Definite Integral of Piecewise Greatest Integer Function

To find the value of the given expression, let us first analyze the properties of the function $f(x)$.

The function $f(x)$ is defined on $[-10, 10]$ as:

$$f(x) = \begin{cases} x - [x] & \text{if } [x] \text{ is odd} \\ 1 + [x] - x & \text{if } [x] \text{ is even} \end{cases}$$

Step 1: Periodicity and Symmetry of $f(x)$

Let us examine the behavior of $f(x)$ for different intervals of $x$:

• For $x \in [0, 1)$, $[x] = 0$ (even):
  $f(x) = 1 + 0 - x = 1 - x$
• For $x \in [1, 2)$, $[x] = 1$ (odd):
  $f(x) = x - 1$

Now consider $f(x+2)$:

• If $[x]$ is even, then $[x+2] = [x] + 2$ is also even:
  $f(x+2) = 1 + [x+2] - (x+2) = 1 + [x] + 2 - x - 2 = 1 + [x] - x = f(x)$
• If $[x]$ is odd, then $[x+2] = [x] + 2$ is also odd:
  $f(x+2) = (x+2) - [x+2] = x + 2 - ([x] + 2) = x - [x] = f(x)$

Thus, $f(x)$ is a periodic function with fundamental period $T = 2$.

Next, let us check for symmetry:

• For $x \in (0, 1)$, $-x \in (-1, 0)$ where $[-x] = -1$ (odd):
  $f(-x) = -x - [-x] = -x - (-1) = 1 - x = f(x)$

Hence, $f(x)$ is an even function.

Step 2: Simplifying the Integral

The integrand is $g(x) = f(x) \cos(\pi x)$.
Since $\cos(\pi(x+2)) = \cos(\pi x + 2\pi) = \cos(\pi x)$, the function $\cos(\pi x)$ also has a period of $2$. Therefore, the product $f(x) \cos(\pi x)$ is periodic with period $T = 2$.

The interval of integration is $[-10, 10]$, which has a total length of $20 = 10 \times 2$ (representing $10$ full periods).

Using the property of periodic functions:
$\int_{-10}^{10} f(x) \cos(\pi x) \, dx = 10 \int_{0}^{2} f(x) \cos(\pi x) \, dx$

Step 3: Evaluating the Integral over One Period

We split the integral $\int_{0}^{2} f(x) \cos(\pi x) \, dx$ into two parts:
$\int_{0}^{2} f(x) \cos(\pi x) \, dx = \int_{0}^{1} (1 - x) \cos(\pi x) \, dx + \int_{1}^{2} (x - 1) \cos(\pi x) \, dx$

For the second integral, substitute $u = x - 1 \implies dx = du$:
$\int_{1}^{2} (x - 1) \cos(\pi x) \, dx = \int_{0}^{1} u \cos(\pi(u+1)) \, du = \int_{0}^{1} u (-\cos(\pi u)) \, du = -\int_{0}^{1} x \cos(\pi x) \, dx$

Combining the two integrals:
$\int_{0}^{2} f(x) \cos(\pi x) \, dx = \int_{0}^{1} (1 - x) \cos(\pi x) \, dx - \int_{0}^{1} x \cos(\pi x) \, dx = \int_{0}^{1} (1 - 2x) \cos(\pi x) \, dx$

Step 4: Integration by Parts

We integrate by parts by setting:

• $u = 1 - 2x \implies du = -2 \, dx$
• $dv = \cos(\pi x) \, dx \implies v = \frac{\sin(\pi x)}{\pi}$

$\int_{0}^{1} (1 - 2x) \cos(\pi x) \, dx = \left[ (1 - 2x) \frac{\sin(\pi x)}{\pi} \right]_{0}^{1} - \int_{0}^{1} (-2) \frac{\sin(\pi x)}{\pi} \, dx$

Since $\sin(0) = \sin(\pi) = 0$, the boundary term evaluates to zero:
$= 0 + \frac{2}{\pi} \int_{0}^{1} \sin(\pi x) \, dx = \frac{2}{\pi} \left[ -\frac{\cos(\pi x)}{\pi} \right]_{0}^{1}$
$= \frac{2}{\pi^2} \Big(-\cos(\pi) + \cos(0)\Big) = \frac{2}{\pi^2} \Big(-(-1) + 1\Big) = \frac{4}{\pi^2}$

Step 5: Final Calculation

Substituting back into the original expression:
$\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos(\pi x) \, dx = \frac{\pi^2}{10} \left( 10 \int_{0}^{2} f(x) \cos(\pi x) \, dx \right) = \pi^2 \left( \frac{4}{\pi^2} \right) = 4$