The value of `int_(0)^(infty)[2e^(-x)] dx` (where ,[.] denotes the greatest integer function of x) is equal to

Options

A. 1
B. `log_(e)2`
C. 0
D. `(1)/(e)`

Answer

To solve the integral \( \int_{0}^{\infty} [2e^{-x}] \, dx \), where \([.]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Analyze the function \( f(x) = 2e^{-x} \)
The function \( f(x) = 2e^{-x} \) is a monotonically decreasing function. As \( x \) increases, \( e^{-x} \) decreases, thus \( f(x) \) will also decrease.

### Step 2: Determine the intervals for the greatest integer function
To find the intervals where \( [2e^{-x}] \) takes constant integer values, we need to find where \( 2e^{-x} \) crosses integer values.

1. When \( x = 0 \):
\[ f(0) = 2e^{0} = 2 \]
So, \( [2e^{-0}] = 2 \).

2. When \( x \) approaches \( \log(2) \):
\[ f(\log(2)) = 2e^{-\log(2)} = 2 \cdot \frac{1}{2} = 1 \]
So, \( [2e^{-\log(2)}] = 1 \).

3. When \( x \) approaches \( \infty \):
\[ f(x) \to 0 \quad \text{as } x \to \infty \]
So, \( [2e^{-x}] = 0 \) for \( x > \log(2) \).

### Step 3: Set up the integral based on intervals
We can now split the integral into three parts based on the intervals:
1. From \( 0 \) to \( \log(2) \), \( [2e^{-x}] = 2 \).
2. From \( \log(2) \) to \( \infty \), \( [2e^{-x}] = 1 \) until \( x \) reaches \( \infty \), where it becomes \( 0 \).

Thus, we can write:
\[ \int_{0}^{\infty} [2e^{-x}] \, dx = \int_{0}^{\log(2)} 2 \, dx + \int_{\log(2)}^{\infty} 1 \, dx \]

### Step 4: Calculate the integrals
1. For the first integral:
\[ \int_{0}^{\log(2)} 2 \, dx = 2 \cdot x \Big|_{0}^{\log(2)} = 2 \cdot \log(2) - 2 \cdot 0 = 2 \log(2) \]

2. For the second integral:
\[ \int_{\log(2)}^{\infty} 1 \, dx = \left[ x \right]_{\log(2)}^{\infty} = \infty - \log(2) = \infty \]

### Step 5: Combine the results
Since the second integral diverges, we conclude:
\[ \int_{0}^{\infty} [2e^{-x}] \, dx = \infty \]

### Final Answer
The value of \( \int_{0}^{\infty} [2e^{-x}] \, dx \) is \( \infty \).

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