Update: Fourier Transform of a shifted and scaled $\operatorname{sinc}$ signal Let $x_N$ be the function given by
$$x_N(t)=A\frac{\sin(M\pi(t-N))}{\pi(t-N)}$$
The Fourier Transform of $x_N$ is
$$\begin{align}
X_N(j\omega)&amp;=\mathscr{F}\{x_N\}(j\omega)\\\\
&amp;=\int_{-\infty}^\infty x_N(t)e^{-j\omega t}\,dt\\\\
&amp;=Ae^{-jN\omega }\int_{-\infty}^\infty \frac{\sin(M\pi t)}{\pi t}e^{-j\omega t}\,dt\tag1
\end{align}$$
Enforcing the substitution $t \rightarrow t/M\pi$ reveals
$$\begin{align}
X_N(j\omega )&amp;=\frac{Ae^{-jN\omega }}\pi\int_{-\infty}^\infty \frac{\sin(t)}{t}e^{-j(\omega/M\pi) t}\,dt\\\\
&amp;=\frac{Ae^{-jN\omega }}2\left(\text{sgn}(-\omega/M\pi +1)-\text{sgn}(-\omega/M\pi -1)\right)\\\\
&amp;=\begin{cases}Ae^{-jN\omega}&amp;,|\omega|&lt;M\pi\\\\0&amp;,\text{elsewhere}\end{cases}
\end{align}$$
where we used the Fourier Transform of the sinc function, $\operatorname{sinc}(t)=\frac{\sin(t)}{t}$
$$\mathscr{\operatorname{rect}}(\omega)=\begin{cases}\pi&amp;,|\omega|&lt;1\\\\0&amp;,\text{elsewhere}\end{cases}$$

Let $\displaystyle x(t):=\frac{\sin(2\pi(t-1))}{\pi(t-1)}$. Then we have $A=1,M=2, N=1$ so we get :
$$
X(j\omega)=\begin{cases} e^{-j\omega}&amp;\text{if $|\omega|&lt;2\pi$}\\
0&amp;\text{if otherwise}\end{cases}
$$

Is this example correct? because in my book it says the answer is $e^{2\omega}$

I would hope for someone to assist me in determining if this is correct or not. I have spent entire day on this problem. I would much appreciate any help and thank you :)
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 $x_N(t)$.