Evaluate `int _(-1) ^(15) Sgn ({x})dx,` (where `{**}` denotes the fractional part function)

Answer

To evaluate the integral \(\int_{-1}^{15} \text{Sgn}(\{x\}) \, dx\), where \(\{x\}\) denotes the fractional part of \(x\), we can follow these steps:

### Step 1: Understand the Signum Function
The signum function, \(\text{Sgn}(x)\), is defined as:
- \(\text{Sgn}(x) = 1\) if \(x > 0\)
- \(\text{Sgn}(x) = 0\) if \(x = 0\)
- \(\text{Sgn}(x) = -1\) if \(x < 0\)

### Step 2: Analyze the Fractional Part Function
The fractional part function \(\{x\}\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
This means that \(\{x\}\) is always between \(0\) and \(1\) for any real number \(x\).

### Step 3: Determine the Behavior of \(\text{Sgn}(\{x\})\)
Since \(\{x\}\) is always non-negative and only equals \(0\) when \(x\) is an integer, we can conclude:
- For \(x \in (-1, 0)\), \(\{x\}\) is positive, hence \(\text{Sgn}(\{x\}) = 1\).
- For \(x \in (0, 15)\), \(\{x\}\) is also positive, hence \(\text{Sgn}(\{x\}) = 1\).
- At integer points, \(\{x\} = 0\), but since these are isolated points, they do not affect the integral.

### Step 4: Rewrite the Integral
Given that \(\text{Sgn}(\{x\}) = 1\) almost everywhere in the interval \([-1, 15]\), we can simplify the integral:
\[ \int_{-1}^{15} \text{Sgn}(\{x\}) \, dx = \int_{-1}^{15} 1 \, dx \]

### Step 5: Evaluate the Integral
Now we can evaluate the integral:
\[ \int_{-1}^{15} 1 \, dx = [x]_{-1}^{15} = 15 - (-1) = 15 + 1 = 16 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-1}^{15} \text{Sgn}(\{x\}) \, dx = 16 \]