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The question asks to show that the leading term of the integral

$$ \int_{-\infty}^\infty (1+t^2)^{-1}\exp\left(ik(t^5/5+t)\right) dt $$

for large $k$ using the method of steepest descent is equal to

$$ \sqrt{\frac{\pi}{k}} e^{\frac{-4k}{5\sqrt2}} \cos\left(\frac{4k}{5\sqrt2} - \frac{3\pi}{8}\right) $$

...I don't even know how to pick my contour for this problem. Thanks for the help!

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  • $\begingroup$ I edited your latex to make the formulas more readable. $\endgroup$ – Antonio Vargas May 2 '14 at 1:42
  • $\begingroup$ Thanks...can you also answer my question? $\endgroup$ – user137302 May 2 '14 at 2:31
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    $\begingroup$ I'm taking a look at it. $\endgroup$ – Antonio Vargas May 2 '14 at 2:32
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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.

enter image description here

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 $\gamma$.

enter image description here

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.

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  • $\begingroup$ Hi @Antonino Vargas, i know this is a rather old question, but might u explain to me why you exactly choose this contour? I somehow don't get it. Thanks! $\endgroup$ – tired Dec 1 '15 at 10:32
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    $\begingroup$ Hi @tired. This contour was chosen because it passes through the two saddle points at $t=e^{i\pi/4}$ and $t=e^{i\pi3/4}$ in the directions of steepest descent from them. This can be seen in the second image: the contour comes from a dark region, passes through the saddle point (where the white curves intersect) at a certain angle and heads to another darker region. this ensures that the contour is as low as it can be on the surface $\operatorname{Re} \varphi(t)$ with the saddles being its highest points (and thus they contribute most to the size of the integral). $\endgroup$ – Antonio Vargas Dec 2 '15 at 2:38
  • $\begingroup$ Hi @AntonioVargas, how do you know the crossing angle at the saddle point, the left one for example, is $pi/8$? I have trouble thinking of an easy way. $\endgroup$ – Taozi Mar 16 '17 at 18:04
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    $\begingroup$ @Taozi, sorry for the late reply. The goal is for the contour to cross this saddle along a path which makes $\re g(t)$ (the important quantity when measuring $|e^{kg(t)}|$) rise and fall the fastest. In other words, we want $\re g(t)$ to vary a lot, and $\DeclareMathOperator{\im}{Im} \im g(t)$ to vary little (or not at all). The simplest method to get the angle, then, is to expand the function $g$ in series around the saddle point. For example near the saddle at $t=e^{i\pi/4}$ we have $$g(t) = -\frac{4}{5} e^{i\pi/4} + 2e^{i3\pi/4} \left(t-e^{i3\pi/4}\right)^2 + \cdots.$$ (...) $\endgroup$ – Antonio Vargas Mar 30 '17 at 12:30
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    $\begingroup$ @AntonioVargas Hi Antonio, your explanation is very clear, thanks a lot. A minor typo is the example you elaborate upon should be at "saddle at $t=e^{i 3 \pi /4}$". $\endgroup$ – Taozi Mar 31 '17 at 0:03
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User Antonio Vargas has already given a very good answer. In this similar answer, we will use slightly different words and stress slightly different things.

I) We are given the integrand

$$ z ~~\mapsto~~ e^{ikS(z)}f(z), \qquad f(z)~:=~ \frac{1}{z^2+1}~=~\frac{1}{(z-i)(z+i)},$$

$$ S(z)~:=~\frac{z^5}{5}+z, \qquad S^{\prime}(z)~=~z^4 + 1, \qquad S^{\prime\prime}(z)~=~4z^3. \tag{1} $$

We are only considering integration contours $\gamma$ that start and end at $|z|=\infty$ via one of the 5 exponentially damped sectors of the $z$-plane. The main point is that the integrand (1) is a meromorphic function in the complex $z$-plane, so the that the contour integral (1) is a sum of the residue at

$$\int_{\gamma}\! \mathrm{d}z~ e^{ikS(z)}f(z) \tag{2} $$

does only depend on monodromy at $|z|=\infty$ and around the poles $z_{\pm i}=\pm i$. There are 4 critical points:

$$z_j ~=~ \exp(\frac{ij\pi}{4}), \qquad j~\in~\{1,3,5,7\}, $$

$$ S(z_j)~=~\frac{4}{5}z_j, \qquad S^{\prime}(z_j)~=~0, \qquad S^{\prime\prime}(z_j)~=~4z_j^3. \tag{3} $$

Each of the corresponding 4 steepest descent contours corresponds to monodromy between two neighboring exponentially damped sectors, cf. e.g. Ref. [W].

II) OP's integration contour is just above the $x$-axis:

$$ I(k)~:=~ \int_{\mathbb{R}+i0^+}\! \mathrm{d}z~ e^{ikS(z)}f(z). \tag{4} $$

It becomes clear that OP's integral (4) is the sum of the following contributions:

  • the 2 steepest descent contours corresponding to the 2 critical points $z_3$ and $z_1$ with angular steepest decent direction $\pm \frac{\pi}{8}$, respectively;

  • $2\pi i$ times the residue at the pole $z_{+i}=+ i$ in the upper half plane.

The method of steeepest descent therefore yields

$$ I(k)~\sim~ \sum_{j\in\{1,3\}} \sqrt{\frac{2\pi}{-ikS^{\prime\prime}(z_j)}} e^{ikS(z_j)}f(z_j) + 2\pi i {\rm Res}\left( z \mapsto e^{ikS(z)}f(z), i\right) $$ $$ ~=~\sqrt{\frac{2\pi}{4e^{\frac{-i\pi}{4}} }}\exp\left(ik\frac{4z_3}{5}\right) f(z_3) + \sqrt{\frac{2\pi}{4e^{\frac{i\pi}{4}} }}\exp\left(ik\frac{4z_1}{5}\right) f(z_1) + \pi e^{ikS(i)} $$ $$ ~=~ \sqrt{\frac{\pi}{k}} \exp\left(-\frac{2\sqrt{2}k}{5}\right) \cos\left(\frac{2\sqrt{2}k}{5} - \frac{3\pi}{8}\right) + \pi e^{-\frac{6k}{5}}\quad\text{for}\quad k~\to~ \infty.\tag{5} $$

References:

  • [W] E. Witten, Analytic Continuation Of Chern-Simons Theory, arXiv:1001.2933; p. 23-29, 48-49. A related 2015 KITP lecture by Witten, A New Look At The Path Integral Of Quantum Mechanics, can be found on YouTube.
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