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Today in class I showed some ways for dealing with the classical integral $\int_{0}^{2\pi}\frac{d\theta}{(A+B\cos\theta)^2}$ under the constraints $A>B>0$, including

  1. Symmetry and the tangent half-angle substitution;
  2. Relating the integral to the area enclosed by an ellipse, via the polar form of an ellipse with respect to a focus and the formula $\text{Area}=\pi a b =\frac{1}{2}\int_{0}^{2\pi}\rho(\theta)^2\,d\theta$;
  3. Computing the geometric-like series $\sum_{n\geq 0} r^n \sin(n\theta)$ and applying Parseval's identity to it;
  4. Applying Feynman's trick (differentiation under the integral sign) to $\int_{0}^{2\pi}\frac{d\theta}{1-R\cos\theta}$ which is an elementary integral due to point 1.

I finished the lesson by remarking that the point $4.$ allows to compute $\int_{0}^{2\pi}\frac{d\theta}{\left(1-R\cos\theta\right)^3}$ almost without extra efforts, while the $L^2$ machinery (point 3.) does not seem to grant the same. Apparently I dug my own grave, since someone readily asked (with the assumption $R\in(-1,1)$)

What is the asymptotic behavior of the coefficients $c_n$ in $$ \frac{1}{\left(1-R\cos\theta\right)^{3/2}}= c_0+\sum_{n\geq 1}c_n \cos(n\theta) $$ ? (Q2) Do we get something interesting by following the "unnatural" approach of applying Parseval's identity to such Fourier (cosine) series?

At the moment I replied that the Paley-Wiener theorem ensures an exponential decay of the $c_n$s, and with just a maybe to the second question. Later I figured out a pretty technical proof (through hypergeometric functions) of

$$ |c_n| \sim K_R\cdot\sqrt{n}\cdot\left(\frac{|R|}{1+\sqrt{1-R^2}}\right)^n \quad \text{as }n\to +\infty$$ and the fact that $c_n$ is given by a linear combination of complete elliptic integrals of the first and second kind. (Q1) I would like to know if there is a more elementary way for deriving the previous asymptotic behaviour. And the outcome of (Q2).

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A sketch how to extract the wanted leading order asymptotics without knowledge of special functions:

The $n$-th Fourier coefficent of the function ${f(x,R)=(1-R\cos(x))^{-3/2}}$ is given by

$$c_n(a)=\frac{ a^{3/2}}{\pi}\text{Re}\int_0^{2 \pi}\frac{e^{ixn}}{(a-\cos(x))^{3/2}}=-\frac{2a^{3/2}}{\pi}\text{Re}\frac{d}{da}\underbrace{\int_0^{2\pi}e^{ixn}F(x,a)dx}_{C_n(a)} $$

here $a=1/R>1$ and $(a-\cos(x))^{-1/2}=F(x,a)$

To get the asypmtotics of interest we integrate $e^{izn}F(z,a)$ over a rectangle $\mathcal{D}$ with verticies $[\epsilon,2\pi-\epsilon,2\pi -\epsilon+i\infty,\epsilon+i\infty]$ where $\epsilon\rightarrow0_+$. Leaving the details to the interested reader, we notice that:

  • the bottom part of the contour integral yields $C_n(a)$
  • the top part vanishs due to the exponential decay at complex infinity
  • the two vertical parts cancel out until we reach the branch point of $F(z,a)$ at $z_{0,1}=i\text{arccosh}(a)+2\pi \delta_{n,1}$
  • there are no singularities contained inside $\mathcal{D}$

Taking these four points into account we get (the remaining parts of the vertical contours are actually two times the left piece due to a factor of $(-1)(i)^2$ stemming from the paramtrisation $z= i y$ and the phaseshift due to the branchcut ) :

$$ \oint_\mathcal{D}e^{izn}F(z,a)dz=0=C_n(a)-2\int_{\text{arccosh(a)}}^{\infty}\frac{e^{-ny}}{\sqrt{\cosh(y)-a}}dy $$

the remaining integral is now pretty simple to estimate in the limit $n\rightarrow \infty$ since it is clearly dominated by the contribution of the branch point due to exponenital decay of the numerator.

$$ C_n(a)\sim\frac{2}{(a^2-1)^{1/4}}\int_{\text{arccosh(a)}}^{\infty}dy\frac{e^{-ny}}{\sqrt{y-\text{arccosh(a)}}}=\\ \frac{4}{(a^2-1)^{1/4}}e^{-n\text{arccosh(a)}}\int_0^{\infty}dye^{-n y^2}=\\ \frac{2\sqrt{\pi}}{(a^2-1)^{1/4}}\frac{e^{-n\text{arccosh(a)}}}{\sqrt{n}} $$

Taking the derivative w.r.t. $a$ gives us a mulitplicative factor of $n$ and OPs result follows by using the the connection between inverse hyperbolics and logarithms.


