Saddle Point Approximation for the Integral ∫0∞xe-ax-b/√xdx

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SUMMARY

The discussion focuses on applying the saddle point approximation to the integral I = ∫0∞xe^(-ax-b/√x) where a, b > 0. Participants analyze the function f(x) = ln(x) - ax - b/√x to identify the maximum point x0 and determine the appropriate parameters for the approximation. The key insight is that the derivative f'(x) must be evaluated correctly to find x', the critical point, and adjustments may be necessary to account for additional terms in the approximation. The conversation emphasizes the importance of correctly expressing x in terms of x' and a new variable z for accurate integration.

PREREQUISITES
  • Saddle point approximation techniques
  • Understanding of integral calculus
  • Knowledge of derivatives and critical points
  • Familiarity with logarithmic functions and their properties
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  • Study the derivation of the saddle point approximation in detail
  • Learn about the Euler integral and its applications
  • Explore the Stirling formula and its connection to saddle point methods
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Mathematicians, physics students, and researchers involved in asymptotic analysis and integral approximations will benefit from this discussion.

derravaragh
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Homework Statement


Apply saddle point approximation to the following integral:

I = ∫0xe-ax-b/√xdx a,b > 0

Recall that to derive Stirling formula from the Euler integral in class we required N >> 1. For the integral defined above, identify in terms of a and b appropriate parameter that justifies the use of the saddle point approximation.

Homework Equations


The Attempt at a Solution


In class, my teacher worked through a simpler problem for the integral xne-x, and started by finding out the maximum x0=n.

Trying to follow my teachers example, I changed the integral to eln(x) - ax - b/√x
and looked at the limits. I set f(x) = ln(x) - ax - b/√x, so that as x→∞, f(x)→-∞ and is almost linear, and as x→0, f(x) depends on ln(x) - b/√x.

I guess my issue is I'm not really sure how to determine the parameters. My intuition tells me that a<x0<b, but I don't know how to show this, or if that's even what is asked of me. I tried to take the derivative of f(x) to determine x0 and came up with f'(0) = (1/x)-a-(b/2)*x-3/2 = 0, which isn't making this easier on me. Am I on the right track or am I completely missing the obvious here?
 
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derravaragh said:
f'(0) = (1/x)-a-(b/2)*x-3/2 = 0
You don't mean f'(0), and you have a sign wrong.
Let x' be the solution of f'(x) = 0. Express x as x' plus some new variable, z. Plug that into the integrand and approximate for small z. You can use the f'(x')=0 equation to get some cancellation. (You may find you get too much cancellation and you need to include smaller terms, like z2.)
 

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