# Inverse DTFT of $H_1(\Omega)=\begin{cases} 10,& \frac{\pi}{3} \leq |\Omega| < \pi\\ 0,& 0 \leq |\Omega| < \frac{\pi}{3}\\ \end{cases}$

What is the inverse DTFT of the $2\pi$-periodic extension of following function:

$$H_1(\Omega)=\begin{cases} 10,& \text{for } \frac{\pi}{3} \leq |\Omega| < \pi\\ 0,& \text{for } 0 \leq |\Omega| < \frac{\pi}{3}\\ \end{cases}$$

I have found out using the definition that it is: $$h[n]=-\frac{10\sin(\frac{\pi}{3}n)}{\pi n}$$

But the problem is that this has a DTFT of

$$H_2(\Omega)=\begin{cases} -10,& \text{for } 0 \leq |\Omega| \leq \frac{\pi}{3}\\ 0,& \text{for } \frac{\pi}{3} < |\Omega| < \pi\\ \end{cases}$$

Now, $H_1(\Omega)$ and $H_2(\Omega)$ seem quite different to me. Am I missing something? Have I done something wrong in my calculations?

Let's use $\omega$ instead of $\Omega$ for the definition of the inverse DTFT and state the following:
$$h[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} H(\omega) e^{j \omega n} d\omega$$ Given that $H(\omega) = 1$ , for $\pi/3 < |\omega| <\pi$ and zero otherwise, the integral becomes:
\begin{align} h[n] &= \frac{1}{2\pi} \int_{-\pi}^{-\pi/3} e^{j \omega n} d\omega + \frac{1}{2\pi} \int_{\pi/3}^{\pi} e^{j \omega n} d\omega \\ &= \frac{1}{2\pi} \left( \frac{ e^{j \omega n} }{j n} |_{-\pi}^{-\pi/3} + \frac{ e^{j \omega n} }{j n} |_{\pi}^{\pi/3} \right)\\ &= \frac{1}{2\pi} \left( \frac{ e^{-j (\pi/3) n} -e^{-j\pi n} }{j n} + \frac{ e^{j \pi n} - e^{j \pi/3 n} }{j n} \right)\\ &= \frac{1}{n\pi} \left( \frac{ e^{-j (\pi/3) n}- e^{j (\pi/3) n} }{2j} + \frac{ e^{j \pi n} - e^{-j\pi n} }{2j} \right)\\ &= \frac{1}{n\pi} \left( -\sin(\frac{\pi}{3}n) + \sin(\pi n) \right)\\ &= \frac{\sin(\pi n) }{n\pi} - \frac{ \sin(\frac{\pi}{3}n)}{n\pi} \\ \end{align}
at this point recognize that $$\frac{\sin(\pi n) }{n\pi} = \begin{cases} 1 &,\text{ for } n=0 \\0 &, \text{ for } n \neq 0 \\ \end{cases} = \delta[n]$$
Hence the I-DFT becomes: $$\boxed{ h[n] = \delta[n] - \frac{\sin( \frac{\pi}{3}n) }{\pi n} }$$
Note that your $H(\omega)$ had a gain of $10$ instead of $1$ so multiply this with $10$.