A similiar technique was used in the answer to this question

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  • $\begingroup$ This is indeed pretty elegant, I haven't thought about applying Laplace's method and the residue theorem, but it comes pretty natural. Appreciated (+1) $\endgroup$ – Jack D'Aurizio Apr 7 '18 at 18:14
  • $\begingroup$ @JackD'Aurizio Thanks, this was a challenging one! Please let me know if i should fill in some details, i have time tomorrow :). I would also be interested in your convolution approach, would you mind positing it ? $\endgroup$ – tired Apr 7 '18 at 18:27
  • $\begingroup$ I haven't typed it yet, but I hope my handwriting and language are understandable: on the Fourier series of $\frac{1}{(1-R\cos\theta)^m}$. $\endgroup$ – Jack D'Aurizio Apr 7 '18 at 18:34
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I don't think this is more elementary, but it may be a little more straightforward. Start with Heine's toroidal identity, $$ \frac{1}{\sqrt{z-\cos\,t}} = \frac{\sqrt{2}}{\pi} \sum_{m=-\infty}^\infty\,Q_{m-1/2}(z)\,\exp{(i \, m\, t}), \, \, |z|\ge1, \, 0\le t \le 2\pi \, .$$ Splitting the summation into positive and negative $m$ and using the fact that for the toroidal function (Gradshteyn & Rhyzhik 8.737.4) $Q_{-(m-1/2)-1}(z) = Q_{m-1/2}(z)$ for integer $m$ then one obtains $$ \frac{1}{\sqrt{z-\cos\,t}} = \frac{\sqrt{2}}{\pi}\Big(2 \sum_{m=1}^\infty\,Q_{m-1/2}(z)\,\cos{( m\, t)}+ Q_{-1/2}(z)\Big) .$$ Differentiate both sides with respect to $z$ and use G&R 8.706.1 (with the fact that $z>1$ to get the phase correct) to get

$$ (z-\cos\,t)^{-3/2} = \frac{-2\sqrt{2}}{\pi\,\sqrt{z^2-1}}\Big(2 \sum_{m=1}^\infty\,Q_{m-1/2}^{\,1}(z)\,\cos{( m\, t)}+ Q_{-1/2}^{\,1}(z)\Big) .$$ From the problem's statement, with $z=1/R,$ we see that the associated toroidal function $Q_{m-1/2}^{\,1}(z)$ must be asymptotically estimated. G&R 8.777.2 has a hypergeometric representation that has an asymptotic nature for large $m,$

$$Q_{m-1/2}^{\,1}(z)=-2\sqrt{\pi}\,\frac{\Gamma(m+3/2)}{\Gamma(m+1)} \,\frac{\sqrt{z^2-1}}{\zeta^{\,m+3/2}} F\big(3/2,m+3/2; m+1; 1/\zeta^2\big)$$ where $\zeta = z+\sqrt{z^2-1} > 1.$ For convenience set $y=1/\zeta^2$ and expand the ration of Pochhamer symbols in the hypergeometric function as $$\frac{(m+3/2)_k}{(m+1)_k} \sim 1 + \frac{k}{2m} + \, ... \,\, m\to \infty.$$ Therefore

$$F\big(3/2,m+3/2; m+1; 1/\zeta^2\big)\sim \sum_{k=0}^\infty \dfrac{(3/2)_k}{k!} \big(1+\frac{k}{2m}\big)y^k= (1-y)^{-3/2}\Big(1+\frac{3y}{4m(1-y)} \Big). $$ For this to be a useful asymptotic expression, $y$ needs to bounded away from 1, which implies $z$ must be bounded from 1, and $R$ must be bounded from 1. Writing $$ (z-\cos\,t)^{-3/2} =d_0(z)+\sum_{m=1}^\infty d_m(z)\,\cos{(m\,t)} $$ then, using only the first term of the expression above and the first in the asymptotic expansion of the ratio of gamma functions,

$$ d_m(z) \sim \frac{8\sqrt{2\, m}}{\sqrt{\pi}} (z+\sqrt{z^2-1})^{-3/2}(1-y)^{3/2}$$

In the proposer's notation it is clear to see that $$c_m = R^{-3/2}d_m(z \to 1/R)$$ and, with some simplification one finds $$c_m \sim \frac{4}{\sqrt{\pi}} \,\sqrt{m}\, (1-R^2)^{-3/4} \Big(\frac{R}{1+\sqrt{1-R^2}}\Big)^m .$$

This expression was derived for $z>1$ implying $0\le R<1.$ To extend to $-1\le R \le 0,$ simply let $t \to t+\pi$ and a factor of $(-1)^m$ is introduced into the summand. It is then clear that the $(-1)^m$ can be incorporated into the final expression as the proposer did, by making the unsigned $R$ an absolute value.

I have checked many of the intermediate steps in Mathematica, but to do so one must use $Q_{m-1/2}^{\,1}(z)=$ LegendreQ[m-1/2, 1, 3, z].

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  • $\begingroup$ I wouldn't call this approach elementary, but I appreciate it (+1). It is pretty close to my original derivation, relying on the Fourier series of $\frac{1}{\sqrt{1-r e^{i\theta}}}$, convolution and the substitution $R=\frac{2r}{1+r^2}$. $\endgroup$ – Jack D'Aurizio Apr 6 '18 at 15:52

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