Key Concepts and Formulas

• Fractional Part Function: The fractional part of a number $x$, denoted by $\{x\}$, is defined as $x - [x]$, where $[x]$ is the greatest integer less than or equal to $x$. The range of $\{x\}$ is $[0, 1)$. A key property is $\{x+n\} = \{x\}$ for any integer $n$, indicating that $\{x\}$ is a periodic function with period 1.
• Definite Integral of a Periodic Function: If $f(x)$ is a periodic function with period $T$, then $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$ for any positive integer $n$.
• Trigonometric Identity: $\sin^2 x = \frac{1 - \cos(2x)}{2}$. This identity is useful for simplifying integrals involving $\sin^2 x$.
• Integral of $e^{ax}$: $\int e^{ax} dx = \frac{1}{a}e^{ax} + C$.

Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx}$

Step 1: Simplify the exponent using the fractional part function. The expression in the exponent is $\frac{x}{\pi} - \left[ \frac{x}{\pi} \right]$, which is the definition of the fractional part of $\frac{x}{\pi}$. We denote this as $\left\{ \frac{x}{\pi} \right\}$. So, the integral becomes: $I = \int\limits_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left\{ {{x \over \pi }} \right\}}}}}}dx$

Step 2: Determine the periodicity of the integrand. Let $f(x) = \frac{\sin^2 x}{e^{\{x/\pi\}}}$. We need to check if $f(x)$ is periodic. Consider $f(x+\pi)$: $f(x+\pi) = \frac{\sin^2 (x+\pi)}{e^{\{(x+\pi)/\pi\}}}$ Using the trigonometric identity $\sin(x+\pi) = -\sin x$, we get $\sin^2(x+\pi) = (-\sin x)^2 = \sin^2 x$. Using the property of the fractional part function, $\{(x+\pi)/\pi\} = \{x/\pi + 1\}$. Since $\{y+1\} = \{y\}$, we have $\{x/\pi + 1\} = \{x/\pi\}$. Therefore, $f(x+\pi) = \frac{\sin^2 x}{e^{\{x/\pi\}}} = f(x)$ This shows that the integrand $f(x)$ is periodic with a period $T = \pi$.

Step 3: Apply the property of definite integrals for periodic functions. The integral is from $0$ to $100\pi$, and the period of the integrand is $\pi$. We can use the property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$, where $n=100$ and $T=\pi$. $I = 100 \int_0^\pi {{{{{\sin }^2}x} \over {{e^{\left\{ {{x \over \pi }} \right\}}}}}}dx$

Step 4: Simplify the integrand over the interval $[0, \pi)$. For $x \in [0, \pi)$, the value of $\frac{x}{\pi}$ lies in the interval $[0, 1)$. For any $y \in [0, 1)$, the fractional part $\{y\}$ is equal to $y$. Thus, for $x \in [0, \pi)$, $\left\{ \frac{x}{\pi} \right\} = \frac{x}{\pi}$. The integral becomes: $I = 100 \int_0^\pi {{{{{\sin }^2}x} \over {{e^{x/\pi}}}}}dx$

Step 5: Use the trigonometric identity to simplify $\sin^2 x$. We use the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$. $I = 100 \int_0^\pi {\frac{1 - \cos(2x)}{2 \cdot e^{x/\pi}}}dx$ $I = 50 \int_0^\pi {\frac{1}{e^{x/\pi}}}dx - 50 \int_0^\pi {\frac{\cos(2x)}{e^{x/\pi}}}dx$ Let $I_1 = \int_0^\pi {\frac{1}{e^{x/\pi}}}dx$ and $I_2 = \int_0^\pi {\frac{\cos(2x)}{e^{x/\pi}}}dx$. Then $I = 50 (I_1 - I_2)$.

Step 6: Evaluate $I_1$. $I_1 = \int_0^\pi {e^{-x/\pi}}dx$ Let $u = -x/\pi$, so $du = -1/\pi \, dx$, which means $dx = -\pi \, du$. When $x=0$, $u=0$. When $x=\pi$, $u=-1$. $I_1 = \int_0^{-1} e^u (-\pi \, du) = \pi \int_{-1}^0 e^u du = \pi [e^u]_{-1}^0 = \pi (e^0 - e^{-1}) = \pi (1 - e^{-1})$

