Let $$F(w) = \int_0^\infty t^{-1/2} \sin (w t)\, dt$$
$$F(w) = w^{-1/2} \int_0^{\infty} w^{-1/2}t^{-1/2} \sin wt \, w dt$$
Let $q=wt$.
$$F(w) = w^{-1/2} \int_0^\infty q^{-1/2} \sin q \, dq = \text{const.}\cdot w^{-1/2}.$$
It is known that $\int_0^\infty q^{-1/2} \sin q \, dq = \sqrt{\frac{\pi}{2}}$, but we don't need to evaluate this integral to show that $t^{-1/2}$ is its own Fourier transform (up-to-a-constant).
UPDATE:
One way to compute $\int_0^\infty q^{-1/2} \sin q\, dq = 2\int_0^\infty \sin p^2 \, dp, $ is to consider the contour integral
$$ \oint_C e^{i z^2} \, dz$$ with C the segment $0$ to $R$ along the $x$-axis, a quarter circle of radius $R$ from $R$ to $z^\star=Re^{i\pi/4}$ and a segment from $z^\star$ to the origin (a slice (1/8) of a pie of radius $R$).
Then $$0=\oint_C e^{i z^2} \, dz = \int_0^\infty e^{ip^2} \, dp - e^{i\pi/4} \int_0^\infty e^{-y^2} \, dy + \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta,$$ and
$$ \begin{aligned} \left| \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta \right| &\le \int_{\Gamma_R} \left| \exp (R^2e^{i2\theta})R i e^{i\theta}\right| \,d\theta \\ &= \int_0^{\pi/4} Re^{-R^2\sin 2\theta} \,d\theta \\&<\int_0^{\pi/4} Re^{- \frac{4}{\pi} R^2\theta} \, d\theta \\&=\left(1-e^{-R^2} \right)\frac{\pi}{4R} \to 0.\end{aligned} $$
So
$$ \int_0^\infty \cos p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$
$$ \int_0^\infty \sin p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$
and the Fourier sine transform of $t^{-1/2}$ is $w^{-1/2}$.
$$t^{-1/2} \iff w^{-1/2}.$$
Best Answer
HINT:
For $m>0$ and $\alpha \ne0$, does the integral $\int_{-\infty }^0 \sin(\alpha x)e^{-mx}\,dx$ converge?
And for $x<0$, is $f(x)=0$?