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**Using Laplace transform, prove that**

\[
\int_0^\infty \frac{\sin t}{t} \, dt = \frac{\pi}{2}
\]

Let's solve it step by step using the Laplace transform approach.

### Solution Outline:

1. **Step 1: Recognize the Integral**  
   The integral \(\int_0^\infty \frac{\sin t}{t} \, dt\) is a classic improper integral known as the Dirichlet integral, which equals \(\frac{\pi}{2}\).

2. **Step 2: Define a function involving the Laplace transform**  
   To use the Laplace transform, we introduce a decaying exponential function, such as \(e^{-st}\), which will allow the use of the Laplace transform and remove the problematic improper limits.

   Consider:
   \[
   I(s) = \int_0^\infty \frac{\sin t}{t} e^{-st} \, dt
   \]
   Here, \(s > 0\) is a parameter that ensures convergence of the integral.

3. **Step 3: Solve the integral \(I(s)\) using the known Laplace transform**  
   The integral involving \(\sin t / t\) has a known result in the Laplace transform tables:
   \[
   \mathcal{L}\left\{ \frac{\sin t}{t} \right\} = \tan^{-1}\left( \frac{1}{s} \right)
   \]
   Thus, we can write:
   \[
   I(s) = \tan^{-1}\left( \frac{1}{s} \right)
   \]

4. **Step 4: Find the limit as \(s \to 0\)**  
   To recover the original integral, we now take the limit as \(s \to 0\):
   \[
   \lim_{s \to 0} I(s) = \lim_{s \to 0} \tan^{-1}\left( \frac{1}{s} \right) = \frac{\pi}{2}
   \]

Thus, we have shown that:
\[
\int_0^\infty \frac{\sin t}{t} \, dt = \frac{\pi}{2}
\]

Let me know if you need further details or have any questions!

Here are 5 related questions:
1. How do we find the Laplace transform of more complex functions like \(\sin t / t\)?
2. What are the properties of Laplace transforms that make them useful in solving integrals?
3. How do we handle improper integrals without the Laplace transform method?
4. Can the Laplace transform be used to solve other types of improper integrals?
5. What other applications do improper integrals have in mathematical physics?

**Tip**: The Laplace transform is especially helpful when dealing with improper integrals and differential equations because it transforms problems into a simpler algebra

Using Laplace transform, prove that \int_0^\infty \frac{\sin t}{t} dt = \frac{\pi}{2}.

Learn how to use the Laplace transform to prove the improper integral \int_0^\infty \frac{\sin t}{t} dt = \frac{\pi}{2}. This problem involves key concepts in calculus, including improper integrals and Laplace transforms, and is suitable for university-level students.

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