Evaluate: `int_0^oo[2e^(-x)]dx ,w h e r e[x]` represents greatest integer function.

Explanation

To evaluate the integral

\[ I = \int_0^\infty \lfloor 2e^{-x} \rfloor \, dx \]

where \(\lfloor x \rfloor\) denotes the greatest integer function, we will follow these steps:

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

The function \(2e^{-x}\) is a decreasing function. We need to find the points where this function takes on integer values.

### Step 2: Determine the range of \(2e^{-x}\)

- At \(x = 0\):
\[ 2e^{-0} = 2 \]
- As \(x \to \infty\):
\[ 2e^{-x} \to 0 \]

### Step 3: Find the points of discontinuity

To find where \(2e^{-x}\) equals integers, we set:
\[ 2e^{-x} = n \quad (n \text{ is an integer}) \]
This gives:
\[ e^{-x} = \frac{n}{2} \implies x = -\ln\left(\frac{n}{2}\right) = \ln(2) - \ln(n) \]

### Step 4: Identify the intervals for the greatest integer function

1. For \(0 \leq x < \ln(2)\):
\[ 2e^{-x} \geq 1 \implies \lfloor 2e^{-x} \rfloor = 1 \]

2. For \(\ln(2) \leq x < \ln(1)\) (which is not applicable since \(\ln(1) = 0\)):
\[ 2e^{-x} < 1 \implies \lfloor 2e^{-x} \rfloor = 0 \]

### Step 5: Set up the integral based on intervals

Thus, we can split the integral into two parts:
\[ I = \int_0^{\ln(2)} 1 \, dx + \int_{\ln(2)}^\infty 0 \, dx \]

### Step 6: Evaluate the integrals

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

2. The second integral evaluates to zero:
\[ \int_{\ln(2)}^\infty 0 \, dx = 0 \]

### Step 7: Combine the results

Thus, the value of the integral \(I\) is:
\[ I = \ln(2) + 0 = \ln(2) \]

### Final Answer
\[ \int_0^\infty \lfloor 2e^{-x} \rfloor \, dx = \ln(2) \]

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