Find the Laplace transform L{f(t)} where f(t) = { sint, 0≤t
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![## Laplace Transform of a Piecewise Function
In this example, we are asked to find the Laplace transform \(\mathcal{L}\{f(t)\}\) of the function \( f(t) \) defined as:
\[
f(t) = \begin{cases}
\sin(t), & 0 \le t < \pi \\
0, & t \ge \pi
\end{cases}
\]
### Step-by-Step Solution:
The Laplace transform of a function \( f(t) \) is given by the integral:
\[ \mathcal{L}\{f(t)\} = \int_0^\infty f(t)e^{-st} \, dt \]
Where \( s \) is a complex number such that the integral converges.
### Given Function Breakdown
1. For \( 0 \le t < \pi \), \( f(t) = \sin(t) \).
2. For \( t \ge \pi \), \( f(t) = 0 \).
### Applying the Laplace Transform to Each Interval
**Interval 1: \( 0 \le t < \pi \)**
\[
\begin{aligned}
&\mathcal{L}\{f(t)\text{ on } [0,\pi)\} = \int_0^\pi \sin(t)e^{-st} \, dt
\end{aligned}
\]
**Interval 2: \( t \ge \pi \)**
Since \( f(t) = 0 \) for \( t \ge \pi \),
\[
\begin{aligned}
&\mathcal{L}\{f(t)\text{ on } [\pi,\infty)\} = \int_\pi^\infty 0 \cdot e^{-st} \, dt = 0
\end{aligned}
\]
Combining both intervals, we integrate only on \( 0 \leq t < \pi \):
\[
\begin{aligned}
\mathcal{L}\{f(t)\} &= \int_0^\pi \sin(t)e^{-st} \, dt
\end{aligned}
\]
### Evaluating the Integral
To solve \( \int_0^\pi \sin(t)e^{-st} dt \), we use integration by parts or recognize it as a standard Laplace transform of \( \sin(t) \) multiplied](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1d700aa-072a-4ff7-be1d-cdca0f10aa9d%2F2c655ce6-386a-4bbd-8350-7b69a415440e%2F8t3u95l_processed.png&w=3840&q=75)
Transcribed Image Text:## Laplace Transform of a Piecewise Function
In this example, we are asked to find the Laplace transform \(\mathcal{L}\{f(t)\}\) of the function \( f(t) \) defined as:
\[
f(t) = \begin{cases}
\sin(t), & 0 \le t < \pi \\
0, & t \ge \pi
\end{cases}
\]
### Step-by-Step Solution:
The Laplace transform of a function \( f(t) \) is given by the integral:
\[ \mathcal{L}\{f(t)\} = \int_0^\infty f(t)e^{-st} \, dt \]
Where \( s \) is a complex number such that the integral converges.
### Given Function Breakdown
1. For \( 0 \le t < \pi \), \( f(t) = \sin(t) \).
2. For \( t \ge \pi \), \( f(t) = 0 \).
### Applying the Laplace Transform to Each Interval
**Interval 1: \( 0 \le t < \pi \)**
\[
\begin{aligned}
&\mathcal{L}\{f(t)\text{ on } [0,\pi)\} = \int_0^\pi \sin(t)e^{-st} \, dt
\end{aligned}
\]
**Interval 2: \( t \ge \pi \)**
Since \( f(t) = 0 \) for \( t \ge \pi \),
\[
\begin{aligned}
&\mathcal{L}\{f(t)\text{ on } [\pi,\infty)\} = \int_\pi^\infty 0 \cdot e^{-st} \, dt = 0
\end{aligned}
\]
Combining both intervals, we integrate only on \( 0 \leq t < \pi \):
\[
\begin{aligned}
\mathcal{L}\{f(t)\} &= \int_0^\pi \sin(t)e^{-st} \, dt
\end{aligned}
\]
### Evaluating the Integral
To solve \( \int_0^\pi \sin(t)e^{-st} dt \), we use integration by parts or recognize it as a standard Laplace transform of \( \sin(t) \) multiplied
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