To solve the integral
\[
I = \int_0^{10} \frac{\lfloor \sin(2 \pi x) \rfloor}{e^{-\lfloor x \rfloor + x}} \, dx,
\]
we will break down the problem step by step.
### Step 1: Understanding the Greatest Integer Function
The greatest integer function, \(\lfloor \sin(2 \pi x) \rfloor\), takes the value of \(\sin(2 \pi x)\) and rounds it down to the nearest integer. The function \(\sin(2 \pi x)\) oscillates between -1 and 1.
- For \(0 \leq x < \frac{1}{2}\), \(\sin(2 \pi x) \geq 0\) and hence \(\lfloor \sin(2 \pi x) \rfloor = 0\).
- For \(\frac{1}{2} \leq x < 1\), \(\sin(2 \pi x) < 0\) and hence \(\lfloor \sin(2 \pi x) \rfloor = -1\).
### Step 2: Splitting the Integral
Given that the function is periodic with a period of 1, we can break the integral from 0 to 10 into 10 intervals of length 1:
\[
I = \sum_{n=0}^{9} \int_n^{n+1} \frac{\lfloor \sin(2 \pi x) \rfloor}{e^{-\lfloor x \rfloor + x}} \, dx.
\]
### Step 3: Evaluating Each Integral
1. For \(n = 0\) to \(n = 0.5\):
\[
\int_0^{0.5} \frac{0}{e^{0}} \, dx = 0.
\]
2. For \(n = 0.5\) to \(n = 1\):
\[
\int_{0.5}^{1} \frac{-1}{e^{0 + x}} \, dx = -\int_{0.5}^{1} e^{-x} \, dx.
\]
The integral of \(e^{-x}\) is \(-e^{-x}\), so:
\[
-\left[-e^{-x}\right]_{0.5}^{1} = -(-e^{-1} + e^{-0.5}) = e^{-1} - e^{-0.5}.
\]
3. This pattern continues for each subsequent interval:
- For \(n = 1\) to \(n = 1.5\): Similar to the previous case, we get \(e^{-2} - e^{-1.5}\).
- For \(n = 1.5\) to \(n = 2\): \(e^{-3} - e^{-2.5}\).
- This continues until \(n = 9\) to \(n = 10\).
### Step 4: Summing the Integrals
Each segment contributes:
\[
\int_{n}^{n+0.5} 0 \, dx + \int_{n+0.5}^{n+1} (-1) e^{-n} \, dx = e^{-n} - e^{-(n+0.5)}.
\]
Thus, for \(n = 0\) to \(9\):
\[
I = \sum_{n=0}^{9} (e^{-n} - e^{-(n+0.5)}) = \sum_{n=0}^{9} e^{-n} - \sum_{n=0}^{9} e^{-(n+0.5)}.
\]
### Step 5: Finding \(\alpha\), \(\beta\), and \(\gamma\)
From the integral, we can express \(I\) in the form:
\[
I = \alpha e^{-1} + \beta e^{-1/2} + \gamma.
\]
By comparing coefficients, we find:
- \(\alpha = 10\),
- \(\beta = -10\),
- \(\gamma = 0\).
### Step 6: Calculating \(\alpha + \beta + \gamma\)
Now, we compute:
\[
\alpha + \beta + \gamma = 10 - 10 + 0 = 0.
\]
Thus, the final answer is:
\[
\boxed{0}.
\]
