I'm looking for asymptotics for an integral of the form:

$$F(n):=\int_{1/2-i\infty}^{1/2+i\infty} e^{\phi(n,z)}dz$$

where $\phi(n,z)=(n-n^3)\log(1-z)+n^2\log(1+z)-n\log(z)$. One can solve for the saddle points of $\phi$:

$$z_{1,2}=\frac{1-n-n^2\pm \sqrt{1-6n+3n^2+2n^3+n^4}}{2(n^2-n)},$$

with taylor series:

$$z_1=-1-\frac{2}{n}+O(1/n^2)$$ $$z_2=\frac{1}{n^2}+O(1/n^3).$$

It looks like $z_2$ is the maximum saddle, and $z_1$ the minimum. So to get asymptotics of $F(n)$, we need to deform the vertical line contour to the steepest-descent contour, passing through $z_2$. The steepest descent contour looks like this:

enter image description here

I have some questions. First, I'm having trouble solving for the contour of steepest-descent through $z_2$, in other words $Im(\phi(n,z))=0$, passing through $z_2$. The equation is rather nasty, and probably transcendental. Is it sufficient to somehow approximate this contour to some order in $n$? The issue is that I'm thinking I need to parametrize the steepest descent contour with some running variable $t$, so that $\Gamma_t=z_2+f_n(t)$, where $f_n(0)=0$. Also, $z_2$ depends on $n$, and so I'm thinking I need a parametrization of $\Gamma_t$ that's somehow uniform in $n,t$ for $t$ near zero and $n$ large.

Ultimately, it looks like I need to perform some sort of rescaling in the integral in terms of $z$. I was thinking of trying $u=z\cdot z_1$, so that the saddle point gets moved to $u=1$. But I'm still stuck in getting some kind of a uniform estimate on $\phi(n,z)$ near $z_1$, along the contour of steepest descent.

Note that the integral has singularities at $z=\pm 1$ and $z=0$. In particular, $z_2\rightarrow 0$, so this further complicates taylor expansions of $\phi(n,z)$ near $z_2$. I've also noticed that $\phi^{(k)}(n,z_1)$ looks like a growing polynomial in $n$, e.g. $n^k$, so I can't seem to truncate taylor series for $\phi(n,z)$ around $z_1$ so that it's uniform in $n,z$.

I've never seen literature on such inhomogenous steepest descent problems. A reference would be sincerely appreciated!

  • 5
    $\begingroup$ If my calculations are correct then another representation for your integral is $$F(n) = 2\pi i \sum_{k=0}^{n-1} \binom{n^3-2-k}{n-1-k}\binom{n^2}{k}.$$ This isn't necessarily easier to estimate rigorously but it allows us to guess the asymptotic $$F(n) \sim i \sqrt{2\pi} e^{n+1} n^{2n-5/2}$$ which agrees numerically (I tried $n = 1000$ and $4000$). I'm very interested to see if this can be approached with saddle point methods. $\endgroup$ – Antonio Vargas Dec 23 '14 at 4:53
  • 1
    $\begingroup$ @Antonio Vargas: You should make this a reply, even if you're waiting for the saddle point methpd. It's a valid methog of approaching the problem, one I suspect OP didn't consider. $\endgroup$ – Raskolnikov Dec 24 '14 at 11:53

The power of the method of steepest decent is the fact that you do not have to get the contour "right". Having the saddle point and the correct "direction" is enough. The reason is that for $n\to \infty$, the integral is exponentially dominated from points close to the saddle point and everything along the contour away from the saddle point is negligible. For a reference, have a look at "Advanced Mathematical Methods" by Orszag and Bender.

For your specific case with $\phi(n,z)=(n-n^3)\log(1-z)+n^2\log(1+z)-n\log(z)$. As you have noted, we have two saddle points at $z_1 \approx -1 $ and $z_2 \approx 0$. For the first saddle point, the direction of steepest decent is along the real axis (as can be checked by calculating the second derivative). The second saddle point suits our needs as its steepest decent is long the imaginary axis.

The saddle point is characterized by the expansion of $\phi$ around $z^*=z_2$ to second order. We obtain to leading order (in $n$) $$ \phi \sim \phi_2= (2 \log n -1) n + \frac{n^5}{2} (z-z^*)^2 .$$

The idea of the method of steepest decent, is to write $\phi = \phi_2 + \delta \phi$. We obtain $$F(n) = \int_{-1/2-i \infty}^{1/2+i \infty}\!dz\, e^{\phi_2 + \delta \phi}.$$ In the next step, we deform the contour such that it passes through $z^* \approx n^{-2}$ (parallel to the imaginary axis). We then know that the integral is dominated by points close to $z^*$ and thus we can expand $\exp(\delta \phi)$ around $z=z^*$. The leading term is (with $z = z^* + i \xi $) $$F(n) \sim i \underbrace{e^{2n+1}}_{\exp(\delta \phi|_{z=z^*})} \int_{-\infty}^{\infty} \!d\xi\, e^{\phi_2} = i \sqrt{2\pi} e^{n+1} n^{2n-5/2},$$ where we have neglected terms that vanish for $n\to \infty$ in $\delta \phi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.