Given that $$ \int_0^\infty xY(x)\sin(xt)dx = F(t) = \begin{cases} 1 & 0<t\leq 1 \\ 2 & 1<t\leq 2 \end{cases} $$ The goal is to find $Y(0)$.
I'm trying to apply the inverse Fourier sine transformation:
$$ F(x) = xY(x)= \frac{2}{\pi}\int_0^1\sin(tx)dt+\frac{2}{\pi}\int_1^2 2\sin(tx)dt\\ = \frac{2\cos(x)-4\cos(2x)+2}{\pi x} $$ Then we can find $Y(x)$. However, it seems like $Y(0)$ is undefined. Am I on the right track to solve this problem? Thanks for the help!
In the integral $\int_0^\infty xY(x)\sin(xt)dx,$ changing $Y(0)$'s value won't affect the integral. So you need to use continuity to define $Y$'s value at zero. (I mean that, if you don't assume continuity, you can never get $Y(0)$ as it can be any value.)
You've proved that $Y(x)=\frac{2\cos(x)-4\cos(2x)+2}{\pi x^2},$ so $$ Y(0)=\lim_{x\rightarrow 0+}\frac{2}{\pi}\frac{\cos(x)+1-2\cos(2x)}{x^2}=\frac{2}{\pi}\frac{-1+8}{2}=\frac{7}{\pi}. $$