Laplace Transform of Integrals & Special Functions

Find the Laplace transforms of

1. \( L\left\{ \int_{0}^{t} \frac{\sin(t)}{t} \, dt \right\} \)

Solution: \[ L\left\{ \int_{0}^{t} \frac{\sin(t)}{t} \, dt \right\} = \frac{1}{s} L\left\{ \frac{\sin(t)}{t} \right\} \] Using the Laplace transform of \( \frac{\sin(t)}{t} \): \[ L\left\{ \frac{\sin(t)}{t} \right\} = \int_{s}^{\infty} \frac{1}{s^2 + 1} \, ds = \tan^{-1}(\infty) - \tan^{-1}(s) = \frac{\pi}{2} - \tan^{-1}(s) \] Therefore: \[ L\left\{ \int_{0}^{t} \frac{\sin(t)}{t} \, dt \right\} = \frac{1}{s} \left( \frac{\pi}{2} - \tan^{-1}(s) \right) = \frac{1}{s} \cot^{-1}(s) \]

2. \( L\left\{ \int_{0}^{t} e^{-t} \cos(t) \, dt \right\} \)

Solution: \[ L\left\{ \int_{0}^{t} e^{-t} \cos(t) \, dt \right\} = \frac{1}{s} L\left\{ e^{-t} \cos(t) \right\} \] Using the Laplace transform of \( e^{-t} \cos(t) \): \[ L\left\{ e^{-t} \cos(t) \right\} = \left[ \frac{s}{s^2 + 1} \right]_{s \to (s + 1)} = \frac{s + 1}{(s + 1)^2 + 1} \] Therefore: \[ L\left\{ \int_{0}^{t} e^{-t} \cos(t) \, dt \right\} = \frac{1}{s} \cdot \frac{s + 1}{(s + 1)^2 + 1} = \frac{s + 1}{s(s^2 + 2s + 2)} \]

3. \( L\left\{ \int_{0}^{t} \frac{e^{t} \sin(t)}{t} \, dt \right\} \)

Solution: \[ L\left\{ \int_{0}^{t} \frac{e^{t} \sin(t)}{t} \, dt \right\} = \frac{1}{s} L\left\{ \frac{e^{t} \sin(t)}{t} \right\} \] Using the Laplace transform of \( \frac{e^{t} \sin(t)}{t} \): \[ L\left\{ \frac{e^{t} \sin(t)}{t} \right\} = \left[ \int_{s}^{\infty} \frac{1}{s^2 + 1} \, ds \right]_{s \to (s - 1)} = \left[ \tan^{-1}(s) \right]_{s \to (s - 1)} \] Therefore: \[ L\left\{ \int_{0}^{t} \frac{e^{t} \sin(t)}{t} \, dt \right\} = \frac{1}{s} \cot^{-1}(s - 1) \]

Model Problems

Evaluate the following Laplace transforms:

1. \[ L\left\{ e^{-t} \int_{0}^{t} \frac{\sin(t)}{t} \, dt \right\} \]
2. \[ L\left\{ t \int_{0}^{t} \frac{e^{-t} \sin(t)}{t} \, dt \right\} \]
3. \[ L\left\{ t \int_{0}^{t} e^{-t} \sin(4t) \, dt \right\} \]
4. \[ L\left\{ t \int_{0}^{t} \frac{1 - e^{-2t}}{t} \, dt \right\} \]
5. \[ L\left\{ t \int_{0}^{t} \int_{0}^{t} \int_{0}^{t} t \sin(t) \, dt \, dt \, dt \right\} \]

Answers

1. \[ \frac{1}{s + 1} \cot^{-1}(s + 1) \]
2. \[ \frac{s + (s^2 + 2s + 2) \cot^{-1}(s + 1)}{s^2 (s^2 + 2s + 2)} \]
3. \[ \frac{8(s + 1)}{s^2 (s^2 + 2s + 17)^2} \]
4. \[ \frac{1}{s} \log\left(1 + \frac{2}{s}\right) \]
5. \[ \frac{2}{s^2 (s^2 + 1)^2} \]

Dirac’s Delta Function (Unit Impulse Function):

Define the function: \[ f_{\epsilon}(t - a) = \begin{cases} 0, & t < a, \\ \frac{1}{\epsilon}, & a \le t \le a + \epsilon, \\ 0, & t > a + \epsilon. \end{cases} \] By definition: \[ L\{ f_{\epsilon}(t - a) \} = \int_{0}^{\infty} e^{-st} f_{\epsilon}(t - a) \, dt = \int_{0}^{a} e^{-st} (0) \, dt + \int_{a}^{a + \epsilon} e^{-st} \frac{1}{\epsilon} \, dt + \int_{a + \epsilon}^{\infty} e^{-st} (0) \, dt \] Simplifying: \[ L\{ f_{\epsilon}(t - a) \} = \frac{1}{\epsilon} \left[ \frac{e^{-st}}{-s} \right]_{a}^{a + \epsilon} = \frac{e^{-as} (1 - e^{-\epsilon s})}{\epsilon s} \] The Dirac Delta function is defined as: \[ \delta(t - a) = \lim_{\epsilon \to 0} f_{\epsilon}(t - a) \] Laplace Transform of Dirac Delta Function: \[ L\{ \delta(t - a) \} = \lim_{\epsilon \to 0} L\{ f_{\epsilon}(t - a) \} = \lim_{\epsilon \to 0} \frac{e^{-as} (1 - e^{-\epsilon s})}{\epsilon s} \] Using L'Hôpital's rule: \[ L\{ \delta(t - a) \} = e^{-as} \lim_{\epsilon \to 0} \frac{0 + s e^{-\epsilon s}}{s} = e^{-as} \lim_{\epsilon \to 0} e^{-\epsilon s} = e^{-as} \]

