Method of steepest descent: why can we relate these two contours?

Question

Let $f,g:\mathbb{C}\to \mathbb{C}$ be two functions which are holomorphic in $\Omega\subset \mathbb{C}$. Consider the integral $$I(\lambda)=\int_{\Gamma}g(z)e^{\lambda f(z)}dz,\quad \lambda \in (0,+\infty)$$

where $\Gamma$ is a contour in $\Omega$. I want to understand the method of steepest descend which allows to approximate $I(\lambda)$ as $\lambda \to +\infty$.

Now, if I understand, the rough idea is to deform the contour into another contour $\Gamma'$ passing through a saddle point of $f(z)$ in the direction of steepest descent of its real part.

To do so we look for a saddle point $f'(z_0)=0$, expand $f(z)$ up to second order around it $$f(z)=f(z_0)+\frac{1}{2}(z-z_0)f''(z_0)+\cdots$$

and we parameterize $z - z_0 = r_1e^{i\theta_1}$. Also letting $\frac{1}{2}f''(z_0)=r_2 e^{i\theta_2}$ we have the changes in the real and imaginary parts of $f$: $$\operatorname{Re}[f(z)-f(z_0)]=r_1^2r_2\cos(2\theta_1+\theta_2),\quad \operatorname{Im}[f(z)-f(z_0)]=r_1^2r_2\sin(2\theta_1+\theta_2).$$

The direction of steepest descent has vanishing change in the imaginary part and negative change in the real part. These two conditions give $2\theta_1+\theta_2$ either $\pi$ or $3\pi$. Therefore the desired contour $\Gamma'$ can be parameterized as $$z(t)=z_0+\frac{t}{\sqrt{r_2}}e^{i\theta_1}$$

Question: why can we deform $\Gamma$ into $\Gamma'$ and not change $I(\lambda)$?

I mean, I do know that from Cauchy's theorem if $\Gamma$ and $\Gamma'$ have the same endpoints then the integral is the same along both.

But in this whole derivation I see no reason why $\Gamma'$ would share endpoints with $\Gamma$.

Answer 1

The rough idea behind the method of steepest descent is as follows.

1. By Cauchy's integral theorem we can deform the integration contour $\Gamma$ into an integration contour $\Gamma^{\prime\prime}$ with the same endpoints that goes though the stationary point $z_0$ in the direction of steepest descent$^1$.

2. Since the $\Gamma^{\prime\prime}$ contour integral is exponentially suppressed in $\lambda$ away from the stationary point $z_0$, we can replace $\Gamma^{\prime\prime}$ with the tangent $\Gamma^{\prime}$ (or an appropriate double-sided line segment thereof) at $z_0$. The endpoints are to a large extent irrelevant.

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$^1$ If the deformation crosses poles and/or branch cuts, their effect (such as, residues) have to be taken into account. Also note that there can be more than one stationary point. Another effect is the Stokes phenomenon.