# Steepest descent of integrand with a movable saddle?

I want to apply the steepest descent method to the following integration: $$\int_0^\infty e^{-x^2 + i \sqrt{x^2 + 1} \cdot \lambda } dx$$

It has movable saddle so I need to transform it into the standard form, something like

$$\int_C g(z) e^{\lambda f(z) } dz$$

I know for Gamma function: $$\Gamma(x+1)= \int_0^\infty e^{-t} t^{x} dt =\int_0^\infty e^{-t + x \ln t} dt$$ letting $t = x s$ transforms it into standard form. And for Airy function:

$$Ai(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i (t^3/3 + x t)} dt$$

letting $t = \sqrt{x} z$ does the trick.

However, no change of variable seems to transform my integration into the standard form. For this reason I can not proceed at all. Any hint or suggestion ? Thanks!

• You need to find the saddle point (in general there can be more than one, in that case there will be one that yields the dominant contribution). If there is a saddle point at $z = z_0$, then the contour is deformed to move through $z_0$. Then write the integrand as $f(z_0 + u)$ and integrate over $u$, you define a new variable $w$ such that $f(z_0 + u) = \exp(-w^2)$, becomes exactly a Gaussian. This then has to be multiplied by the Jacobian $\frac{du}{dw}$. – Count Iblis Mar 15 '17 at 20:12
• defining $g(x)=e^{-x^2}$? Maybe this is too simple but definitly worth a try – tired Mar 15 '17 at 21:09
• @AntonioVargas i came away with this cheap trick a few times, but i have no idea when it works...:( – tired Mar 15 '17 at 22:25
• Set $x^2+1=y$. The integral in question transforms into $$I=e\int_1^{\infty}dy\frac{y}{\sqrt{y^2-1}}e^{-y^2}e^{i\lambda y}$$ after a rescaling $y\rightarrow\lambda y$ this reads $$I=e\lambda^2\int_{1/\lambda}^{\infty}dy\frac{y}{\sqrt{\lambda^2y^2-1}}e^{-\lambda^2(y^2+iy)}$$ which is not too far away from a form where classic steepest descent method can be applied (i think @AntonioVargas has elaborated somewhere on this site on integrals of this type, where the prefactor of the exponential is a slowly varying fiunction of $\lambda$) – tired Mar 15 '17 at 22:35
• @tired This does not work. If you take $exp(-x^2)$ in separation, the next step of asymptotics is essentially stationary phase, and the stationary point would be $x = 0$, then $exp(-x^2) = exp(0) = 1$, which is saying the term $exp(-x^2)$ has no effect. However, this conclusion is not right. – Taozi Mar 16 '17 at 0:42

Let us here sketched a derivation.

1. OP's integral has a saddle point at $z=0$ with angular steepest descent direction $\frac{\pi}{4}$. Therefore, inspired by the method of steepest descent, we rewrite OP's integral as $$I(\lambda) ~:=~\int_{\mathbb{R}_+}\!\mathrm{d}z~ \exp\left\{-z^2+i\lambda\sqrt{1+z^2}\right\} ~=~ \frac{e^{i\lambda}}{2}J(\lambda),\qquad \lambda~>~0, \tag{1}$$ where $$J(\lambda) ~:=~\int_{\mathbb{R}}\!\mathrm{d}z~ \exp\left\{-z^2+i\lambda\left(\sqrt{1+z^2}-1\right)\right\}~\stackrel{z=\frac{u}{\sqrt{\lambda}}\exp\left(\frac{i\pi}{4}\right)}{=}~\frac{1}{\sqrt{\lambda}}\exp\left(\frac{i\pi}{4}\right) K(\lambda),\tag{2}$$ where $$K(\lambda)~:=~\int_{\mathbb{R}}\!\mathrm{d}u~ \exp\left\{-\frac{iu^2}{\lambda}-f_{\lambda}(u)\right\},\tag{3}$$ and where $$-f_{\lambda}(u)~:=~i\lambda\left(\sqrt{1+\frac{iu^2}{\lambda}}-1\right) ~=~-\frac{u^2}{2} + \frac{iu^4}{8\lambda}+ O(u^6\lambda^{-2}).\tag{4}$$

2. Let us define the non-negative function $$f(u)~:=~\frac{u^2}{2}. \tag{5}$$ Clearly $$f_{\lambda}(u)~\longrightarrow~ f(u)\quad\text{for}\quad\lambda~\to~ \infty \tag{6}$$ $u$-pointwise.

