# 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 $X\rightarrow \infty$.

• Do you know how to apply Cauchy's Integral Theorem? – Mark Viola Mar 11 '17 at 18:31
• Yes, but this isn't a closed contour. – Hogg Mar 11 '17 at 18:34
• It's not a closed contour. But the integrand is analytic and you can deform the contour using the theorem. – Mark Viola Mar 11 '17 at 18:37
• Could you explain further? – Hogg Mar 11 '17 at 18:38
• $\int_C=\int_{C_1}+\int_{C_2}$ where $C_1+C_2-C$ forms a closed contour. – Mark Viola Mar 11 '17 at 18:40

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.