Can someone help me with the example of Fourier transform
$$\mathcal{F}[\sin(ln(t))u(t)](\omega)\ = \ \ ?$$
I know that there is a formula in the Table of Fourier Transform Pairs but it is only for $\sin(\omega_0 t)u(t)$, but how I deal with the natural logarithm inside the $\sin$ ?
yes sorry for my bad english is not my first language, thanks for the corrections
It is complicated. For any $a \in \mathbb{C}, Re(a) > -1$ we have the Laplace transform : $$\mathcal{L}[e^{a \ln t} 1_{t > 0}](s) =\mathcal{L}[t^a 1_{t > 0}](s)=\int_0^\infty t^a e^{-st}dt = s^{-a-1}\int_0^\infty u^a e^{-u}du=s^{-a-1} \Gamma(a+1)$$ where for $s > 0$ I applied the change of variable $u = st$ and I extended to $s \in \mathbb{C}, Re(s) > 0$ by analytic continuation, and $\Gamma$ is the Gamma function.
Hence $$\mathcal{L}[\sin(\ln t) 1_{t > 0}](s) =\mathcal{L}[\frac{e^{i \ln t}-e^{-i \ln t}}{2i} 1_{t > 0}](s)=\frac{s^{-i-1}\Gamma(1+i)-s^{i-1}\Gamma(1-i)}{2i} $$ and in the sense of distributions $$\mathcal{F}[\sin(\ln t) 1_{t > 0}](\omega)=\lim_{\sigma\to 0^+}\mathcal{L}[\sin(\ln t) 1_{t > 0}](\sigma+i\omega)$$ $$=p.v.\frac{(i\omega)^{-i-1}\Gamma(1+i)-(i\omega)^{i-1}\Gamma(1-i)}{2i}$$ where $p.v.$ is the principal value