Laplace Transforms

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.2 – Definition of the Laplace Transforms

Expected Educational Results

• Objective 4–1: I can evaluate an indefinite trigonometric integral.
• Objective 4–2: I can evaluate a definite trigonometric integral using FTC-II.

Improper Integrals

The Tanzalin Method for Integration by Parts

Example 01a: Integrate $\displaystyle\int{t^2 e^{-st} dt}$

Solution:

Multiply along the diagonals and multiply with sign along diagonals.

$\displaystyle\int{t^2 e^{-st} dt}= -t^2 \dfrac{e^{-st}}{s} - 2t \dfrac{e^{-st}}{s^2} - 2 \dfrac{e^{-st}}{s^3} + C$

Improper Integrals

Example 01b: Integrate $\displaystyle\int_0^{\infty}{t^2 e^{-st} dt}$

Solution:

$\displaystyle\int_0^{\infty}{t^2 e^{-st} dt}=\lim_{b\rightarrow\infty}{\int_0^{b}{t^2 e^{-st} \,dt}}= \lim_{b\rightarrow\infty}{\left(-b^2 \dfrac{e^{-sb}}{s} - 2b \dfrac{e^{-sb}}{s^2} - 2 \dfrac{e^{-sb}}{s^3}-\left[-0^2\dfrac{e^0}{s}-2(0)\dfrac{e^0}{s^3}-2\dfrac{e^0}{s^3}\right]\right)}$

Rewrite all terms:

$\displaystyle\Rightarrow\lim_{b\rightarrow\infty}{\left(-\dfrac{b^2}{se^{sb}} - \dfrac{2b}{s^2e^{sb}} - 2 \dfrac{1}{s^3e^{-sb}}-\left[0-0-\dfrac{2}{s^3}\right]\right)}$

Use l’Hopital’s Rule to evaluate the first two limits:

$\displaystyle\Rightarrow\lim_{b\rightarrow\infty}{\left(-\dfrac{2}{s(s^2e^{sb})} - \dfrac{2}{s^2(se^{sb})} - 2 \dfrac{1}{s^3e^{-sb}}\right)+\dfrac{2}{s^3}}$

Evaluate the limits:

$\displaystyle\Rightarrow \left(0 - 0 - 0\right)+\dfrac{2}{s^3}=\dfrac{2}{s^3}$

Example 02a: Integrate $\displaystyle\int{\cos{(t)} e^{-st} dt}$

Multiply along the diagonals and multiply with sign along diagonals. Multiply along the the last horizontal and integrate the product.

$\displaystyle\int{\cos{(t)} e^{-st} dt}= -\cos{(t)} \dfrac{e^{-st}}{s} + \sin{(t)} \dfrac{e^{-st}}{s^2} - \dfrac{1}{s^2}\int{\cos{(t)} e^{-st}\,dt}$

Combine the integrals:

$\displaystyle \left(1+\dfrac{1}{s^2}\right)\int{\cos{(t)} e^{-st} dt} = \left(\dfrac{s^2+1}{s^2}\right)\int{\cos{(t)} e^{-st} dt} = -\cos{(t)} \dfrac{e^{-st}}{s} + \sin{(t)} \dfrac{e^{-st}}{s^2}$

Solve for the integral:

$\displaystyle \int{\cos{(t)} e^{-st} dt} = \left(\dfrac{s^2}{s^2+1}\right) \left(-\cos{(t)} \dfrac{e^{-st}}{s} + \sin{(t)} \dfrac{e^{-st}}{s^2}\right) + C$

Improper Integrals

Example 02b: Integrate $\displaystyle\int_0^{\infty}{\cos{(t)} e^{-st} dt}$

Solution:

$\displaystyle\int_0^{\infty}{\cos{(t)} e^{-st} dt}=\lim_{b\rightarrow\infty}{\int_0^{b}{\cos{(t)} e^{-st} \,dt}}$

$\displaystyle\Rightarrow \lim_{b\rightarrow\infty}{\left[\left(\dfrac{s^2}{s^2+1}\right) \left(-\cos{(b)} \dfrac{e^{-sb}}{s} + \sin{(b)} \dfrac{e^{-sb}}{s^2}-\left(-\cos{(0)} \dfrac{e^{0}}{s} + \sin{(0)} \dfrac{e^{0}}{s^2}\right)\right)\right]}$

Rewrite all terms:

$\displaystyle\Rightarrow \left(\dfrac{s^2}{s^2+1}\right)\lim_{b\rightarrow\infty}{\left(\dfrac{\cos{(b)}}{se^{sb}} + \dfrac{\sin{(b)}}{s^2e^{sb}} + \dfrac{1}{s} + 0\right)}$

Evalaute the limit:

$\displaystyle\Rightarrow \left(\dfrac{s^2}{s^2+1}\right)\lim_{b\rightarrow\infty}{\left(0 + 0 + \dfrac{1}{s}\right)}=\dfrac{s^2}{s^2+1}\left(\dfrac{1}{s}\right)=\dfrac{s}{s^2+1}$

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Last Modified: Sunday, November 2020 8:59 EDT