complex-analysismathematical physics
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.
Let's write your integral as $\DeclareMathOperator{re}{Re}$
$$ \int_{-\infty}^{\infty} f(t) e^{kg(t)}\,dt, $$
where
$$ f(t) = \frac{1}{1+t^2} \qquad \text{and} \qquad g(t) = \frac{i}{5}t^5 + it. $$
The lay of the land.
The critical points of the exponent function $g$ occur at $t = e^{i\pi (2k+1)/4}$, $k=0,1,2,3$.
Here's a plot showing these critical points in yellow with the paths of constant altitude (of $\re g$) passing through them shown in white. The real axis is shown in black. The background is colored according to the value of $\re g(t)$, with higher points colored lighter and lower points darker.
Note that the function $\re g$ has ten "hills" and "valleys" radiating away from the origin. Since
$$ g(t) \sim \frac{i}{5}t^5 $$
as $t \to \infty$ we deduce that these hills and valleys lie approximately on the rays
$$ t = s e^{i\pi(2k+1)/10}, \quad s > 0,\,k = 0,1,\ldots,9, $$
with even $k$ corresponding to valleys and odd $k$ corresponding to hills.
In order for the contour to pass through either of the two saddle points in the lower half-plane we would need to deform at least one tail of the current contour (the real axis) over one of these hills. This doesn't seem feasible, so we'll instead focus on the two saddle points in the upper half-plane.
The new contour.
With a little work it's possible to show that we can deform the contour to the one shown in black in the following image. We'll call this new contour $\gamma$.
Here lines of constant altitude on the surface $\re g(t)$ are again shown in white. The point $t=i$ is shown in yellow.
The new contour $\gamma$ consists of two curves. The first originates at $t = e^{i\pi 9/10} \infty$ then passes through the saddle point at $t = e^{i\pi 3/4}$ at an angle of $\pi/8$ before terminating at $t = i \infty$. The second originates at $t = i \infty$, passes through the saddle point at $t = e^{i\pi/4}$ at an angle of $-\pi/8$, then terminates at $e^{i\pi/10} \infty$.
Note that to deform the contour from the real axis to $\gamma$ we must enclose the pole of $f$ located at $t=i$. Ultimately we have
$$ \begin{align} \int_{-\infty}^{\infty} f(t) e^{kg(t)}\,dt &= \int_\gamma f(t) e^{kg(t)}\,dt + 2\pi i\operatorname{Res}\left(f(t) e^{kg(t)},t=i\right) \\ &= \int_\gamma f(t) e^{kg(t)}\,dt + \pi e^{-6k/5}. \tag{1} \end{align} $$
We will show in the next section that the term $\pi e^{-6k/5}$ is negligible compared to the integral $\int_\gamma$.
Estimating the new integral.
Now we will estimate the integral
$$ I(k) = \int_\gamma f(t) e^{kg(t)}\,dt. $$
We've chosen the contour $\gamma$ in such a way that the largest values of $\re g(t)$ for $t \in \gamma$ occur at the saddle points $t = e^{i\pi 3/4}, e^{i\pi/4}$. Further,
$$ \re g\left(e^{i\pi 3/4}\right) = \re g\left(e^{i\pi/4}\right) = -\frac{4}{5\sqrt{2}}. $$
Consequently we'll need to take both saddle points into account. For the first we have
$$ g\left(e^{i\pi 3/4} + se^{i\pi/8}\right) = \frac{4}{5} e^{-i\pi 3/4} - 2s^2 + O(s^3) $$
and for the second we have
$$ g\left(e^{i\pi/4} + se^{-i\pi/8}\right) = \frac{4}{5} e^{i\pi 3/4} - 2s^2 + O(s^3) $$
as $s \to 0$, so applying the Laplace method yields, to leading order,
$$ \begin{align} I(k) &\approx e^{i\pi/8} f\left(e^{i\pi 3/4}\right) \int_{-\infty}^{\infty} \exp\left[k\left(\frac{4}{5} e^{-i\pi 3/4} - 2s^2\right)\right]\,ds \\ &\qquad + e^{-i\pi/8} f\left(e^{i\pi/4}\right) \int_{-\infty}^{\infty} \exp\left[k\left(\frac{4}{5} e^{i\pi 3/4} - 2s^2\right)\right]\,ds \\ &= \sqrt{\frac{\pi}{k}} \exp\left(-\frac{4}{5\sqrt2}k\right) \cos\left(\frac{4}{5\sqrt2}k - \frac{3\pi}{8}\right) \end{align} $$
as $k \to \infty$. Combining this with equation $(1)$ we conclude that, to leading order,
$$ \int_{-\infty}^{\infty} f(t) e^{kg(t)}\,dt \approx \sqrt{\frac{\pi}{k}} \exp\left(-\frac{4}{5\sqrt2}k\right) \cos\left(\frac{4}{5\sqrt2}k - \frac{3\pi}{8}\right) $$
as $k \to \infty$, as desired.
To start, recognize that
$$ I(\lambda) = \operatorname{Re} \int_0^1 e^{i\lambda x^3}\,dx. $$
Now there are several questions that you can ask to get yourself going:
Where is the saddle point?
What are the paths of steepest descent away from the saddle point?
How can I deform my contour so that it follows this path of steepest descent?
The last one is a bit tricky since the endpoints of the contour are finite. You only need to follow a portion of the path of steepest descent though; you can have the contour return to its start/endpoint afterwards.
Let me know if you get stuck on any of these.
Best Answer
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:
$$\sqrt{s}\int\limits_{0}^{1}e^{isz^2}dz$$
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 $$\sqrt{\frac{\pi}{8}}$$