= 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 is
$$
V(x)=-\frac{\sin(\pi\nu)}\pi\int_0^\infty e^{-x\sinh y-\nu y}\,dy.
$$
Set $v=\sinh y$. The <integral> becomes $\int_0^\infty e^{-xv}a(v)\,dv$, where $a(v)=e^{-\nu\operatorname{arsinh}v}/\sqrt{1+v^2}$, $a(0)=1$ and $a'(0)=-\nu$. Successive <integration by parts>, with exponentially vanishing terms at infinity, gives
$$
\int_0^\infty e^{-xv}a(v)\,dv=\frac1x-\frac\nu{x^2}+O(x^{-3}).
$$
Hence 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 $x$.
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