Key Concepts and Formulas

1. Fractional Part Function: The fractional part of $x$ is defined as $\{x\} = x - [x]$. This function is periodic with period 1, i.e., $\{x+1\} = \{x\}$.
2. Greatest Integer Function: $[x]$ denotes the greatest integer less than or equal to $x$. Its value is constant over intervals of the form $[n, n+1)$, where $n$ is an integer.
3. Periodicity Property of Definite Integrals: If a function $f(x)$ is periodic with period $T$, then for any integer $n$, $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$.
4. Properties of Sine Function: $\sin(2\pi x)$ is periodic with period 1. The value of $[\sin(2\pi x)]$ is 0 for $x \in [0, 1/2)$ and -1 for $x \in [1/2, 1)$ within the interval $[0, 1)$.

Step-by-Step Solution

Let the given integral be $I$. $I = \int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx$

Step 1: Simplify the Integrand using the Fractional Part Function We use the definition of the fractional part function, $x - [x] = \{x\}$. This simplifies the denominator. $I = \int_0^{10} {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}} dx$ Explanation: By expressing $x - [x]$ as $\{x\}$, we make the periodic nature of the denominator explicit, which is key to analyzing the integrand's overall periodicity.

Step 2: Determine the Periodicity of the Integrand Let $f(x) = \frac{[\sin 2\pi x]}{e^{\{x\}}}$. We examine the periodicity of the numerator and the denominator separately.

• The term $\sin 2\pi x$ has a period of $T_1 = \frac{2\pi}{2\pi} = 1$. Consequently, $[\sin 2\pi x]$ is also periodic with period 1.
• The fractional part function $\{x\}$ has a period of 1, i.e., $\{x+1\} = \{x\}$. Therefore, $e^{\{x\}}$ is also periodic with period 1. Since both the numerator and the denominator are periodic with period 1, the entire integrand $f(x)$ is periodic with period $T=1$. Explanation: Identifying the period of the integrand is crucial. It allows us to use the property of definite integrals over multiple periods, significantly simplifying the problem.

Step 3: Apply the Periodicity Property of Definite Integrals Given that the integrand is periodic with period $T=1$ and the interval of integration is from 0 to 10 (which is $10T$), we can apply the property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$. $I = 10 \int_0^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}} dx$ Explanation: This step reduces the integral from a range of 10 periods to the integral over a single period, multiplied by the number of periods. This makes the evaluation manageable.

Step 4: Analyze the Numerator $[\sin 2\pi x]$ in the Interval $[0, 1]$ We need to determine the values of $[\sin 2\pi x]$ for $x \in [0, 1]$.

• For $x \in [0, 1/2)$:

  • $2\pi x \in [0, \pi)$.
  • In this range, $\sin(2\pi x)$ is between 0 and 1 (inclusive of 0, exclusive of values greater than 1). Specifically, $\sin(2\pi x) \in [0, 1]$.
  • Thus, $[\sin 2\pi x] = 0$ for $x \in [0, 1/2)$.

• For $x \in [1/2, 1)$:

  • $2\pi x \in [\pi, 2\pi)$.
  • In this range, $\sin(2\pi x)$ is between -1 and 0 (inclusive of 0, exclusive of values less than -1). Specifically, $\sin(2\pi x) \in [-1, 0)$.
  • Thus, $[\sin 2\pi x] = -1$ for $x \in [1/2, 1)$. Explanation: Understanding the behavior of the greatest integer function applied to $\sin(2\pi x)$ within the fundamental period is essential for splitting the integral correctly.

Step 5: Split the Integral and Evaluate the Denominator Using the analysis from Step 4, we split the integral from 0 to 1 into two parts: $I = 10 \left( \int_0^{1/2} {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}} dx + \int_{1/2}^1 {{{[\sin 2\pi x]} \over {{e^{\{ x\} }}}}} dx \right)$ Substitute the values of $[\sin 2\pi x]$: $I = 10 \left( \int_0^{1/2} {{{0} \over {{e^{\{ x\} }}}}} dx + \int_{1/2}^1 {{{ - 1} \over {{e^{\{ x\} }}}}} dx \right)$ The first integral is 0. For the second integral, in the interval $[1/2, 1)$, $\{x\} = x$. $I = 10 \left( 0 + \int_{1/2}^1 {{{ - 1} \over {{e^x}}}} dx \right)$ $I = 10 \int_{1/2}^1 {-e^{-x}} dx$ Explanation: The integral is split based on the intervals where the numerator changes its value. The first interval contributes zero to the integral. In the second interval, we use $\{x\}=x$ to simplify the exponential term, making it ready for integration.

Step 6: Perform the Integration Now, we evaluate the definite integral: $I = 10 \left[ -(-e^{-x}) \right]_{1/2}^1$ $I = 10 \left[ e^{-x} \right]_{1/2}^1$ $I = 10 (e^{-1} - e^{-1/2})$ $I = 10 e^{-1} - 10 e^{-1/2}$ Explanation: The integration of $e^{-x}$ is straightforward, and applying the limits of integration yields the value of the integral in terms of exponential functions.

Step 7: Compare with the Given Form and Find Coefficients The problem states that the integral is equal to $\alpha e^{-1} + \beta e^{-1/2} + \gamma$. Comparing our result $10 e^{-1} - 10 e^{-1/2}$ with the given form, we can identify the integer coefficients:

• $\alpha = 10$
• $\beta = -10$
• $\gamma = 0$ Explanation: By directly comparing the obtained expression with the target form, we extract the integer values of $\alpha$, $\beta$, and $\gamma$.

Step 8: Calculate $\alpha + \beta + \gamma$ Finally, we compute the sum of these coefficients: $\alpha + \beta + \gamma = 10 + (-10) + 0 = 0$

Common Mistakes & Tips

• Incorrect Periodicity Analysis: Ensure you correctly identify the period of the entire integrand, not just parts of it. The period of $\sin(2\pi x)$ is 1, not $2\pi$.
• Handling the Greatest Integer Function: Pay close attention to the intervals where $[\sin 2\pi x]$ takes different integer values. Remember that the integral over a single point is zero.
• Fractional Part in Different Intervals: Be mindful that $\{x\} = x$ only for $x \in [0, 1)$. For other intervals, $\{x\}$ will be different, though the periodicity property handles this implicitly when splitting the integral.

Summary The problem was solved by first simplifying the integrand using the fractional part function. We then established the integrand's periodicity with period 1. Using the periodicity property of definite integrals, the integral over 10 periods was reduced to 10 times the integral over a single period. By analyzing the behavior of $[\sin 2\pi x]$ in the interval $[0, 1]$, the integral was split into two parts. One part evaluated to zero, and the other was integrated directly after using $\{x\}=x$. Finally, comparing the result with the given form allowed us to find the integer coefficients $\alpha$, $\beta$, and $\gamma$, and their sum was calculated.

The final answer is \boxed{0}.