Math Problem Statement

(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

Solution

To show that L[cosh(t)]=ss21\mathcal{L}[\cosh(t)] = \frac{s}{s^2 - 1} using the definition of the Laplace transform, we start with the definition:

L{f(t)}=0estf(t)dt\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt

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

cosh(t)=et+et2\cosh(t) = \frac{e^t + e^{-t}}{2}

So, we need to compute:

L{cosh(t)}=0estcosh(t)dt=0est(et+et2)dt\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:

L{cosh(t)}=120estetdt+120estetdt\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:

=120e(1s)tdt+120e(s+1)tdt= \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 0eatdt\int_0^\infty e^{at} \, dt, which evaluates to 1a\frac{1}{-a} for (a)<0\Re(a) < 0. So, we need to evaluate the integrals:

  1. For the first integral 120e(1s)tdt\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)}

  1. For the second integral 120e(s+1)tdt\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:

L{cosh(t)}=12(1s1+1s+1)\mathcal{L}\{\cosh(t)\} = \frac{1}{2} \left( \frac{1}{s-1} + \frac{1}{s+1} \right)

Combining the fractions:

L{cosh(t)}=12(s+1+s1(s1)(s+1))=12(2ss21)=ss21\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:

L[cosh(t)]=ss21\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:

(s)>1\Re(s) > 1

Thus, the region of convergence is (s)>1\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)\sinh(t)?
  2. How do you find the inverse Laplace transform of ss21\frac{s}{s^2 - 1}?
  3. Can you show the Laplace transform of eatcosh(bt)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.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Hyperbolic Functions
Integral Calculus

Formulas

Definition of Laplace transform
Integral of exponential functions

Theorems

Region of convergence for Laplace transforms

Suitable Grade Level

Advanced Undergraduate