Helsaint wrote:

Sin^2(t) = 1/2 - cos2t

That is not correct.

]]>Sin^2(t) = 1/2 - cos2t

L{sin^2(t)} = 1/2s - s/(s^2+4)]]>

Welcome to the forum. That is not correct.

]]>I get that the laplace transform of sin^2t = -(sin^2te^-st)/s + 2/s^3+4s evaluated from 0...infinity.

when I evaluate the limit from 0..infinity I get that the transform to equal 0. Did I evaluate that right?

]]>There are diiferent definitions for a fourier transform, that page will partly explain that.

]]>I just tried the Fourier transform of f(x) = 1 and got

... is that correct? I'll check on Wolfram.]]>There are FFT and DFT's. Wikipedia can be a horror story at times. To me that is exactly what that is saying.

I have never seen their notation. They are using small f with a cap ( borrowed from statistics)

]]>Does this mean that if I put in some function of x, such as sin(x), I'll get f(ξ) where ξ is a real number? Not sure, I'll post my working in a second. Sorry if I sound stupid...

]]>Zeroing the LHS will leave you with just the Laplace term. That should be your answer.

I was just asking to see what you thought about it. Since t approaches infinity it will drown out s no matter how small as long as s > 0.

That is nice, spotting the Laplace Transform there.

In addition zetafunc, welcome to the forum! Why not consider becoming a member here?

]]>What I meant is that I get

Then evaluate RHS at 0 and subtract that from the evaluation at infinity. I got 0... so then we have

Therefore

assuming s > 0.I also tried the Laplace transformation for sin[sup]a[/sup](t) and got

.]]>I am glad to help but we are not done yet.

The LHS has to be evaluated at infinity and then you subtract the evaluation of it at 0. The RHS is untouched.

How are you getting 0 for the LHS?

If s is very small then the LHS is not zero. Were you given some interval for s?

Thanks for the response again and confirming that my IBP was correct -- I think I get it now - subtract the rightmost term from both sides to get y(s) - 2/(s[sup]3[/sup] + 4s), evaluate the RHS at 0 and infinity to get 0 (0 - 0 = 0), then add 2/(s[sup]3[/sup] + 4s) to both sides to get the completed Laplace transform? Is that correct? Phew, thanks for your help.

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