3. We note that the real part is given by $${\rm Re}~ f_{\lambda}(u)~=~\lambda\sqrt{\frac{\sqrt{1+\frac{u^4}{\lambda^2}}-1}{2}} .\tag{7}$$ Define a non-negative function $$g(u)~:=~ \frac{|u|}{2}~\theta(|u|\!-\!\sqrt{2}) , \tag{8}$$ where $\theta$ denotes the Heaviside step function. One may show that $$\forall\lambda\geq\frac{1}{\sqrt{2}}\forall u\in\mathbb{R}:~~ {\rm Re}~ f_{\lambda}(u) ~\geq~ g(u),\tag{9}$$ because $$\forall\lambda\geq\frac{1}{\sqrt{2}}\forall u\geq\sqrt{2}:~~ \sqrt{1+\frac{u^4}{\lambda^2}}-1 ~\geq~ \frac{u^2}{2\lambda^2}.\tag{10}$$

4. It follows that $\exp\left\{-g(u)\right\}$ is a majorant function for the integrand (3). Lebesgue's dominated convergence theorem then shows that the corresponding integral (3) satisfies $$K(\lambda)~\longrightarrow~ \int_{\mathbb{R}}\!\mathrm{d}u~ \exp\left\{-f(u)\right\}~=~\sqrt{2\pi} \quad \text{for} \quad\lambda~\to~ \infty. \tag{11}$$

5. Next we can estimate corrections to any order in $1/\lambda$ we like. E.g. $$K(\lambda)~\stackrel{(3)+(4)}{=}~\int_{\mathbb{R}}\!\mathrm{d}u~ \exp\left\{-\left(\frac{i}{\lambda}+\frac{1}{2}\right)u^2\right\} \left(1 +\frac{iu^4}{8\lambda} + O(\lambda^{-2})\right)$$ $$~=~\sqrt{\frac{\pi}{\frac{i}{\lambda}+\frac{1}{2}}}+\frac{i}{8\lambda}\underbrace{\int_{\mathbb{R}}\!\mathrm{d}u~\exp\left\{-\frac{u^2}{2}\right\}u^4}_{=3\sqrt{2\pi}} + O(\lambda^{-2}) ~=~\sqrt{2\pi}\left(1 -\frac{5i}{8\lambda} + O(\lambda^{-2})\right) ,\tag{12}$$ and so forth. Hence OP's integral (1) has an asymptotic expansion

$$I(\lambda)~\sim~ e^{i\lambda}\frac{1+i}{2}\sqrt{\frac{\pi}{\lambda}}\left(1 -\frac{5i}{8\lambda} + O(\lambda^{-2})\right)\quad \text{for}\quad \lambda~\to~\infty .\tag{13}$$

• Hi @Qmechanic Your solution is very beautiful. I still have two questions. 1> what's the rational behind the form of substitution in equation 2, esp. the $\exp(i \pi/4)?$ 2> In equation 12 did you separate the exponent into two groups and Taylor expand the second group? If so, why $\int_R du \exp \{-(i/\lambda + 1/2)u^2 \} u^4 = -5 \sqrt{2\pi}$? – Taozi Mar 23 '17 at 14:57
• I updated the answer. – Qmechanic Mar 23 '17 at 16:19

Not a full answer but too long for a comment.

It appears that the saddle point at $x=0$ is dominant. There are two more but they can be avoided (and they possibly lie on the branch cuts of the integrand anyway---that would be bad).

If we make the substitution $x = u/\sqrt{\lambda}$ we get

\begin{align} \int_0^\infty \exp\left\{-x^2+i\lambda\sqrt{x^2+1}\right\}dx &= \frac{1}{\sqrt{\lambda}} \int_0^\infty \exp\left\{-\frac{u^2}{\lambda}+i\lambda\sqrt{\frac{u^2}{\lambda}+1}\right\}du \\ &= \frac{e^{i\lambda}}{\sqrt{\lambda}} \int_0^\infty \exp\left\{-\frac{u^2}{\lambda}-i\lambda \left( 1-\sqrt{\frac{u^2}{\lambda}+1} \right) \right\}du. \end{align}

Numerically,

$$\int_0^\infty \exp\left\{-\frac{u^2}{\lambda}-i\lambda \left( 1-\sqrt{\frac{u^2}{\lambda}+1} \right) \right\}du \to \int_0^\infty \exp\{iu^2/2\}\,du = \frac{1+i}{2}\sqrt{\pi}$$

as $\lambda \to \infty$, but I'm not sure how to show it at the moment. This would imply that

$$\int_0^\infty \exp\left\{-x^2+i\lambda\sqrt{x^2+1}\right\}dx \sim \frac{(1+i)e^{i\lambda}}{2} \sqrt{\frac{\pi}{\lambda}}$$

as $\lambda \to \infty$.

• I did a quick check of your solution in Mathematica, it appears remarkably accurate, I need more time to understand how and why the substitution works. – Taozi Mar 16 '17 at 19:20
• Unfortunately, the result appears to be the same if you throw away the $e^{-x^2}$ term right at the beginning. – Taozi Mar 16 '17 at 19:28
• @Taozi, I used this substitution because near $x=0$ we have $$-x^2 + i\lambda\sqrt{x^2+1} = i \lambda +\left(-1+\frac{i \lambda }{2}\right) x^2 + O(x^3),$$ and substituting $x = u/\sqrt{\lambda}$ asymptotically removes the $\lambda$-dependence from the coefficient of $x^2$. – Antonio Vargas Mar 16 '17 at 20:06