Step 7: Evaluate $I_2$. $I_2 = \int_0^\pi {\cos(2x) e^{-x/\pi}}dx$ This integral is of the form $\int e^{ax} \cos(bx) dx$, where $a = -1/\pi$ and $b = 2$. The formula for this integral is $\frac{e^{ax}}{a^2+b^2} (a \cos(bx) + b \sin(bx))$. Here, $a^2+b^2 = (-1/\pi)^2 + 2^2 = \frac{1}{\pi^2} + 4 = \frac{1+4\pi^2}{\pi^2}$. So, $\int {\cos(2x) e^{-x/\pi}}dx = \frac{e^{-x/\pi}}{\frac{1+4\pi^2}{\pi^2}} \left( -\frac{1}{\pi} \cos(2x) + 2 \sin(2x) \right)$ $= \frac{\pi^2 e^{-x/\pi}}{1+4\pi^2} \left( -\frac{1}{\pi} \cos(2x) + 2 \sin(2x) \right)$ Now, we evaluate this from $0$ to $\pi$: $I_2 = \left[ \frac{\pi^2 e^{-x/\pi}}{1+4\pi^2} \left( -\frac{1}{\pi} \cos(2x) + 2 \sin(2x) \right) \right]_0^\pi$ At $x=\pi$: $\frac{\pi^2 e^{-\pi/\pi}}{1+4\pi^2} \left( -\frac{1}{\pi} \cos(2\pi) + 2 \sin(2\pi) \right) = \frac{\pi^2 e^{-1}}{1+4\pi^2} \left( -\frac{1}{\pi}(1) + 2(0) \right) = \frac{\pi^2 e^{-1}}{1+4\pi^2} \left( -\frac{1}{\pi} \right) = -\frac{\pi e^{-1}}{1+4\pi^2}$ At $x=0$: $\frac{\pi^2 e^0}{1+4\pi^2} \left( -\frac{1}{\pi} \cos(0) + 2 \sin(0) \right) = \frac{\pi^2}{1+4\pi^2} \left( -\frac{1}{\pi}(1) + 0 \right) = \frac{\pi^2}{1+4\pi^2} \left( -\frac{1}{\pi} \right) = -\frac{\pi}{1+4\pi^2}$ So, $I_2 = \left( -\frac{\pi e^{-1}}{1+4\pi^2} \right) - \left( -\frac{\pi}{1+4\pi^2} \right) = \frac{\pi}{1+4\pi^2} (1 - e^{-1})$

Step 8: Combine $I_1$ and $I_2$ to find $I$. $I = 50 (I_1 - I_2)$ $I = 50 \left( \pi (1 - e^{-1}) - \frac{\pi}{1+4\pi^2} (1 - e^{-1}) \right)$ $I = 50 \pi (1 - e^{-1}) \left( 1 - \frac{1}{1+4\pi^2} \right)$ $I = 50 \pi (1 - e^{-1}) \left( \frac{1+4\pi^2 - 1}{1+4\pi^2} \right)$ $I = 50 \pi (1 - e^{-1}) \left( \frac{4\pi^2}{1+4\pi^2} \right)$ $I = \frac{200 \pi^3 (1 - e^{-1})}{1+4\pi^2}$

Step 9: Compare with the given expression to find $\alpha$. The given expression is: $I = \frac{\alpha \pi^3}{1 + 4\pi^2}$ By comparing our result with the given expression: $\frac{200 \pi^3 (1 - e^{-1})}{1+4\pi^2} = \frac{\alpha \pi^3}{1 + 4\pi^2}$ This implies: $\alpha = 200 (1 - e^{-1})$

Common Mistakes & Tips

• Misinterpreting the fractional part: Ensure you correctly identify $x - [x]$ as $\{x\}$ and understand its properties, especially its periodicity.
• Periodicity of the integrand: Carefully check if the entire integrand is periodic, not just a part of it. In this case, both $\sin^2 x$ and $e^{\{x/\pi\}}$ contribute to the overall periodicity.
• Integration of exponential and trigonometric functions: Be precise when using the formula for $\int e^{ax} \cos(bx) dx$, as errors in signs or coefficients can lead to incorrect results.

Summary

The problem involved evaluating a definite integral with a complex integrand. We first simplified the integrand by recognizing the fractional part function and establishing the periodicity of the integrand. By applying the property of definite integrals for periodic functions, we reduced the integral to a simpler form over one period. We then used a trigonometric identity and standard integration techniques to evaluate the resulting integrals. Finally, by comparing the computed value with the given expression, we determined the value of $\alpha$.

The final answer is \boxed{200 (1 - e^{-1})}. This corresponds to option (A).