Laplace transform, with integral

Click For Summary
SUMMARY

The discussion focuses on evaluating the definite Laplace transform of the function f(t) = sin(t) over the interval [0, π]. The integral to solve is ∫₀^π e^(-st) sin(t) dt, which requires the application of integration by parts (IBP). The user initially sets u = sin(t) and dV = e^(-st) dt, leading to a recursive situation where the integration by parts alternates between sine and cosine. The key insight provided is to rearrange the integral after returning to sine, allowing for a solvable equation.

PREREQUISITES
  • Understanding of Laplace transforms
  • Proficiency in integration by parts (IBP)
  • Familiarity with definite integrals
  • Knowledge of exponential functions and their properties
NEXT STEPS
  • Study the method of integration by parts in detail
  • Learn how to solve Laplace transforms of trigonometric functions
  • Explore techniques for solving recursive integrals
  • Investigate the properties of the Laplace transform, specifically linearity and shifting
USEFUL FOR

Students studying differential equations, mathematicians working with Laplace transforms, and educators teaching integral calculus concepts.

arenaninja
Messages
25
Reaction score
0

Homework Statement


I'm trying to solve a definite Laplace transform. The function is defined as [tex]f(t) = sin(t)[/tex] over the interval [tex][0, \pi][/tex]

Homework Equations


The integrate to evaluate is:
[tex]\int_0^\pi e^{-st} sin(t) dt[/tex]


The Attempt at a Solution


To evaluate, first use integration by parts (IBP).
[tex] \begin{matrix}<br /> u = sin(t) \quad dV = e^{-st} dt\\<br /> du = cos(t) dt \quad V = \frac{e^{-st}}{-s}<br /> \end{matrix}[/tex]

[tex]\int_0^\pi e^{-st} sin(t) dt = \frac{sin(t) e^{-st}}{-s} + \frac{1}{s} \int_0^\pi e^{-st} cos(t) dt[/tex]
However, I'm stuck here. I can try to keep evaluating by parts, but it looks to me like I'm stuck in a loop. Integrating by parts will alternate me between sine and cosine, and the only thing that will change will be the increasing power for the "s" in the denominator.

Any help is greatly appreciated.
 
Physics news on Phys.org
The trick is when you get back to the sine, move the integral to the LHS, so you end up with

[tex](\textrm{some stuff}) \int_0^\pi e^{-st}\sin t\, dt = \textrm{other stuff}[/tex]
 

Similar threads

Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K