 \) (since \( e^x < 2 \)).
- Therefore, \( \left\lfloor \frac{2}{e^x} \right\rfloor = 1 \).

- For \( x = \ln(2) \):
- \( \frac{2}{e^{\ln(2)}} = 1 \).
- Therefore, \( \left\lfloor \frac{2}{e^{\ln(2)}} \right\rfloor = 1 \).

- For \( x > \ln(2) \):
- \( \frac{2}{e^x} < 1 \) (since \( e^x > 2 \)).
- Therefore, \( \left\lfloor \frac{2}{e^x} \right\rfloor = 0 \).

### Step 3: Break the integral into two parts

Now we can break the integral into two parts:
\[ I = \int_{0}^{\ln(2)} \left\lfloor \frac{2}{e^x} \right\rfloor dx + \int_{\ln(2)}^{\infty} \left\lfloor \frac{2}{e^x} \right\rfloor dx \]

### Step 4: Evaluate the first integral

For \( x \) in the range \( [0, \ln(2)] \):
\[ \int_{0}^{\ln(2)} \left\lfloor \frac{2}{e^x} \right\rfloor dx = \int_{0}^{\ln(2)} 1 \, dx = \left[ x \right]_{0}^{\ln(2)} = \ln(2) - 0 = \ln(2) \]

### Step 5: Evaluate the second integral

For \( x \) in the range \( [\ln(2), \infty) \):
\[ \int_{\ln(2)}^{\infty} \left\lfloor \frac{2}{e^x} \right\rfloor dx = \int_{\ln(2)}^{\infty} 0 \, dx = 0 \]

### Step 6: Combine results

Combining the results from both integrals:
\[ I = \ln(2) + 0 = \ln(2) \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\infty} \left\lfloor \frac{2}{e^x} \right\rfloor dx = \ln(2) \]