Calculate the integral $\int_{C_1} e^{X(z+iz-z^2/2)} dz$ as $X \rightarrow \infty$ I need help calculating the following integral as $X \rightarrow \infty$
$$\int_{C_1} e^{X(z+iz-z^2/2)} dz$$
along the path
$$z=x+i\frac{x+1}{x-1},$$
for $x\in (-\infty,-1]$.
It seems very difficult.
$C_2: z=x+i,x\in\mathbb{R}$, $C_3: z=\infty+iy, y\in [0,1]$, and $C: z=x, \in [-1,\infty)$.
So the closed contour is $C=-C_1+C_2-C_3$.
The integral along the path $C_2$ I calculated as
$$e^{iX}\sqrt{\frac{2\pi }{X}}$$
as $X\rightarrow \infty$.
 A: I am going to assume that the contour $C_1$ is oriented so that it starts at $-1$ and ends at $i-\infty$. Define $\varphi(z) = z+iz-z^2/2$.
The highest point of the contour $C_1$ on the surface $\operatorname{Re} \varphi(z)$ is at the endpoint $z=-1$, so this is where the largest contribution to the integral will come from.
Using Cauchy's theorem we can deform the contour to the path of steepest descent from the point $z=-1$ on the surface $\operatorname{Re}\varphi(z)$. Call this contour $\gamma$, shown in red below over the level sets of this surface (tan = higher, blue = lower).

Near $z=-1$ we have
$$
\varphi(z) = -\left(\frac{3}{2}+i\right) + (2+i) (z+1) + O\!\left((z+1)^2\right),
$$
so there are sets $U,V \subset \mathbb C$ with $0 \in U$ and $-1 \in V$ and a biholomorphic map $\psi : U \to V$ such that
$$
(\varphi \circ \psi)(w) = -\left(\frac{3}{2}+i\right) - w.
$$
The function $\psi$ maps a segment of the real axis to a segment of the steepest descent path on the surface $\operatorname{Re} \varphi(z)$ going through the point $z=-1$ with $\psi(0) = -1$ and $\psi'(0) = -1/(2+i)$.
Define the number $a > 0$ by $[0,a) = U \cap \mathbb R_{\geq 0}$. We then have $\psi([0,a)) = \gamma \cap V$.
So, write
$$
\begin{align*}
\int_{C_1} e^{X\varphi(z)}\,dz &= \int_\gamma e^{X\varphi(z)}\,dz \\
&= \int_{\gamma \cap V} e^{X\varphi(z)}\,dz + \int_{\gamma \cap V^c} e^{X\varphi(z)}\,dz.
\end{align*}
$$
In the first integral make the substitution $z = \psi(w)$ to get
$$
\int_{\gamma \cap V} e^{X\varphi(z)}\,dz = e^{-(3/2+i)X} \int_0^a e^{-Xw} \psi'(w)\,dw.
$$
By Watson's lemma we have
$$
\int_0^a e^{-Xw} \psi'(w)\,dw \sim \frac{\psi'(0)}{X} = -\frac{1}{(2+i)X}
$$
as $X \to \infty$, so that
$$
\int_{\gamma \cap V} e^{X\varphi(z)}\,dz \sim -\frac{\exp\{-(3/2+i)X\}}{(2+i)X}
$$
as $X \to \infty$.
The remaining integral $\int_{\gamma \cap V^c}$ is exponentially small compared to this one. Indeed, there is a constant $c < \operatorname{Re} \varphi(-1)$ such that $\operatorname{Re} \varphi(z) < c$ for all $z \in \gamma \cap V^c$, so it follows that
$$
\int_{\gamma \cap V^c} e^{X\varphi(z)}\,dz = O(\exp\{cX\})
$$
as $X \to \infty$. Putting everything together, this shows that
$$
\int_{C_1} e^{X\varphi(z)}\,dz \sim -\frac{\exp\{-(3/2+i)X\}}{(2+i)X}
$$
as $X \to \infty$.
We can test this numerically. Setting $X = 5$ I get
$$
\int_{C_1} e^{5\varphi(z)}\,dz \approx -0.0000319283 - 0.0000362014i
$$
and
$$
-\frac{\exp\{-(3/2+i)5\}}{(2+i)5} \approx -0.0000337658 - 0.0000361537i
$$
and the relative error gets smaller as $X$ is increased further.
If you want more terms of the approximation then you can use more terms of the power series for $\psi'(w)$ in the application of Watson's lemma above.
