I have an issue with application of steepest descent, especially in the presence of more than one stationary point, where it seems that deformation of the integration contour could take one through an arbitrary number of such saddles.

For example consider $$I(x) = \int_{-2}^{6}\textrm{e}^{x \sin(t)} dt$$ in the two limits $x \rightarrow \pm \infty$ along the real axis.

We can identify the stationary points of the function $h(\zeta) = \sin(\zeta)$ as $\pm \pi/2$, $\pm 3\pi/2$ etc. By my reckoning the contours of steepest descent cross the stationary points parallel to the real axis (i.e. their imaginary part vanishes).

As $x \rightarrow + \infty$ the natural stationary point along the original contour of integration is at $x = \pi/2$ and steepest descent gives $$I(x) \sim \sqrt{2\pi} \textrm{e}^{x}/ \sqrt{x}.$$ For $x \rightarrow -\infty$ I find two stationary points along the original contour, at $x = 3\pi/2$ and $x = -\pi/2$ which leads to (counting them both) $$I(x) \sim 2\sqrt{2\pi} \textrm{e}^{-x}/ \sqrt{-x}.$$

This seems to match well with numerical integration in the corresponding limits.

Here is my doubt

The integrand is entire so we can deform the contour arbitrarily between the fixed endpoints. In this way I could choose to deform the contour to include an arbitrary number of stationary points of $h(\zeta)$, such as extending the contour along the negative axis to include $-3\pi/2$ or along the positive axis to pass through $5\pi/2$, then returning along along an arbitrary curve to get back to the endpoints. I should then also count these stationary points to estimate the asymptotic behaviour, changing the numerical pre-factors in the asymptotic results above.

In short: Is it necessary that the stationary points used in steepest descent be on the original contour of integration? If not, why can I not deform my contour to include an arbitrary number of stationary points for functions with multiple saddles?

  • $\begingroup$ Depends on the sign of x $\endgroup$ – lux Jun 1 at 20:18
  • $\begingroup$ Same as here, find the region $\{t: \operatorname {Re} \phi(t) < 1\}$ for $\phi(t) = \sin t$ and $\phi(t) = -\sin t$. $\endgroup$ – Maxim Jun 2 at 4:20
  • $\begingroup$ Are there not multiple which intervals where this holds evenly spaced along the real axis? $\endgroup$ – lux Jun 2 at 5:04
  1. Note first of all that OP's result follows directly from Laplace's method.

  2. However, OP asks specifically about the method of steepest descent, where it in principle is possible to deform the integration contour $[2,6]$ into the complex plane.

  3. Now anti-Stokes lines emanate from the stationary points in the direction where the argument to the exponential function in the integrand is imaginary.

  4. It turns out that the steepest descent contour should only cross anti-Stokes lines at stationary points. The long story short (when one try to draw all the anti-Stokes lines) is that it is therefore not possible to include (with non-zero multiplicity) the stationary points outside the interval $[-2,6]$ in the steepest descent contour.

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