(t) = ! t
−∞
g(ξ)dξ , g(t) = " cos πt
T per |t| ≤ T /2
0 altrove., calcola y(t)

To calculate \( y(t) \) based on the given integral and functions, let's break down the problem step by step. We are given two functions:

1. \( y(t) = \int_{-\infty}^{t} g(\xi) \, d\xi \)
2. \( g(t) = 
\begin{cases}
\cos\left( \frac{\pi t}{T_{\text{per}}} \right) & \text{if } |t| \leq \frac{T}{2}, \\
0 & \text{elsewhere}.
\end{cases} \)

### Steps to calculate \( y(t) \):

1. **Understand the function \( g(t) \)**:
   - \( g(t) = \cos\left( \frac{\pi t}{T_{\text{per}}} \right) \) only when \( |t| \leq \frac{T}{2} \), i.e., the function is a cosine wave restricted to the interval \( -\frac{T}{2} \leq t \leq \frac{T}{2} \).
   - For \( t \) outside this range, \( g(t) = 0 \).

2. **Integrating \( g(\xi) \)**:
   The integral \( y(t) = \int_{-\infty}^{t} g(\xi) \, d\xi \) can be split into cases depending on the value of \( t \).

3. **Case 1: \( t < -\frac{T}{2} \)**:
   - In this range, \( g(\xi) = 0 \) for all \( \xi \), since the cosine function is zero outside \( |t| \leq \frac{T}{2} \).
   - Therefore, \( y(t) = 0 \) for \( t < -\frac{T}{2} \).

4. **Case 2: \( -\frac{T}{2} \leq t \leq \frac{T}{2} \)**:
   - In this range, we need to integrate the function \( g(\xi) = \cos\left( \frac{\pi \xi}{T_{\text{per}}} \right) \) from \( -\frac{T}{2} \) to \( t \).
   - The integral is:
     \[
     y(t) = \int_{-\frac{T}{2}}^{t} \cos\left( \frac{\pi \xi}{T_{\text{per}}} \right) d\xi
     \]

     The antiderivative of \( \cos\left( \frac{\pi \xi}{T_{\text{per}}} \right) \) is:
     \[
     \int \cos\left( \frac{\pi \xi}{T_{\text{per}}} \right) d\xi = \frac{T_{\text{per}}}{\pi} \sin\left( \frac{\pi \xi}{T_{\text{per}}} \right)
     \]

     Applying the limits of integration, we get:
     \[
     y(t) = \frac{T_{\text{per}}}{\pi} \left[ \sin\left( \frac{\pi t}{T_{\text{per}}} \right) - \sin\left( -\frac{\pi}{2} \right) \right]
     \]

     Since \( \sin\left( -\frac{\pi}{2} \right) = -1 \), we have:
     \[
     y(t) = \frac{T_{\text{per}}}{\pi} \left[ \sin\left( \frac{\pi t}{T_{\text{per}}} \right) + 1 \right]
     \]

5. **Case 3: \( t > \frac{T}{2} \)**:
   - For \( t > \frac{T}{2} \), the function \( g(\xi) \) is zero outside the interval \( |t| \leq \frac{T}{2} \). Therefore, the integral is the same as the value at \( t = \frac{T}{2} \).
   - So, \( y(t) = y\left( \frac{T}{2} \right) \).

     This is just the result from Case 2 evaluated at \( t = \frac{T}{2} \):
     \[
     y\left( \frac{T}{2} \right) = \frac{T_{\text{per}}}{\pi} (1 + 1) = \frac{2 T_{\text{per}}}{\pi}
     \]

### Final result for \( y(t) \):

- For \( t < -\frac{T}{2} \): \( y(t) = 0 \)
- For \( -\frac{T}{2} \leq t \leq \frac{T}{2} \): \( y(t) = \frac{T_{\text{per}}}{\pi} \left[ \sin\left( \frac{\pi t}{T_{\text{per}}} \right) + 1 \right] \)
- For \( t > \frac{T}{2} \): \( y(t) = \frac{2 T_{\text{per}}}{\pi} \)

---

Would you like me to provide more details or clarify any part of the solution?

### Related Questions:
1. How would the solution change if the periodic function was sine instead of cosine?
2. What happens to \( y(t) \) if the period \( T_{\text{per}} \) becomes very large?
3. Can you explain how to compute the integral for different types of periodic functions?
4. What is the physical interpretation of integrating a periodic signal over time?
5. How does the result change if the interval of nonzero \( g(t) \) is shifted?

**Tip**: When integrating trigonometric functions, always remember to adjust the antiderivative to account for the coefficient inside the argument of the sine or cosine function.

! t -∞ g(ξ)dξ, g(t) = " cos πt / T per |t| ≤ T /2 0 altrove., calcola y(t)

Explore the solution for finding y(t) by integrating a cosine function over a piecewise-defined domain. This problem involves concepts from integral calculus and trigonometric functions, suitable for undergraduate calculus students.

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