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The provided image contains a mathematical problem involving the sine integral function, \( \text{si}(x) \), defined as:

\[
\text{si}(x) = \int_0^x \frac{\sin t}{t} \, dt \quad \text{for} \quad x \geq 0
\]

The task is to compute \( \text{si}(1) \) with an accuracy of \( 6D \) (6 decimal places), using a few terms from the Laurent series expansion of the function \( f(t) = \sin(t) \).

### Steps:
1. **Laurent Series of \( \frac{\sin t}{t} \):** The sine function can be expanded as a series around \( t = 0 \). For small \( t \), we use the expansion:
   \[
   \frac{\sin t}{t} = 1 - \frac{t^2}{6} + \frac{t^4}{120} - \dots
   \]
2. **Substitute into the integral:**
   To find \( \text{si}(1) \), we substitute this series into the integral:
   \[
   \text{si}(1) = \int_0^1 \left( 1 - \frac{t^2}{6} + \frac{t^4}{120} - \dots \right) dt
   \]
3. **Integrate term by term:**
   \[
   \int_0^1 1 \, dt = 1, \quad \int_0^1 \frac{t^2}{6} \, dt = \frac{1}{18}, \quad \int_0^1 \frac{t^4}{120} \, dt = \frac{1}{600}
   \]
   Summing the first few terms:
   \[
   \text{si}(1) \approx 1 - \frac{1}{18} + \frac{1}{600} = 0.945833\ldots
   \]

Thus, \( \text{si}(1) \approx 0.945833 \) to 6 decimal places.

Would you like further details or have any questions?

### Related Questions:
1. How can we derive the Laurent series for more complex functions?
2. What are the properties of the sine integral function \( \text{si}(x) \)?
3. How does the convergence of the Laurent series affect the accuracy?
4. Can this series expansion be used for larger values of \( x \)?
5. How is the sine integral function applied in physics and engineering?

**Tip:** Using the first few terms of a series expansion often provides an excellent approximation, especially when evaluating integrals at small limits.

تمرین (1) تابع si(x) = ∫ (sint/t) dt برای x ≥ 0 را در نظر بگیرید. دست کم چند جمله از بسط مک لورن تابع f(t) = sint لازم است تا si(1) با دقت 6D مشخص شود؟

Learn how to use the Laurent series expansion of sin(t) to compute the sine integral si(1) with 6 decimal place accuracy. This solution involves integral calculus and the series expansion of the sine function.

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