(Fourier Transform) I'm trying to take the Fourier Sine Transform of a simple function, $$ te^{-t} $$, and I'm wondering whether there are any elementary identities I should be using or any clever "tricks", because right now, the process just looks like brutal integration by parts. Am I missing something?

Question

mendicant_bias · Answer

@dan815

mendicant_bias · Answer

$$F_s(\omega)=\int\limits_{0}^{\infty}f(x)\sin(\omega x)dx=\int\limits_{0}^{\infty}te^{-t}\sin{(\omega t)}dt$$

anonymous · Answer

You can try this equivalent expression for sine:
$$\sin\omega t=\frac{e^{i\omega t}-e^{-i\omega t}}{2i}$$
That won't prevent integrating by parts, but it lightens the load a bit.

mendicant_bias · Answer

Yeah, that's a good idea. That still appears to simplify things by quite a bit, thanks. I also feel like I haven't seen that formula before. It looks like a hyperbolic identity, but this is just sine. Hm.

anonymous · Answer

It's derived from Euler's identity.
$$\large e^{\pm i kx}=\cos kx\pm i\sin kx$$

anonymous · Answer

*formula, not identity

irishboy123 · Answer

that's a LP transform:  s = 1, so

$$\frac {2 \omega}{(1 + \omega ^ 2)^2}$$

mendicant_bias · Answer

LP transform? I don't know what that is. Is that some short sign for Laplace Transform? I don't know how the Laplace Transform and the Fourier Transform relate, but they are not identical, so their formulas can't be treated the same, right?

irishboy123 · Answer

they are related but that's irrelevant here...

...because what you have is the "formula" for the Laplace transform of t sin wt, and you can look that up in table -- you said you wished to  avoid "brutal integration by parts".

mendicant_bias · Answer

I can sense something seething underneath that post, but yeah, that's an excellent point, thank you.

irishboy123 · Answer

@Mendicant_Bias 
my sincere apols if i came across badly. my last post was not my best here.  mea culpa.