Periodic Function:

A function \( f(t) \) is said to be periodic with period \( T > 0 \) if: \[ f(t) = f(T + t) = f(T + 2t) = \dots = f(T + nt) \] Examples:

• \( \sin(t) \) and \( \cos(t) \) are periodic functions with period \( 2\pi \)
• \( \tan(t) \) and \( \cot(t) \) are periodic functions with period \( \pi \)

Laplace Transform of Periodic Function: If \( f(t) \) is a periodic function with period \( T > 0 \), then: \[ L\{ f(t) \} = \frac{1}{1 - e^{-sT}} \int_{0}^{T} f(t) e^{-st} \, dt \]

Example 1: Laplace Transform of a Square-Wave Function.
Find the Laplace transform of the square-wave function of period \( 2a \) defined as: \[ f(t) = \begin{cases} k, & 0 < t < a, \\ -k, & a < t < 2a. \end{cases} \]

Solution: Given that \( f(t) \) is a periodic function with period \( T = 2a \), the Laplace transform is: \[ L\{ f(t) \} = \frac{1}{1 - e^{-sT}} \int_{0}^{T} f(t) e^{-st} \, dt \] Substituting \( T = 2a \): \[ L\{ f(t) \} = \frac{1}{1 - e^{-2as}} \left[ \int_{0}^{a} k e^{-st} \, dt + \int_{a}^{2a} (-k) e^{-st} \, dt \right] \] Evaluating the integrals: \[ L\{ f(t) \} = \frac{1}{1 - e^{-2as}} \left[ \left( \frac{k e^{-st}}{-s} \right)_{0}^{a} - \left( \frac{k e^{-st}}{-s} \right)_{a}^{2a} \right] \] Simplifying: \[ L\{ f(t) \} = \frac{k}{s(1 - e^{-2as})} \left[ 1 - 2e^{-as} + e^{-2as} \right] \] Further simplification: \[ L\{ f(t) \} = \frac{k(1 - e^{-as})}{s(1 + e^{-as})} = \frac{k}{s} \tanh\left( \frac{as}{2} \right) \]

Example 2: Laplace Transform of a Full-Wave Rectifier.
Find the Laplace transform of the full-wave rectifier: \[ f(t) = E \sin(wt), \quad 0 < t < \frac{\pi}{w}, \] which has period \( T = \frac{\pi}{w} \)

Solution: Given that \( f(t) = E \sin(wt) \) is a periodic function with period \( T = \frac{\pi}{w} \), the Laplace transform is: \[ L\{ f(t) \} = \frac{1}{1 - e^{-sT}} \int_{0}^{T} e^{-st} f(t) \, dt \] Substituting \( T = \frac{\pi}{w} \): \[ L\{ f(t) \} = \frac{1}{1 - e^{-\pi s / w}} \int_{0}^{\pi / w} e^{-st} E \sin(wt) \, dt \] Evaluating the integral: \[ L\{ f(t) \} = \frac{E}{1 - e^{-\pi s / w}} \left[ \frac{e^{-st}}{s^2 + w^2} (-s \sin(wt) - w \cos(wt)) \right]_{0}^{\pi / w} \] Simplifying: \[ L\{ f(t) \} = \frac{E}{1 - e^{-\pi s / w}} \left[ \frac{e^{-\pi s / w} w}{s^2 + w^2} + \frac{w}{s^2 + w^2} \right] \] Further simplification: \[ L\{ f(t) \} = \frac{E w}{s^2 + w^2} \left[ \frac{1 + e^{-\pi s / w}}{1 - e^{-\pi s / w}} \right] = \frac{E w}{s^2 + w^2} \coth\left( \frac{\pi s}{2w} \right) \]

Model Problems

1. Find the Laplace transform of the rectified semi-wave function defined by: \[ f(t) = \begin{cases} \sin(wt), & 0 < t < \frac{\pi}{w}, \\ 0, & \frac{\pi}{w} < t < \frac{2\pi}{w} \end{cases} \]
2. Draw the graph of the periodic function: \[ f(t) = \begin{cases} t, & 0 < t < \pi, \\ \pi - t, & \pi < t < 2\pi. \end{cases} \] and find its Laplace transform.
3. Find the Laplace transform of the triangular wave function of period \( 2a \) given by: \[ f(t) = \begin{cases} t, & 0 < t < a, \\ 2a - t, & a < t < 2a. \end{cases} \]

Answers

1. \[ \frac{w}{(1 - e^{-\pi s / w})(s^2 + w^2)} \]
2. \[ \frac{1 - e^{-\pi s} - \pi s e^{-\pi s}}{s^2 (1 - e^{-\pi s})} \]
3. \[ \frac{1}{s^2} \tanh\left( \frac{as}{2} \right) \]