Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there's no longer a singularity then, we can split the integral in two:
$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega
&=&
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\
&=&
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega -
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;.
\end{eqnarray}$$
Now we can apply different substitutions to the two parts, $\omega'=\omega(x+\frac{1}{2})$ in the first part and $\omega'=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega'}/(i\omega')$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there's a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get
$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega' +
\int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'\;,$$
which is (again using the sufficient decay at infinity)
$$\oint \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'$$
on a contour that encloses the pole at the origin, and hence the value is $2\pi$.
Your question is a bit ambiguous, since you don't state what you mean by fourier integral and fourier transform.
One possible source of confusion is that, while the fourier transform is indeed a linear isometry on $L^2$, the integral $$
\int_{-\infty}^\infty f(t)e^{-i2\pi\omega t} \,dt
$$
does not converge for every $f \in L^2$. It does, however converge for every $f \in L^1 \cap L^2$, and the fourier transform on the full space $L^2$ can therefore be defined as the unique extension of the transform defined by the integral on $L^1 \cap L^2$. The result is then sometimes called the Fourier-Plancherel-Transform, but sometimes also simply the fourier transform on $L^2$.
Or you could simply be referring to the difference between the integral one uses to compute the coefficients of a fourier series and the integral used to define the fourier transform (on $L^2 \cap L^2$).
Best Answer
Let $f(x) = \text{sinc}(x)$. We can rewrite
$$ f(x) = \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} = \frac{1}{2\pi}\frac{e^{i \pi x}-e^{-i \pi x}}{i x} = \frac{1}{2 \pi}\int \limits_{- \pi}^{\pi}e^{i \omega x} \,d \omega = \mathcal{F}^{-1}(1_{[-\pi, \pi]}).$$