How to calculate DTFT of cosine function divided by n
The given impulse response is a scaled version of the impulse response of a highpass Hilbert transformer with delay $\tau=N/2$ ($N$ odd) with frequency response
$$H(e^{j\omega})=\begin{cases}-j\,\textrm{sgn}(\omega)\,e^{-j\omega\tau},&amp;\omega_c&lt;\omega&lt;\pi\\j\,\textrm{sgn}(\omega)\,e^{-j\omega\tau},&amp;-\pi&lt;\omega&lt;-\omega_c\\0,&amp;-\omega_c&lt;\omega&lt;\omega_c\end{cases}\tag{1}$$
This can be shown as follows:
$$\begin{align}h[n]&amp;=\frac{1}{2\pi}\int_{-\pi}^{\pi}H(e^{j\omega})e^{jn\omega }d\omega\\&amp;=\frac{1}{2\pi}\left[-j\int_{\omega_c}^{\pi}e^{j\omega(n-\tau)}d\omega+j\int_{-\pi}^{-\omega_c}e^{j\omega(n-\tau)}d\omega\right]\\&amp;=\frac{1}{2\pi (n-\tau)}\left[-e^{j\pi(n-\tau)}+e^{j\omega_c(n-\tau)}+e^{-j\omega_c(n-\tau)}-e^{-j\pi(n-\tau)}\right]\\&amp;=\frac{\cos[\omega_c(n-\tau)]-\cos[\pi(n-\tau)]}{\pi(n-\tau)}\tag{2}\end{align}$$
Since for $\tau=(2k+1)/2$, $k\in\mathbb{Z}$, we have $\cos[\pi(n-\tau)]=0$ we arrive at the following inverse DTFT of $(1)$:
$$h[n]=\frac{\cos[\omega_c(n-\tau)]}{\pi(n-\tau)}\tag{3}$$