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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 $\Gamma$.

Gold
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  • It's a prerequisite that it should be possible to deform $\Gamma$ into $\Gamma'$ s.t. $\Gamma'$ locally goes through a saddle point in the direction of the steepest descent and that $\operatorname {Re} f$ has its maximum (globally on $\Gamma'$) at the saddle point. The choice of a saddle point and of $\Gamma'$ depends on the geometry of the level curves of $\operatorname {Re} f$ (see examples here). – Maxim Feb 14 '20 at 23:38

1 Answers1

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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.

Qmechanic
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  • Thanks @QMechanic. Regarding point (1), let $a,b$ be the endpoints of $\Gamma$ and let $z_0$ be the stationary point. Then $\Gamma''$ has the same endpoints $a,b$ and passess through $z_0$ in the direction of steepest descent. My doubt is exactly why such a contour exists. If the region $\Omega$ is path connected, then we can always find $\Gamma''{a,z_0}$ conecting $a$ to $z_0$ and $\Gamma''{z_0,b}$ connecting $z_0$ to $b$. Therefore we can always find a path from $a$ to $b$ through $z_0$. But why we can always choose it so that it passess through $z_0$ in the steepest descent direction? – Gold Feb 15 '20 at 18:37
  • I updated the answer. – Qmechanic Feb 15 '20 at 18:56