I'm helping a student out with determining the transform above. His instructor apparently offered a hint about applying the convolution theorem, but I can't seem to get anywhere with that suggestion.
Instead, I tried a different approach using the inverse transform definition, given as $$f(t)=\mathcal{F}^{-1}_t\{\hat{f}(\xi)\}=\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(\xi)e^{i\xi t}\,d\xi$$ My attempt: $$\begin{align*} \frac{\sin2\pi t-\sin\pi t}{\pi t}&=\frac{\dfrac{e^{2\pi it}-e^{-2\pi it}}{2i}-\dfrac{e^{\pi i t}-e^{-\pi it}}{2i}}{\pi t}\\\\ &=\frac{\left(e^{2\pi it}-e^{\pi it}\right)+\left(e^{-\pi i t}-e^{-2\pi it}\right)}{2\pi it}\\\\ &=\frac{1}{2\pi}\left(\int_\pi^{2\pi} e^{i\xi t}\,d\xi+\int_{-2\pi}^{-\pi}e^{i\xi t}\,d\xi\right)\\\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(\xi)\,e^{i\xi t}\,d\xi \end{align*}$$ where I'm defining $$\hat{f}(\xi)=\begin{cases}1&\text{for }\pi<|\xi|<2\pi\\0&\text{otherwise}\end{cases}$$ However, the answer should be $\dfrac{1}{\sqrt{2\pi}}$ in place of $1$ in $\hat{f}(\xi)$. (I have also checked the result with Mathematica; the provided solution is correct. Edit 1: After consulting the documentation center, I noticed the definition Mathematica uses has $\dfrac{1}{\sqrt{2\pi}}$, so perhaps it's a problem with the definition).
Where have I made a mistake? It's been some time since I first learned about the transform, so maybe I'm missing a key property of the transform?
Edit 2: If it make any difference, the forward transform is defined in the student's text as $$\mathcal{F}_\xi\{f(t)\}=\hat{f}(\xi)=\int_{-\infty}^\infty f(t)e^{-i\xi t}\,dt$$ I had taken the course with the same professor two years ago, and he's using the same set of lecture notes, which I recall containing typos here and there, so it's certainly possible that he made a mistake with his definition. At first glance, everything seems to check out, but I'm probably missing something obvious.