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

  • Thread starter Thread starter Jerimy240
  • Start date Start date
  • Tags Tags
    Interval
Jerimy240
Messages
2
Reaction score
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?
 
Physics news on Phys.org
Yes, you are integrating
\int_0^{\pi} e^{-st}sin(t)dt
and if you use integration by parts with you will, after a couple of integrations get something like
\int_0^{\pi} e^{-st}sin(t)dt= F(s)- C\int_0^\pi e^{-st}sin(t)dt

Now add that integral to both sides:
(1+ C)\int_0^{\pi} e^{-st}sin(t)dt= F(s)
 
Thanks HOI, the final answer is (1+e^-spie)/(s^2 + 1) I don't know how to get there
 
Do one more integration by parts and then look at HallsofIvy's suggestion again.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top