Applying the method of steepest descent to $ \int_{-i \infty}^{i \infty} \frac{\exp{x (t^2 - 2t)}}{t - a}\,dt$ as $x \to \infty$ Consider the integral $I(x) = \int_{-i \infty}^{i \infty} \dfrac{\exp{x (t^2 - 2t)}}{t - a} dt = \exp(-x) \int_{-i \infty}^{i \infty} \dfrac{\exp{x (t - 1)^2}}{t - a} dt$. 
I would like to find the leading order of $I(x)$ as $x \to \infty$ by the method of steepest descent., whereby I set $\phi(t) = t^2 - 2t = u + i v$ and then deform the contour from $C : is, s \in [-\infty, \infty]$ to a collection of $\Gamma_j$ on which $v$ is constant or (equivalently) which forms the contour of steepest ascent/descent for $u$.
I tried to consider the integral $\int_{-i R}^{i R} \dfrac{\exp{x (t^2 - 2t)}}{t - a} dt$ with the goal of taking $R \to \infty$. Writing $t = \xi + i \eta$, it follows that $\phi(\xi, \eta) = \xi^2 - \eta^2 - 2 \xi + 2 i \eta( \xi - 1)$, and so $u(\xi, \eta) =  \xi^2 - \eta^2 - 2 \xi = (\xi - 1)^2 - (\eta^2 + 1)$ and $v(\xi, \eta) = 2 \eta (\xi - 1)$. Thus, the contour of steepest descent in $u$ or constant in $v$ that passes through $-i R$ is given by $R = \eta (\xi - 1)$ and can be parameterized by $t = \dfrac{R}{\zeta} + 1 + i \zeta, \zeta \in  [-R, -\infty]$. Rewriting the integral along this contour we get $$\exp(-x) \cdot \exp(2 R i x) \cdot \int_{-R}^{-\infty} \exp\Bigg(-x \Big(\zeta^2 - \dfrac{R^2}{\zeta^2}\Big)\Bigg) \dfrac{d t}{d \zeta} \dfrac{1}{t - a} d \zeta.$$ Presumably this integral could then be approximated by Laplace's method, but this seems like such a brutal, messy calculation. Can anybody see a better way?
 A: Suposse $\Re a<0$ or $\Re a>1$.
The phase function $f(t)=t^2-2t$ has a unique saddle point at $t=1$ and he steepest descent contour $\gamma$ (that passes through $t=1$) is $$\gamma\equiv\{t=1+i\,\alpha\,|\,\alpha\in\mathbb{R}\,\}$$
We can deform the original path into $\gamma$ joining both of them with $\gamma_{\pm}=\{\alpha\pm i\,R\,|\,\alpha\in[0,1]\,\}$ with $R\rightarrow\infty$ forming a rectangle. It's clear that integral over $\gamma_{\pm}$ is $0$ when $R\rightarrow\infty$ and then
$$ \int_{-i \infty}^{i \infty} \frac{\exp{x (t^2 - 2t)}}{t - a}\,dt=\int_{\gamma}\frac{\exp{x (t^2 - 2t)}}{t - a}\,dt=e^{-x}\int_{-\infty}^{\infty}\frac{e^{-x\,\alpha^2}}{\alpha-i\,(1-a)}dt$$
Now, you can apply Laplace Method: the mayor contribution to value of integral comes from the minimum of phase function $f(\alpha)=\alpha^2$, that is, from $\alpha=0$ and then, we can consider the Taylor aproximation of $\frac{1}{\alpha-i\,(1-a)}$ near $\alpha=0$
$$\frac{1}{\alpha-i\,(1-a)}=\sum_{n=0}^{\infty}\frac{-\alpha^n}{i^{n+1}(1-a)^{n+1}}$$
Inserting this series into integral and interchanging sum and integral we obtain
$$\boxed{\int_{-i \infty}^{i \infty} \frac{\exp{x (t^2 - 2t)}}{t - a}\,dt\sim e^{-x}\sum_{k=0}^{\infty}\frac{i\,(-1)^k\,\Gamma(k+1/2)}{(1-a)^{2k+1}}\frac{1}{x^{k+1/2}}}$$
