I really don't understand how we generally choose the contour for the steepest descent method in complex analysis?

I approximate the Fresnel integral $$ \int_{0}^{\infty}\cos{x^2}dx$$ and I found it to be $$I(s=1)=\sqrt{\frac{\pi}{2}}$$

Is that really a right steepest method approximation?

Thanks in advance.

  • $\begingroup$ any ideas anyone? :) $\endgroup$ Nov 4, 2014 at 21:52
  • $\begingroup$ What is $s$? The upper limit? $\endgroup$
    – Qmechanic
    Oct 16, 2016 at 19:03
  • $\begingroup$ Where is the large parameter needed for the method of steepest descent? $\endgroup$
    – Qmechanic
    Oct 22, 2016 at 8:23

3 Answers 3


From a steepest descent approach:

Recast the integral into the following form:

$$ \int\limits_{0}^{\infty} cos(x^2)dx = Re\int\limits_{0}^{\infty}e^{ix^2}dx$$

From here on, I'll drop the Real operator and it will be implicit that you will take the real part at the end. Performing a change of variables and making x a complex variable, the above integral can be recast in the following format:


which is in the expected form for a steepest descent method:

$$g(z)\int e^{sf(z)}dz$$

with $g(z) = \sqrt{s}$ and $f(z) = iz^2$. It can be shown that the path of steepest descent cuts through the origin at an angle of $\frac{\pi}{4}$ degrees. Using Arfken and Weber's notation, $z_0=0$ and $\alpha = \frac{\pi}{4}$. Now the tricky part is drawing the contour. This is where I believe the OP made an error. In order to draw a contour that crosses the origin at $\pi/4$, part of that contour would have to come from the bottom left quadrant. However we also need the contour to run along the real axis from 1 to the origin. You can try it for yourself, you'll find drawing this contour would be very difficult. One way to get around this is to take only 1/2 of the path of steepest descent. In other words when drawing your contour start at the origin then proceed in the $\pi/4$ direction rather than start in the bottom left quadrant and move to the top right. This leads to the OP's missing factor of 1/2. Once you've done this, you can close the contour by coming back to the origin from 1. This leads you to the correct answer:

$$\int\limits_{C}(\cdot)= \frac{1}{2}\int\limits_{S.D.}(\cdot) - \int\limits_{0}^{1}(\cdot) = 0$$ $$ \sqrt{s}\int\limits_{0}^{1}e^{isz^2}dz = \frac{1}{2}\frac{\sqrt{2\pi}g(z_0)e^{sf(z_0)}e^{i\alpha}}{|sf''(z_0)|^{1/2}}$$

plugging in the values from earlier and taking the real part, you should get the correct answer of $$\sqrt{\frac{\pi}{8}}$$

  • $\begingroup$ Aren't saddle point method supposed to give a function $I(s)$ instead of a scalar? I saw $s$ got cancelled, is this a coincidence? Does the cancellation always imply an accurate result instead of an approximation? $\endgroup$
    – Taozi
    Mar 25, 2016 at 4:47

An easy way to compute the Fresnel is not to use a steepest descent but simply Cauchy formula. The function $e^{iz^2}$ is analytic in the whole complex plane, so $$ \int_{\Gamma_N} e^{iz^2} dz =0 $$ where the contour $\Gamma_N$ is a wedge between $[0,N]$ and $[0,iN]$ positively oriented.

You can prove that the integral along the arc from $N$ to $e^{i\pi/4}N$ converges to 0 as $N\to\infty$ using e.g. Jordan's lemma. Moreover

$$ \int_{[0,e^{i\pi/4}N]} e^{iz^2} dz = e^{i\pi/4}\int_0^N e^{-x^2} dx \to \frac{\sqrt{\pi}}{2}e^{i\pi/4} $$ as $N\to\infty$. It follows that

$$\lim_{N\to\infty} \int_0^N e^{ix^2} dx = \frac{\sqrt{\pi}}{2}e^{i\pi/4} $$ Finally taking real part of both sides, we get

$$ \int_0^\infty \cos(x^2) dx = \sqrt{\frac{\pi}{8}} $$ which is the correct answer.


I found that what must be done is to make the change of variables $x→\sqrt{sz}$ and notice that since this is a trivial operation we can just compute in the limit s → ∞. This ‘localizes’ the integral at the saddle point. So now we can do the steepest descent and come to the right solution $$I= \frac{\sqrt{2}}{2\sqrt{2}}$$


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