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

$$H(e^{jomega})=begin{cases}-j,textrm{sgn}(omega),e^{-jomegatau},&omega_c<omega<pi\j,textrm{sgn}(omega),e^{-jomegatau},&-pi<omega<-omega_c\0,&-omega_c<omega<omega_cend{cases}tag{1}$$

This can be shown as follows:

$$begin{align}h[n]&=frac{1}{2pi}int_{-pi}^{pi}H(e^{jomega})e^{jnomega }domega\&=frac{1}{2pi}left[-jint_{omega_c}^{pi}e^{jomega(n-tau)}domega+jint_{-pi}^{-omega_c}e^{jomega(n-tau)}domegaright]\&=frac{1}{2pi (n-tau)}left[-e^{jpi(n-tau)}+e^{jomega_c(n-tau)}+e^{-jomega_c(n-tau)}-e^{-jpi(n-tau)}right]\&=frac{cos[omega_c(n-tau)]-cos[pi(n-tau)]}{pi(n-tau)}tag{2}end{align}$$

Since for $tau=(2k+1)/2$, $kinmathbb{Z}$, we have $cos[pi(n-tau)]=0$ we arrive at the following inverse DTFT of $(1)$:

$$h[n]=frac{cos[omega_c(n-tau)]}{pi(n-tau)}tag{3}$$