Asymptotic expansion of an integral

[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!

 

[sunjin09]

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.