How to find the Inverse Fourier Transform or $g(t)$? 
How to find the $g(t)$? where $F(g(t))=\sqrt{a}$ $sinc (a\omega /2)$


While solving a problem I am stuck at the end.
How to find the $g(t)$? where $F(g(t))=\sqrt{a}$ $sinc (a\omega /2)$
Or how to find inverse Fourier Transform of the $\sqrt{a}$ $sinc(a\omega /2)$
Please help to solve this. Thanks in advance.
 A: Using Fourier transform
$$
\mathcal{F}\{f(t)\} = \int_{-\infty}^{\infty} f(t) \, e^{-i\omega t} \, dt
$$
and
$$
\operatorname{sinc} x = \frac{\sin x}{x}.
$$
I happen to know that
$$
\mathcal{F}\{\chi_{[-1,1]}(t)\} = 2 \frac{\sin\omega}{\omega}
$$
so to get $\operatorname{sinc}\frac{a\omega}{2}$ we should just modify $\chi_{[-1,1]}(t).$ Therefore, we calculate the Fourier transform of $C\,\chi_{[-R,R]}(t)$:
$$
\mathcal{F}\{C\,\chi_{[-R,R]}(t)\}
= \int_{-\infty}^{\infty} C\,\chi_{[-R,R]}(t) \, e^{-i\omega t} \, dt
= C \int_{-R}^{R} e^{-i\omega t} \, dt
= C \left[ \frac{1}{-i\omega} e^{-i\omega t} \right]_{-R}^{R}
= C \frac{1}{-i\omega} \left( e^{-i\omega R} - e^{-i\omega (-R)}  \right)
= C \frac{2}{\omega} \frac{e^{i\omega R} - e^{-i\omega R}}{2i}
= C \frac{2}{\omega} \sin\omega R \\
= 2CR \frac{\sin\omega R}{\omega R}
= 2CR \operatorname{sinc}\omega R.
$$
Thus, to get $\sqrt{a} \operatorname{sinc}\frac{a\omega}{2}$ we shall take $R=\frac{a}{2}$ and $2CR=\sqrt{a},$ i.e. $C=\frac{1}{\sqrt{a}}.$ So,
$$
\mathcal{F}\{\frac{1}{\sqrt{a}}\,\chi_{[-a/2,a/2]}(t)\}
= \sqrt{a} \operatorname{sinc}\frac{a\omega}{2}.
$$
