How Does Asymptotic Expansion Apply to Integrals with Oscillatory Behavior?

[Clausius2]

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.

 

[Clausius2]

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

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

 

[tacman]

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