 Solution

Use the Gaussian pole transition for a large gamma parameter. With $t=1+s/\sqrt{n}$ and $\lambda =n+i\mu \sqrt{n}+\nu +o(1)$, Taylor expansion givesThe upper semicircle is traversed from left to right, so $\int ds/s=-i\pi$ on the indentation. The minus sign in $dt/(1-t)$ therefore makes its contribution $+i\pi$. The remaining leading integral isAt $\mu =0$ its Cauchy principal value vanishes by oddness. Differentiating with respect to $\mu$ removes the pole, leaving the Fourier transform of a Gaussian:Integrating from zero givesThis is uniform for bounded real $\mu ,\nu$ (and extends analytically over bounded complex values where the local contour is continued consistently). Contributions away from the saddle are exponentially small after localization. The bounded shift $\nu$ enters at the next order, not the leading error function. For example, retaining the displayed cubic correction gives $\sqrt{2\pi /n}{e}^{-{\mu }^2/2}[\nu +(2+{\mu }^2)/3]$ as the next term. This also checks the constant and contour sign.