I am trying to compute the Fourier transform of $\operatorname{sinc}(a \cdot + b)$ (where $a,b \in \mathbb{R}$ with $a > 0$). Here, $f^\wedge$ denotes the Fourier transform
$$f^\wedge(x) = \int_{-\infty}^{\infty} f(t) \exp(-ixt) dt$$
and $\operatorname{sinc}(x) = sin(x)/x$ for $x \neq 0$ and $\operatorname{sinc}(0)=1$.
My exercise problem says that
$$(\operatorname{sinc}(a \cdot + b))^\wedge (x) = \begin{cases} \frac{\pi}{a} \exp(i \frac{b}{a} x) & , |x|< a, \\ \frac{\pi}{2a} \exp(i \frac{b}{a} x) & , |x| = a, \\ 0 & , |x|>a. \end{cases}$$
First, I used the shifting rule $(f(\cdot - h))^\wedge(x) = \exp(-ihx)f^\wedge(x)$ from our lecture notes to obtain
$$(\operatorname{sinc}(a \cdot + b))^\wedge (x) = \exp(i \frac{b}{a} x) (\operatorname{sinc}(a \cdot)^\wedge (x).$$ (here, $f(x) = \operatorname{sinc}(ax)$).
Next, I used the scaling rule $(f(a \cdot))^\wedge(x) = \frac{1}{a} f^\wedge(\frac{x}{|a|})$ from our lecture notes to get
$$(\operatorname{sinc}(a \cdot)^\wedge (x) = \frac{1}{a} \operatorname{sinc}^\wedge(\frac{x}{a})$$
(here, $f(x) = \operatorname{sinc}(x)$).
Now I tried to imitate this computation and obtained
$$\operatorname{sinc} = \pi (\mathbb{1}_{[-1,1]})^\vee$$
where
$$f^\vee(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(t) \exp(ixt) dt$$
denotes the inverse Fourier transform. Note that the definition of $\operatorname{sinc}$ in the cited source differs from my definition.
If my computation is correct and if we can assume $(f^\vee)^\wedge = f$ , then we get
$$(\operatorname{sinc})^\wedge(\frac{x}{a}) = \pi ((\mathbb{1}_{[-1,1]})^\vee)^\wedge(\frac{x}{a}) = \pi \mathbb{1}_{[-1,1]}(\frac{x}{a}) = \begin{cases} \pi & |x| \leq a \\ 0 & |x| > a. \end{cases}$$
However, as you can see, the case $|x|=a$ is wrong then. Furthermore, I cannot use $(f^\vee)^\wedge = f$ without any further proof (in our lecture, we have only shown that $(f^\wedge)^\vee = f$ is true if $f$ is continuous and $f(x) = 0$ for $|x|>c$ for a fixed $c$).
Can anyone explain to me why my solution is wrong and how I can get the desired result?