Two things
1. the deniminator of your antiderivative should be s^2 + a^2
2. You nedd to evaluate this between 0 and infinity
See attachment
Compute the Laplace transform of , where are real numbers.
I've tried this several times and I am ending up with an answer that I know is wrong. I seem to be making the same mistake over and over.
I set up the problem like this:
and then use integration by parts.
Judging by WolframAlpha my answer should be
but I keep getting answers similar to this:
Help would be greatly appreciated.
Thanks so much for the help!
I'm confused about two things with your attached document.
When evaluating from 0 to infinity:
1. Why does the first part of the evaluation (with infinity) simply get discarded?
2. When evaluating 0 with , why does it become and not
I attatched a picture to help explain where I am confused.
neglecting the constants lim e^(-t) cos(t) or lim e^(-t) sin(t) are 0
This comes from squeezing theorem - e^(-t) < e^(-t) cos(t) < e^(-t)
lim -e^(-t) = lime^(-t) = 0
Remember your under damped harmonic oscillator
2. you have - s cos(at-b)
a. because you are evaluating at the lower limit you subtract
you have scos(-b)
b cos(-b) = cos(b) you have scos(b)
Well, that can't be right because the Laplace transform is a function of s only, not t. Did you not evaluate at s= 0 and ?
To integrate by parts, let and . Then and . .
To do [/b]that[/b] integral, let again and let . Then and [tex]v= -\frac{1}{a}cos(at- b). .
Putting that into the first integral, .
Adding that integral on the right to the left side gives so .
That's the basic idea- better check for errors.
Alternatively
cos(at-b) = cos(at)cos(b) + sin(at)sin(b)
We can use the basic formulas for cos(at) and sin(at)
L{cos(at-b)} = L{cos(at)cos(b)} + L {sin(at)sin(b)}
= cos(b)s/(s^2+a^2) + asin(b)/(s^2+a^2)
Gives the correct Laplace transform -- the attachment of the post
previous to this one does so using the definition of laplace transform