Asymptotic expansion of an integral

• Elec Man
In summary, the problem at hand is to find the leading order term in the expansion of the given integral. The method of stationary phase is used, where the stationary point is at t = 0. However, complications arise due to the fact that f(0) = 0 and a third derivative is needed for a nonzero value. Integration by parts is suggested as a possible solution, with the endpoint contribution from 1 also taken into account.
Elec Man

Homework Statement

Find the leading order term in the expansion of

$$\int_0^1\ \sin(x(t-\sin t))\ \sinh^a t\ dt$$

Homework Equations

For a general Fourier integral, the method of stationary phase gives

$$\int_a^b\ f(t) e^{i x \psi(t)}\ dt ~ f(a)e^{ix\psi(a) \pm i\pi/2p}(p!/(x|\psi^{(p)}(a)))^(1/p) \Gamma(1/p)/p$$

where p is the order of the first nonzero derivative of $\psi$

The Attempt at a Solution

The first thing to do is to put it into an easier form:

$$\int_0^1\ e^{ix(t-\sin t)} \sinh^a t\ dt$$

Now it seems like it'd just be an ordinary Fourier integral. To evaluate a Fourier integral, you search for the point of stationary phase (where the derivative of the imaginary part is zero). That point is clearly t = 0. From here, you'd normally expand around t=0 and integrate. Unfortunately, it has two complications that I don't know how to deal with. The first is that $\sinh 0 = 0$, so in the above formula, f(a) = 0. The other problem is that you have to go out to the third derivative of (t-\sin t) to find a nonzero value, which makes brute forcing via Taylor expansion harder.

If you expand around t = 0, the integral becomes

$$\int_0^1\ t^a e^{ix(t^3/6)}$$

which doesn't seem any easier to integrate. If you do integration by parts, you get another integral which increases with x, which is the opposite of what we want. Also, in this method, you usually expand the region of integration to 0 to ∞ on the basis that the largest contribution to the integral comes from the area near the stationary point (everything else oscillates rapidly and cancels). I'm not even sure that works here; there's a warning about that if f(a) = 0.

Does anyone have any experience with this that can help me out? Thanks!

Since your f(0)=0 at the stationary point, what you can do is integrate by part
∫ f(t)exp(ix g(t))dt = f(t)/(ix g'(t))-1/(ix)*∫ (f(t)/g'(t))' exp(ix*g(t)) dt
and hopefully in the new integrand, (f(t)/g'(t))' is non-zero at the stationary point. Also note your stationary point contribution is only half, since your range of integral starts at 0. Also account for the endpoint contribution from 1.

1. What is an asymptotic expansion of an integral?

An asymptotic expansion of an integral is a mathematical technique used to approximate the value of an integral when the limits of integration are large or infinite. It involves expressing the integral as a series of terms, with each term becoming increasingly smaller as the limit of integration increases.

2. When is an asymptotic expansion of an integral used?

An asymptotic expansion of an integral is often used in situations where the exact value of the integral is difficult or impossible to calculate. It is also used when the limits of integration are large, making it impractical to calculate the integral using traditional methods.

3. How is an asymptotic expansion of an integral calculated?

The asymptotic expansion of an integral is calculated using a variety of techniques, such as the method of steepest descent or the method of stationary phase. These methods involve manipulating the integrand and performing a series of approximations to obtain a simplified expression for the integral.

4. What are the advantages of using an asymptotic expansion of an integral?

An asymptotic expansion of an integral allows for a quick and accurate approximation of the integral, especially when traditional methods are not feasible. It can also provide insight into the behavior of the integral as the limits of integration increase, and can be used to prove the convergence or divergence of the integral.

5. Are there any limitations to using an asymptotic expansion of an integral?

While an asymptotic expansion of an integral can provide a good approximation, it is not an exact solution and may introduce some error. It also requires a certain level of mathematical skill and understanding to perform the necessary manipulations and approximations. Additionally, it may not be applicable in all situations, such as when the integrand is highly oscillatory or when the limits of integration are not large enough.

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