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I need to calculate an integral by the saddle point method.

$I=\int_0^\infty dt\, f(t)\, e^{i S(t)}$.

Notice that in my case $S(t)$ has not the typical form $S(t)=\lambda g(t)$, with a parameter $\lambda \rightarrow \infty$ . Instead, it has the form $S(t)=\int_0^t h(t′)dt′$, and grows continuously with the parameter $t$.

I can find analytically the point $t_s$ for which $\frac{\partial S(t_s)}{\partial t}=0$.

Nevertheless $\frac{\partial^2 S(t_s)}{\partial t^2}=0$ , as well. This is a problem because the method is only valid for $\frac{\partial^2 S(t_s)}{\partial t^2}\neq0$. If I think of the saddle point method as a simple Taylor expansion of the argument of the exponential to second order, I feel tempted to write the saddle point solution as

$I^{sp}\approx f(t_s)e^{i S(t_s)}$,

instead of

$I^{sp}\approx \frac{1}{\sqrt{\partial^2 S(t_s)/\partial t^2}} f(t_s)e^{i S(t_s)}$

i.e., just evaluate the phase at $t_s$ but do not use the second derivative with respect to $t$ in the denominator. Is it correct? Or I should go to order $>2$ when the second derivative is zero?

Francisco
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  • Can you give a concrete example and specify what kind of approximation you're looking for (is there a large parameter)? – Maxim Mar 03 '19 at 01:55
  • No, in this case S(t) has not the typical form $S(t)=\lambda g(t)$, with $\lambda \rightarrow \infty$ . It has the form $S(t)=\int_0^t h(t') dt'$. And grows continuously with the parameter $t$. It is not a numerical problem, but an analytical derivation what I am aiming at. – Francisco Mar 03 '19 at 13:32
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    That's not exactly how the saddle point method works. Suppose $f$ is oscillatory as well, so you have something like $\int_0^\infty \sin(5 t) \exp(i t^3) ,dt$. You won't get a useful approximation from replacing the integrand with $5 t \exp(i t^3)$, however you deform the contour around zero. You can take higher order terms, but without knowing what $f$ and $h$ are, it's hard to say whether or not that would be useful/feasible. – Maxim Mar 03 '19 at 17:19
  • In the example you mention $S(t)=\int_0^t \sin(5t') exp(it'^3) dt'$, then you should find the zeroes $t_s$ of $\partial S(t)/\partial t=\sin(5t) exp(it^3)=0$, then $S(t_s)=\int_0^{t_s} \sin(5t') exp(it'^3) dt'$. And the second derivative would be $\partial^2 S(t_s)/\partial t^2=e^{i t_s^3} \left(5 \cos (5 t_s)+3 i t_s^2 \sin (5 t_s)\right)$. In that case, I think you would not find the type of problems I mention and you could just use the approximation whatever good or bad. – Francisco Mar 04 '19 at 03:52
  • I took it as an example for $I$. – Maxim Mar 04 '19 at 04:16
  • In the example you mention $t^2$ (the derivative of the phase) has a zero at $t=0$, and I want to evaluate an integral from $0$ to $\infty$. A more explicit example of the type of integrals you could solve with the method in the way I mention appear very often in the field of Physics, they are of the type $I=\int_0^{\infty} f(t) \text{e}^{i \int_0^t \cos[\sin(t')]dt'}$. There you see that the exponential grows continuously, but in an oscillatory way. – Francisco Mar 04 '19 at 15:21
  • What is the solution for $I$ from your last comment? – Maxim Mar 04 '19 at 17:26
  • Look, this is going nowhere. I understand your skepticism, and how does the Socratic method work. Maybe the way I use this method comes from the field of physics, not math. Could you please tell me what bibliography would you read for getting an explicit formula for order >2? That would answer my question. I am reading the book from R. Wong "Asymptotic Approximations of Integrals", but I couldn't find it. Thank you. – Francisco Mar 05 '19 at 12:06

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I think the saddle point solution should be just

$I^{sp}\approx f(t_s)e^{i S(t_s)}$

Because the original formulation can be thought of as a Taylor expansion of the exponential to second order, and the square root of the denominator would come from the gaussian integral involving the second order, which is null, and therefore that integral is not performed for getting the result. See, for instance How is the Saddle point approximation used in physics?

Francisco
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