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1. Laplace Transform of f(t)=∣sin⁡t∣f(t) = |\sin t|:

The given function f(t)=∣sin⁡t∣f(t) = |\sin t| is periodic with a period T=πT = \pi. The Laplace transform for a periodic function f(t)f(t) is given by:

L{f(t)}=11−e−sT∫0Tf(t)e−st dt\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sT}} \int_0^T f(t)e^{-st} \, dt

Here, T=πT = \pi. Substitute ∣sin⁡t∣|\sin t| for f(t)f(t) into the formula:

L{∣sin⁡t∣}=11−e−sπ∫0π∣sin⁡t∣e−st dt\mathcal{L}\{|\sin t|\} = \frac{1}{1 - e^{-s\pi}} \int_0^\pi |\sin t| e^{-st} \, dt

Since sin⁡t\sin t is non-negative on [0,π][0, \pi], ∣sin⁡t∣=sin⁡t|\sin t| = \sin t on this interval. So:

L{∣sin⁡t∣}=11−e−sπ∫0πsin⁡t e−st dt\mathcal{L}\{|\sin t|\} = \frac{1}{1 - e^{-s\pi}} \int_0^\pi \sin t \, e^{-st} \, dt

To compute ∫0πsin⁡t e−st dt\int_0^\pi \sin t \, e^{-st} \, dt, we use integration by parts. Let:

  • u=sin⁡tu = \sin t, so du=cos⁡t dtdu = \cos t \, dt,
  • dv=e−st dtdv = e^{-st} \, dt, so v=e−st−sv = \frac{e^{-st}}{-s}.

The integration becomes:

∫0πsin⁡t e−st dt=[sin⁡t e−st−s]0π−∫0πcos⁡t e−st−s dt\int_0^\pi \sin t \, e^{-st} \, dt = \left[ \frac{\sin t \, e^{-st}}{-s} \right]_0^\pi - \int_0^\pi \frac{\cos t \, e^{-st}}{-s} \, dt

Solve this iteratively (details can be expanded upon request). Finally, substitute the result back into the Laplace transform expression.

2. Express f(t)f(t) in terms of the unit step function and find the Laplace Transform:

Given f(t)f(t):

\begin{cases} 0, & 0 < t < \pi, \\ \cos(2t), & \pi < t < 2\pi, \\ 0, & t > 2\pi \end{cases}$$ We use the unit step function $$u(t - a)$$, which is defined as: $$u(t - a) = \begin{cases} 0, & t < a, \\ 1, & t \geq a \end{cases}$$ Rewriting $$f(t)$$ in terms of $$u(t)$$: $$f(t) = \cos(2t)[u(t - \pi) - u(t - 2\pi)]$$ #### Laplace Transform: Using the shifting property of Laplace transform for $$u(t - a) \cdot g(t - a)$$: $$\mathcal{L}\{u(t - a)g(t - a)\} = e^{-as} \mathcal{L}\{g(t)\}$$ The Laplace transform of $$f(t)$$ is: $$\mathcal{L}\{f(t)\} = \mathcal{L}\{\cos(2t) \cdot u(t - \pi)\} - \mathcal{L}\{\cos(2t) \cdot u(t - 2\pi)\}$$ Using the property above for each term and the Laplace transform of $$\cos(2t)$$, $$\mathcal{L}\{\cos(2t)\} = \frac{s}{s^2 + 4}$$, compute the Laplace transform. --- ### **3. Prove $$\int_0^\infty e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}$$:** The integral $$\int_0^\infty e^{-t^2} \, dt$$ is related to the Gaussian integral: $$\int_{-\infty}^\infty e^{-t^2} \, dt = \sqrt{\pi}$$ Since the function $$e^{-t^2}$$ is symmetric about $$t = 0$$, the integral over $$[0, \infty)$$ is half of the integral over $$(-\infty, \infty)$$. Thus: $$\int_0^\infty e^{-t^2} \, dt = \frac{1}{2} \int_{-\infty}^\infty e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}$$ --- ### Do you want detailed calculations for any part of this? Here are **5 related questions** to deepen your understanding: 1. How can periodic functions be simplified using symmetry for Laplace transforms? 2. What are the steps to compute the Laplace transform of piecewise functions? 3. How does integration by parts help in solving Laplace transform integrals? 4. Can we generalize the Gaussian integral for different powers of $$t$$? 5. What is the connection between Laplace transforms and Fourier transforms for periodic functions? **Tip:** Always check the domain of your function carefully when applying properties like unit step functions or symmetry.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Periodic Functions
Piecewise Functions
Unit Step Function
Integration by Parts
Gaussian Integral

Formulas

\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sT}} \int_0^T f(t)e^{-st} \, dt
\mathcal{L}\{u(t - a)g(t - a)\} = e^{-as} \mathcal{L}\{g(t)\}
\int_0^\infty e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}

Theorems

Laplace Transform of Periodic Functions
Shifting Property of Laplace Transforms
Gaussian Integral

Suitable Grade Level

Grades 11-12

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