5

Based on the asymptotic analysis of correlation functions at large distence in Physics, now I get a math question. Let the function $$f(x)=\int_{-1}^{1}\sqrt{1-k^2}e^{ikx}dk.$$

Without working out the explicit form of this function, how do we know the asymptotic behaviour of $f(x)$ at large distence $\left |x \right |\rightarrow+\infty$?

Andrews
  • 3,953
Kai Li
  • 581
  • 1
    Probably you can use this: http://math.stackexchange.com/questions/318804/relation-between-function-discontinuities-and-fourier-transform-at-infinity – leonbloy Oct 19 '13 at 18:46
  • 1
    The Maple code $$asympt(int(sqrt(-k^2+1)exp(Ik*x), k = -1 .. 1), x, 2) $$ outputs $$ -\sqrt {\pi }\sqrt {2}\cos \left( x+1/4,\pi \right) \left( {x}^{-1} \right) ^{3/2}+O \left( \left( {x}^{-1} \right) ^{5/2} \right) .$$ The closed form of the integral is expressed in terms of the Bessel function. – user64494 Oct 19 '13 at 19:13
  • @ leonbloy Thanks for your good suggestion. – Kai Li Oct 20 '13 at 09:31
  • @ user64494 I can't understand what are you talking about... – Kai Li Oct 20 '13 at 09:32

1 Answers1

8

I going to manipulate the integral into a form that I can analyze the method of stationary phase. Let $k=\cos{\theta}$ and the integral becomes

$$\begin{align}I(x) &= \int_0^{\pi} d\theta \, \sin^2{\theta} \, e^{i x \cos{\theta}}\\ &= \int_0^{\pi} d\theta \, e^{i x \cos{\theta}} - \int_0^{\pi} d\theta \, \cos^2{\theta} \, e^{i x \cos{\theta}}\\ &= \int_0^{\pi} d\theta \, e^{i x \cos{\theta}} + \frac{d^2}{dx^2} \int_0^{\pi} d\theta \, e^{i x \cos{\theta}}\end{align}$$

Now, note that

$$\int_0^{\pi} d\theta \, e^{i x \cos{\theta}} = \int_0^{\pi/2} d\theta \, e^{i x \cos{\theta}} + \int_{\pi/2}^{\pi} d\theta \, e^{i x \cos{\theta}} = 2 \Re{\left [ \int_0^{\pi/2} d\theta \, e^{i x \cos{\theta}}\right]}$$

Now, we may apply stationary phase. The stationary point of the integrand is at $\theta=0$; there, we may approximate the argument of the exponential by its Taylor expansion. Further, because of the oscillatory cancllations, we may simply draw the upper limit of the integral to infinity to first order:

$$\int_0^{\pi/2} d\theta \, e^{i x \cos{\theta}} \sim e^{i x} \int_0^{\infty} d\theta \, e^{-(i x/2) \theta^2} = \frac12 e^{i(x-\pi/4)} \sqrt{\frac{2 \pi}{x}} \quad (x\to\infty)$$

Therefore

$$\int_0^{\pi} d\theta \, e^{i x \cos{\theta}} \sim \sqrt{\frac{2 \pi}{x}} \cos{\left ( x-\frac{\pi}{4}\right)}\quad (x\to\infty)$$

To get the asymptotic behavior of $I(x)$, it looks like we need to take the second derivative of the above result. It turns out that

$$\frac{d^2}{dx^2} \left[x^{-1/2} \cos{\left ( x-\frac{\pi}{4}\right)}\right ]= \left (\frac{3}{4} x^{-5/2} - x^{-1/2}\right ) \cos{\left ( x-\frac{\pi}{4}\right)} + x^{-3/2}\sin{\left ( x-\frac{\pi}{4}\right)} $$

Note that the $x^{-1/2}$ piece drops out when combined with the original integral, and the $x^{-5/2}$ piece is subdominant. Thus, the leading asymptotic behavior of the integral $I(x)$ is finally

$$\int_{-1}^1 dk \, \sqrt{1-k^2} \, e^{i k x} \sim \sqrt{2 \pi} x^{-3/2} \sin{\left ( x-\frac{\pi}{4}\right)}\quad (x\to\infty)$$

Here is a plot illustrating this behavior against the exact value of the integral (the lower plot is the asymptotic result derived here):

enter image description here

ADDENDUM

One objection to the above derivation might be that I neglected the next term in the asymptotic expansion of the first integral of $I(x)$, which we know goes as $A x^{-3/2} \sin{(x-\pi/4)}$. How could I ignore that? It turns out $I$ is the sum of this term and its second derivative, and the only term that remains $O(x^{-3/2})$ vanishes because it is a simple sine term, the sum of itself and its second derivative vanishing. So the derivation of the asymptotic approximation above remains valid.

Ron Gordon
  • 138,521
  • @ Ron Gordon Thanks for your excellent demonstration and detailed answer. – Kai Li Oct 20 '13 at 09:34
  • @ Ron Gordon I just read your question http://math.stackexchange.com/questions/318804/relation-between-function-discontinuities-and-fourier-transform-at-infinity, and I'm confused that according to the conclusion in it, the decaying power here should be $x^{-2}$ which contradicts $x^{-3/2}$?? Thanks a lot. – Kai Li Oct 20 '13 at 13:08
  • 1
    @K-boy: my guess is that it has something to do with the branch point singularity at the integration endpoints. The question I posed was not mean to be anything rigorous but was just more of an observation. Clearly there is something else going on here involving branch points about which I wasn't ready to speculate. – Ron Gordon Oct 20 '13 at 13:14
  • @ Ron Gordon Since we can define a function $g(k)=\sqrt{1-k^2}$ for $k\in[-1,1]$ and $g(k)=0$ otherwise. Then $f(x)$ is the Fourier transform of $g(k)$, and the first derivative of $g(k)$ has discontinuities at $k=-1,1$, which means that $m=1$ in http://math.stackexchange.com/questions/318804/relation-between-function-discontinuities-and-fourier-transform-at-infinity. – Kai Li Oct 20 '13 at 13:15
  • 1
    @K-boy: the derivative of $g$ also blows up at the points of discontinuity. Again, I am not claiming to have nailed a provable theorem, so if there is an apparent contradiction from my conjecture (which I could hardly claim to be mine anyway), it means that there are holes in that formulation. – Ron Gordon Oct 20 '13 at 13:19
  • @ Ron Gordon Thanks. Anyway, is the asymptotic behaviour of $f(x)$ at large distance dominated by the singularities of $g(k)$? And if $g(k)$ has no singularities, i.e., any order of its derivative is continuous, then what about the asymptotic behaviour of $f(x)$ at large distance ? – Kai Li Oct 20 '13 at 13:27
  • 1
    @K-boy: think about a Gaussian; the FT of a Guassian is a Gaussian. This would indicate (not prove!) that functions having all continuous derivatives die faster than any polynomial (e.g., exponential). – Ron Gordon Oct 20 '13 at 13:37
  • @ Ron Gordon Thank you very much. – Kai Li Oct 20 '13 at 15:22
  • What is the justification for "it looks like we need to take the second derivative of the above result"? The derivative of an asymptotics may not be the asymptotics of a derivative. It calls for extra justification. Can you just use the stationary phase method directly? – Hans Jul 15 '18 at 09:03