Update: Fourier Transform of a shifted and scaled $\operatorname{sinc}$ signal
HINT:
Instead of having someone else confirm your result, it would be more instructive to confirm it yourself by computing the inverse Fourier transform of the expression for $X(j\omega)$ that you obtained:
$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega\tag{1}$$
In the definition $(1)$ it is assumed that in the forward transform you use a negative exponent and no scaling constant, i.e.,
$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{2}$$
After having derived and confirmed the result yourself, it would be useful to remember the formula for the impulse response of an ideal low pass filter with cut-off $\omega_c$. Combined with the time shifting property of the Fourier transform it is straightforward to immediately write down the transform of your original function $x_N(t)$.