 Solution

Keep $\nu$ fixed as $x\to +\infty$. On both vertical lines $\sin (\pm \pi +iy)=-i\sinh y$. The right side is traversed upward and the left downward, so their combined contribution isSet $v=\sinh y$. The integral becomes ${\int }_0^{\infty }{e}^{-xv}a(v)dv$, where $a(v)=e^{-\nu \mathrm{arsinh}v}/\sqrt{1+v^2}$, $a(0)=1$ and $a^{\prime }(0)=-\nu$. Successive integration by parts, with exponentially vanishing terms at infinity, givesHence the leading vertical contribution is $\boxed{V(x)=-\sin (\pi \nu )/(\pi x)+O(x^{-2})}$. It is exactly zero for integer $\nu$. These estimates hold for fixed real $\nu$; they are not uniform when the order grows with $x$.