If `varphi(t)={{:(1","," for "0letlt1),(0",","otherwise"):}`, then `int_(-3000)^(3000)(sum_(r'=2014)^(2016)varphi(t-r')varphi(t-2016))dt=`

Options

A. a real number
B. 1
C. 0
D. does not exist

Answer

To solve the given integral

\[ \int_{-3000}^{3000} \left( \sum_{r' = 2014}^{2016} \varphi(t - r') \varphi(t - 2016) \right) dt, \]

we start by analyzing the function \(\varphi(t)\):

\[ \varphi(t) = \begin{cases} 1 & \text{for } 0 \leq t < 1 \\ 0 & \text{otherwise} \end{cases}. \]

### Step 1: Break the Integral into Parts

We can break the integral into three parts based on the values of \(r'\):

\[ \int_{-3000}^{3000} \left( \varphi(t - 2014) \varphi(t - 2016) + \varphi(t - 2015) \varphi(t - 2016) + \varphi(t - 2016) \varphi(t - 2016) \right) dt. \]

### Step 2: Analyze Each Component

1. **For \(r' = 2014\)**:
\[ \int_{-3000}^{3000} \varphi(t - 2014) \varphi(t - 2016) dt. \]
Here, \(\varphi(t - 2014) = 1\) when \(2014 \leq t < 2015\) and \(\varphi(t - 2016) = 1\) when \(2016 \leq t < 2017\). Since there is no overlap in these intervals, this integral evaluates to 0.

2. **For \(r' = 2015\)**:
\[ \int_{-3000}^{3000} \varphi(t - 2015) \varphi(t - 2016) dt. \]
Here, \(\varphi(t - 2015) = 1\) when \(2015 \leq t < 2016\) and \(\varphi(t - 2016) = 1\) when \(2016 \leq t < 2017\). Again, there is no overlap, so this integral also evaluates to 0.

3. **For \(r' = 2016\)**:
\[ \int_{-3000}^{3000} \varphi(t - 2016) \varphi(t - 2016) dt. \]
Here, \(\varphi(t - 2016) = 1\) when \(2016 \leq t < 2017\). Thus, this integral evaluates to:
\[ \int_{2016}^{2017} 1 \, dt = 1. \]

### Step 3: Combine the Results

Now, we combine the results from all three parts:

\[ 0 + 0 + 1 = 1. \]

### Final Answer

Thus, the value of the integral is

\[ \int_{-3000}^{3000} \left( \sum_{r' = 2014}^{2016} \varphi(t - r') \varphi(t - 2016) \right) dt = 1. \]