How Can I Integrate (sin t)(e^-st) by Parts?

  • Thread starter Jerimy240
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In summary, after integrating (sin t)(e^-st) by parts, the final answer is (1+e^-spie)/(s^2 + 1). However, there is still one more integration by parts step needed to reach this answer.
  • #1
Jerimy240
2
0
f(t)={sin t, 0<t<pie
0, t>pie}

After integrating (sin t)(e^-st) by parts I get

-1/s(sin t)e^-st + 1/s Integral[(e^-st (cos t)dt]

Looks like I'll be integrating forever. I'm I missing something?

Also, is there software you can install to help you type math symbols so I can interact on this forum more efficiently?
 
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  • #2
Yes, you are integrating
[tex]\int_0^{\pi} e^{-st}sin(t)dt[/tex]
and if you use integration by parts with you will, after a couple of integrations get something like
[tex]\int_0^{\pi} e^{-st}sin(t)dt= F(s)- C\int_0^\pi e^{-st}sin(t)dt[/tex]

Now add that integral to both sides:
[tex](1+ C)\int_0^{\pi} e^{-st}sin(t)dt= F(s)[/tex]
 
  • #3
Thanks HOI, the final answer is (1+e^-spie)/(s^2 + 1) I don't know how to get there
 
  • #4
Do one more integration by parts and then look at HallsofIvy's suggestion again.
 

1. What is the purpose of evaluating L{sin t} on the interval 0 to π?

The purpose of evaluating L{sin t} on the interval 0 to π is to find the Laplace transform of the sine function over a specific range of values. This can be useful in solving differential equations and analyzing the behavior of systems in engineering and physics.

2. How is L{sin t} on the interval 0 to π calculated?

The Laplace transform of sin t on the interval 0 to π can be calculated using the integral formula L{sin t} = ∫0π sin t e-st dt, where s is a complex variable.

3. What is the result of evaluating L{sin t} on the interval 0 to π?

The result of evaluating L{sin t} on the interval 0 to π is a complex function that represents the Laplace transform of the sine function over that specific range of values. This result can be used to solve differential equations and analyze the behavior of systems.

4. Can L{sin t} on the interval 0 to π be simplified?

Yes, L{sin t} on the interval 0 to π can be simplified using trigonometric identities and properties of Laplace transforms. For example, using the identity sin t = (eit - e-it) /2, the integral can be rewritten and solved using tables or a computer program.

5. Are there any real-world applications of L{sin t} on the interval 0 to π?

Yes, there are many real-world applications of L{sin t} on the interval 0 to π in fields such as electrical engineering, control systems, and signal processing. For example, it can be used to model the behavior of a mechanical system with a sinusoidal input, or to analyze the response of an electronic circuit to a periodic signal.

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