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The question asks for the leading term in the asymptotic expansion of

$$I(x) = \int_0^{\pi/2} \biggr ( 1-\frac{2t}{\pi} \biggr )^k \cos(x\cos{t}) \, dt, \;\; x \to \infty$$

for $k = 0, -1/2, -3/4$.

My attempt:

Let $$J(x) = \int_0^{\pi/2} f_k(t) e^{ixu(t)} \, dt$$ where $f_k(t)=( 1-\frac{2t}{\pi})^k$ and $u(t)=\cos{t}$, so that $I(x) = Re (J(x))$

Applying the stationary phase method, I know the dominant $1/\sqrt{x}$ contribution will come from stationary points of $u(t)$, in our case at $t=0$.

So

$$J(x) \sim f_k(0) \int_0^\epsilon e^{ixu(t)} \, dt$$

We approximate $u(t) \approx 1 - \frac{t^2}{2}$ near $t=0$ so that

$$J(x) \sim f_k(0) e^{ix} \int_0^\epsilon e^{-ixt^2/2} \, dt$$

Asserting the $[\epsilon, \infty)$ contribution to be exponentially small and applying a Fresnel integral we obtain

$$J(x) \sim f_k(0) e^{i(x-\pi/4)} \sqrt{\frac{\pi}{2x}}$$

and so $$I(x) \sim f_k(0) \cos{(x-\pi/4)} \sqrt{\frac{\pi}{2x}}$$

But this is incorrect, (for one note $f_k(0)=1$ at all our values of $k$ of interest), yet I cannot see where I have made a mistake.

I did wonder if perhaps taking the Real part was not justified here, so i repeated the calculation instead using $\cos{\theta}=(e^{i\theta}+e^{-i\theta})/2$, but arrived at the same answer.

Any help is appreciated!

Gary
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Mr_Random_Guy
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  • Why do you think the result is not correct at first order? – Gregory Mar 07 '22 at 19:53
  • I find it unlikely that the question would ask for three specific cases in $k$ if they were all trivially identical. – Mr_Random_Guy Mar 07 '22 at 20:03
  • Perhaps more convincingly, I've numerically experimented to see that my result only seems to hold for $k=0$. As $k$ decreases, my result decays too quickly compared to the true asymptotic – Mr_Random_Guy Mar 07 '22 at 20:31
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    For the other values of $k$, you have an algebraic singularity (a blow-up type) at the upper endpoint of integration. That will also contribute to the asymptotics. – Gary Mar 08 '22 at 02:38
  • I see I've missed the singularity's contribution, but I do not know how to determine its contribution – Mr_Random_Guy Mar 08 '22 at 22:11
  • We can choose $[0, \pi/2 - \pi i/2, \pi/2]$ as the contour of integration and, applying the method of the steepest descent, show that, with the principal branch of $z^k$, $$\operatorname {Re} \int_0^{\pi/2} \left(1 - \frac {2 t} \pi \right)^{! k} e^{i x \cos t} dt = \ \operatorname {Re} \int_0^{e^{-\pi i/4} \infty} e^{i x (1 - t^2/2)} dt + \operatorname {Re} \int_{\pi/2 - i \infty}^{\pi/2} \left( 1 - \frac {2 t} \pi \right)^{! k} e^{i x (\pi/2 - t)} dt + o(x^{-\min(k + 1, 1/2)}).$$ – Maxim Mar 10 '22 at 17:19

1 Answers1

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Thanks to Maxim's comment for this answer. I've slightly rewritten it to try and do this without steepest descent, but the contour integration we'll need is essentially the same thing anyway.

The contribution missing from the singularity at $t=\pi/2$ is

$$K = \int_{\pi/2 - \epsilon}^{\pi/2}f_k(t)e^{ix\cos{t}} dt \approx \int_0^\epsilon (\frac{2t}{\pi})^k e^{ixt} dt$$

Integration by parts once and applying Riemann Lebesgue's lemma shows the $[\epsilon,\infty)$ contribution to be $O(1/x)$ (assuming $k<0$), which we will neglect.

So

$$K \sim \int_0^\infty(\frac{2t}{\pi})^k e^{ixt} dt$$

Following the same idea as a Fresnel integral you can deform the contour of integration to up the imaginary axis (show the quarter circle contribution vanishes) to get

$$K \sim i \int_0^\infty(\frac{2it}{\pi})^k e^{-xt} dt = (\frac{2}{\pi})^k i^{k+1} \frac{\Gamma(k+1)}{x^{k+1}}$$

(note $0<k+1<1$ so we were safe to neglect $O(1/x)$ terms). So all in all

$$I(x) \sim \cos{(x-\pi/4)} \sqrt{\frac{\pi}{2x}} + (\frac{2}{\pi})^k \frac{\Gamma(k+1)}{x^{k+1}} \cos{\frac{\pi(k+1)}{2}}$$

Mr_Random_Guy
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