(a) Use the definition of a Laplace transform to show that
L[cosh(t)] = s
s
2 − 1
and state the region of convergence of this transform

To show that \( \mathcal{L}[\cosh(t)] = \frac{s}{s^2 - 1} \) using the definition of the Laplace transform, we start with the definition:

\[
\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt
\]

For \( f(t) = \cosh(t) \):

\[
\cosh(t) = \frac{e^t + e^{-t}}{2}
\]

So, we need to compute:

\[
\mathcal{L}\{\cosh(t)\} = \int_0^\infty e^{-st} \cosh(t) \, dt = \int_0^\infty e^{-st} \left( \frac{e^t + e^{-t}}{2} \right) dt
\]

Breaking this into two separate integrals, we get:

\[
\mathcal{L}\{\cosh(t)\} = \frac{1}{2} \int_0^\infty e^{-st} e^t \, dt + \frac{1}{2} \int_0^\infty e^{-st} e^{-t} \, dt
\]

Simplify the exponents:

\[
= \frac{1}{2} \int_0^\infty e^{(1-s)t} \, dt + \frac{1}{2} \int_0^\infty e^{-(s+1)t} \, dt
\]

These integrals are of the form \( \int_0^\infty e^{at} \, dt \), which evaluates to \( \frac{1}{-a} \) for \( \Re(a) < 0 \). So, we need to evaluate the integrals:

1. For the first integral \( \frac{1}{2} \int_0^\infty e^{(1-s)t} \, dt \):

\[
\int_0^\infty e^{(1-s)t} \, dt = \left[ \frac{e^{(1-s)t}}{1-s} \right]_0^\infty = \frac{1}{s-1} \quad \text{(since \(\Re(s) > 1\))}
\]

2. For the second integral \( \frac{1}{2} \int_0^\infty e^{-(s+1)t} \, dt \):

\[
\int_0^\infty e^{-(s+1)t} \, dt = \left[ \frac{e^{-(s+1)t}}{-(s+1)} \right]_0^\infty = \frac{1}{s+1} \quad \text{(since \(\Re(s) > -1\))}
\]

Adding these results together:

\[
\mathcal{L}\{\cosh(t)\} = \frac{1}{2} \left( \frac{1}{s-1} + \frac{1}{s+1} \right)
\]

Combining the fractions:

\[
\mathcal{L}\{\cosh(t)\} = \frac{1}{2} \left( \frac{s+1 + s-1}{(s-1)(s+1)} \right) = \frac{1}{2} \left( \frac{2s}{s^2 - 1} \right) = \frac{s}{s^2 - 1}
\]

Therefore, we have shown that:

\[
\mathcal{L}[\cosh(t)] = \frac{s}{s^2 - 1}
\]

The region of convergence for this Laplace transform is determined by the need for the integrals to converge. This requires:

\[
\Re(s) > 1
\]

Thus, the region of convergence is \( \Re(s) > 1 \).

Would you like any more details or have any questions?

Here are some related questions that might interest you:

1. What is the Laplace transform of \(\sinh(t)\)?
2. How do you find the inverse Laplace transform of \(\frac{s}{s^2 - 1}\)?
3. Can you show the Laplace transform of \(e^{at}\cosh(bt)\)?
4. What are the properties of the Laplace transform?
5. How can the Laplace transform be used to solve differential equations?
6. What is the significance of the region of convergence in Laplace transforms?
7. How does the Laplace transform handle discontinuous functions?
8. What are some common applications of the Laplace transform in engineering?

**Tip:** Remember that the Laplace transform is particularly useful for converting differential equations into algebraic equations, which are easier to solve.

Explore the Laplace transform of cosh(t) = s/(s^2 - 1) derived using the definition of Laplace transform. Understand the region of convergence for this transform and its significance in integral calculus. Suitable for advanced undergraduate students studying Laplace transforms and hyperbolic functions.

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