## Mellin transform

Can someone give me some help on this problem?
Show that $\displaystyle \mu(cos)(z)=\int_0^{\infty}cos(t)t^{z-1}dt=\Gamma(z)cos(\pi z/2)$ for $\displaystyle 0<Re(z)<1$.
By definition, $\displaystyle \mu(f)(z)=F(z)=\int_0^{\infty}f(t)t^{z-1}dt$
If I know that $\displaystyle \mu(sin)(z)=\int_0^{\infty}sin(t)t^{z-1}dt=\Gamma(z)sin(\pi z/2)$ for $\displaystyle 0<Re(z)<1$, how can I show that this identity still holds in the larger strip $\displaystyle -1<Re(z)<1$