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 $L^2 \cap L^2$).
Note that we can write
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\frac{1}{2i}\int_{-\infty}^\infty\frac{e^{ik}-e^{-ik}}{k}e^{ikx}\,dk \tag1$$
Observe that for each of the principal value integrals
$$I_{\pm}(x)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{\pm ik}}{k}e^{ikx}\,dk\right)=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x\pm 1)}}{k}\,dk\right)$$
the integrand has a pole at $k=0$. To evaluate these integrals, we analyze the contour integrals
$$\begin{align}
\oint_{C}\frac{e^{iz(x\pm 1)}}{z}\,dz&=\int_{-R}^{-\epsilon} \frac{e^{ik(x\pm 1)}}{k}\,dk+\int_{\epsilon}^{R} \frac{e^{ik(x\pm 1)}}{k}\,dk\\\\
&+\int_{\text{sgn}(x\pm 1)\pi}^0 \frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}\,i\epsilon e^{i\phi}\,d\phi\\\\
&+\int_0^{\text{sgn}(x\pm 1)\pi}\frac{e^{i R e^{i\phi}}}{R e^{i\phi}}\,iR e^{i\phi}\,d\phi \tag 2
\end{align}$$
As $R\to \infty$, the last integral on the right-hand side of $(2)$ tends to $0$. As $\epsilon\to 0$, the third integral on the right-hand side of $(2)$ tends to $-\text{sgn}(x\pm 1)\pi$. Therefore, we have
$$\text{PV}\left(\int_{-\infty}^\infty\frac{e^{ ik(x+ 1)}-e^{ik(x-1)}}{k}\,dk\right)=\text{sgn}(x+1)\pi-\text{sgn}(x-1)\pi=\begin{cases}2\pi &,|x|<1\\\\0&,|x|>1\\\\\pi&,|x|=1\end{cases} \tag 3$$
Substituting $(3)$ into $(1)$ yields
$$\int_{-\infty}^\infty\frac{\sin(k)}{k}e^{ikx}\,dk=\begin{cases}\pi &,|x|<1\\\\0&,|x|>1\\\\\pi/2&,|x|=1\end{cases} $$
Best Answer
There is a much simpler way by using the Fourier inversion theorem!
Then the result just follows from the fact that $$ {\cal F}^{-1}(\mathbf 1_{[-1,1]})(x) = \frac{1}{2\pi}\int_{-1}^1 e^{i\,\omega\,x}\,\mathrm d \omega = \frac{e^{ix}-e^{-ix}}{2i\pi\,x} = \frac{\sin x}{\pi\,x}. $$ and so $$ {\cal F}\left(\frac{\sin x}{x}\right) = \pi\,{\cal F}\left(\frac{\sin x}{\pi\,x}\right) = \pi\,\mathbf 1_{[-1,1]}(\omega) $$
Let me add details if you are not familiar with the Fourier transform: the Fourier inversion theorem tells that you can find the function from its Fourier transform by writing $$ f(x) = \frac{1}{2\pi} \int_{\Bbb R} \widehat f(\omega) \, e^{i\omega\,x}\,\mathrm d \omega = \frac{1}{2\pi} \cal F\big(\hat f\big)(-x) $$ where $\widehat f = {\cal F}(f)$. Equivalently, it means that the inverse operation to the Fourier transform is given by $$ {\cal F}^{-1}(g)(x) = \frac{1}{2\pi} \int_{\Bbb R} g(\omega) \, e^{i\omega\,x}\,\mathrm d \omega = \frac{1}{2\pi} \cal F(g)(-x) $$