# How Does Asymptotic Expansion Apply to Integrals with Oscillatory Behavior?

• Clausius2
In summary, the speaker is seeking help with the expansion of an integral involving functions Z(x) and J_o(\lambda x). They mention that Z(x) behaves like x^\sqrt{2} for small x and exponentially for large x. They note that for other examples, the main contribution to the integral comes from the region x\sim 1/\lambda, and for larger x the integral is cancelled due to rapid oscillations. The speaker suggests a change of variable, t=\lambda x, but it leads to a divergent integral instead. They are looking for hints on how to approach the original convergent integral.
Clausius2
Gold Member
Hey guys,
I need help with the expansion of this integral:

$$\int_0^\infty Z(x) J_o(\lambda x)dx$$ for $$\lambda \rightarrow \infty$$

where I know that $$Z(x)\sim x^\sqrt{2}$$ for small $$x$$ and
exponentially small for large $$x$$

It seems with other examples that I have done that the major contribution to the integral comes from the region $$x\sim 1/\lambda$$. For larger $$x$$ the integrand oscillates rapidly and the integration cancels. One change of variable (re-scaling) that you may try is $$t=\lambda x$$. But if you do it you end up with a divergent integral. And at first glance the original integral is convergent. Any hints?

Thanks.

Last edited:
with divergent integral I mean that I arrive to this integral after the rescalement:

$$\frac{1}{\lambda^{\sqrt{2}+1}}\int_0^\infty t^{\sqrt{2}} J_o(t)dt$$

The asymptotic expansion of an integral involves finding an approximation for the integral as the variable of integration approaches a large value. In this case, we are looking for an expansion of the integral \int_0^\infty Z(x) J_o(\lambda x)dx as \lambda \rightarrow \infty.

One approach to this problem is to use the method of steepest descent, which involves finding the stationary points of the integrand and then deforming the integration path to pass through those points. However, in this case, it may not be feasible to find the stationary points analytically.

Another approach is to use the method of stationary phase, which involves expanding the integrand in terms of its Fourier transform and then evaluating the integral using a saddle point approximation. This method is more suitable for functions with oscillatory behavior, which is the case for the integrand \int_0^\infty Z(x) J_o(\lambda x)dx.

In order to apply the method of stationary phase, we first need to find the stationary points of the integrand. This can be done by setting the derivative of the integrand with respect to x equal to zero. In this case, we get the equation Z'(x)J_o(\lambda x) + \lambda Z(x)J_o'(\lambda x) = 0. Solving for x, we get the stationary points as x = \frac{2m+1}{2\lambda} where m = 0, 1, 2, ...

Next, we expand the integrand in terms of its Fourier transform using the stationary points as the center of the expansion. This gives us \int_0^\infty Z(x) J_o(\lambda x)dx \approx \sum_{m=0}^\infty \frac{Z(x_m)}{\sqrt{\lambda}} J_o(\lambda x_m) e^{i\phi(x_m)} where x_m = \frac{2m+1}{2\lambda} and \phi(x_m) = \int_0^{x_m} \sqrt{Z(x)}dx.

Finally, we can evaluate this sum using the saddle point approximation, which involves keeping only the first term in the sum and evaluating it at the stationary point x_m. This gives us the asymptotic expansion \int_0^\infty Z(x) J_o(\lambda x)dx \approx \frac{Z(x

## 1. What is an asymptotic expansion integral?

An asymptotic expansion integral is a mathematical technique used to approximate the value of an integral when the limits of integration approach infinity. It involves breaking down the integral into simpler terms and taking the dominant term as the approximation.

## 2. How is an asymptotic expansion integral calculated?

An asymptotic expansion integral is calculated by first finding the dominant term in the integral and then finding its corresponding series expansion. The approximation is then made by taking the first few terms of the series.

## 3. What is the purpose of using an asymptotic expansion integral?

The purpose of using an asymptotic expansion integral is to approximate the value of an integral that may be difficult or impossible to solve exactly. It is often used in physics, engineering, and other fields to simplify complex integrals and obtain quick and accurate estimates.

## 4. Are there any limitations to using an asymptotic expansion integral?

Yes, there are limitations to using an asymptotic expansion integral. It is only accurate when the limits of integration are very large, and may not work when the limits are close to each other. Additionally, it may not work for integrals with oscillating or rapidly changing functions.

## 5. Can an asymptotic expansion integral be used for any type of integral?

No, an asymptotic expansion integral can only be used for integrals that have a dominant term that dominates the other terms as the limits of integration approach infinity. It may not work for integrals with multiple dominant terms or for integrals with rapidly changing functions.

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