Originally Posted by
Hollysti This problem looks a lot like the one I posted yesterday, but I am still not sure that I have the hang of how to do these.
Using only the definition fo the Laplace transform, and showing your work, find the laplace transform of the following:
e^(t)sin((pi)t)
Thanks in advance for any help!
L{e^(t)(sin(pit))} = INT{0,inf} e^(st)*e^(t)(sin(pi*t)) dt
= INT{0,inf} e^[(s + 1)t]sin(pi*t) dt
For this, we need to use byparts integration. I'm going to do it the quick way by using a table:
Code:
u = e^[(s + 1)t]  dv = sin(pi*t) dt

(s + 1)e^[(s + 1)t]  (1/pi)cos(pi*t)

(s + 1)^2e^[(s + 1)t]  (1/pi^2)sin(pi*t)
Now, we multiply each function on each downwardsloping diagnal, adding each diagnal while switching the signs of each diagnal from + to , starting with +. For now I'm going to ignore the limits of integration.
INT e^[(s + 1)t]sin(pi*t) dt = ((1/pi)cos(pi*t))e^[(s + 1)t]  ((1/pi^2)sin(pi*t))((s + 1)e^[(s + 1)t]) + INT (s + 1)^2e^[(s + 1)t]((1/pi^2)sin(pi*t)) dt
INT e^[(s + 1)t]sin(pi*t) dt = (1/pi)cos(pi*t)*e^[(s + 1)t]  ((s + 1)/pi^2)sin(pi*t)e^[(s + 1)t]  ((s + 1)/pi)^2*INT e^[(s + 1)t]sin(pi*t) dt
[((s + 1)/pi)^2 + 1]*INT e^[(s + 1)t]sin(pi*t) dt = (1/pi)cos(pi*t)*e^[(s + 1)t]  ((s + 1)/pi^2)sin(pi*t)e^[(s + 1)t]
Plugging back in the limits of integration {0,inf}:
(1/pi)cos(pi*t)*e^[(s + 1)t]  ((s + 1)/pi^2)sin(pi*t)e^[(s + 1)t]
[0 + 0]  [1/pi  0] = 1/pi
[((s + 1)/pi)^2 + 1]*INT e^[(s + 1)t]sint dt = 1/pi
INT e^[(s + 1)t]sin(pi*t) dt = 1/{pi[((s + 1)/pi)^2 + 1]} = pi/[(s + 1)^2 + pi^2]