Asymptotic expansion of $\int_0^{2\pi} \mathrm{d}\theta\, \sqrt{k^2\cos^2\theta - \cos\theta + 1}$ as $k\to0$

Question

I have the integral

$$I = \int_0^{2\pi} \mathrm{d}\theta\, \sqrt{k^2\cos^2\theta - \cos\theta + 1}$$

and I would need the asymptotic expansion of the integral for small values of $k$. For $k=0$ we get quite easily $I=4\sqrt{2}$, while for $k\gg 1$, we have $I\sim 4|k|$.

Now, trying to perform the integration using special functions (at first glance, one would think that it could be done in terms of elliptic functions) yields no result. Gradštejn and Ryžik are of no help, nor Mathematica or other softwares.

We can Taylor expand the integrand and integrate each term of the expansion, but the $2n$-th term, for $n>1$, reads

$$ \frac{k^{2n}}{(2n)!}\frac{ (-1)^{n-1} (2n-3)!! }{2^n} \frac{\cos^{2n}\theta}{(1-\cos \theta)^{n-1/2}}$$

having set $k=0$ in the derivative. Of course, integrating each term would mean performing

$$\int_0^{2\pi}\mathrm{d} \theta\,\frac{\cos^{2n}\theta}{(1-\cos \theta)^{n-1/2}}$$

which is divergent for $n>1$.

Any ideas about how to proceed? Are there some kind of 'regularization' techniques that could help?

Answer 1

Surprisingly, this integral can be written in closed form using a linear combination of complete elliptic functions. This doesn't exactly answer the question but at least the integral disappears.

Step 1: Let $u=\cos\theta$ so that $$I=2\int_{-1}^1\sqrt{\frac{1-u+k^2u^2}{1-u^2}}\,du.$$

Step 2: Notice that an appropriate substitution of the form $u=(Av+B)/(Cv+D)$ allows the linear term to be eliminated in the numerator. Choosing $A=D=1$ and $B=C=K$ where $$K=1+k^2-k\sqrt{2+k^2}$$ yields (after extensive simplification) $$I=\frac{MN}{K^2}\int_{-1}^1\frac1{(v+1/K)^2}\sqrt{\frac{1-v^2/N^2}{1-v^2}}\,dv$$ where $$M=2\sqrt{(1-K^2)(K-K^2-k^2)}\quad\text{and}\quad N=\sqrt{\frac{k^2K^2-K+1}{K-K^2-k^2}}.$$

Step 3: The complete elliptic integral of the first, second and third kinds are, respectively: \begin{align}K\left(\frac1N\right)&=\int_0^1\frac1{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\E\left(\frac1N\right)&=\int_0^1\frac{1-v^2/N^2}{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\\Pi\left(n\mid\frac1N\right)&=\int_0^1\frac1{(1-nv^2)\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv.\end{align} Partial fraction decomposition gives \begin{align}\frac{K^2}{MN}I&=\int_{-1}^1\frac{-\frac1{N^2}+\frac2{KN^2}\cdot\frac1{v+1/K}+\left(1-\frac1{K^2N^2}\right)\cdot\frac1{(v+1/K)^2}}{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\&=-\frac2{N^2}K\left(\frac1N\right)+\frac2{KN^2}J_1+\left(1-\frac1{K^2N^2}\right)J_2.\end{align} Notice that \begin{align}aJ_2&=-bJ_1-c\int_{-1}^1\frac{v^2+d}{\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv\\&=-bJ_1+2cN^2E\left(\frac1N\right)-2c(N^2+d)K\left(\frac1N\right)\end{align} where $$a=-\frac{K^{4}N^{2}+1-K^{2}(N^{2}+1)}{K^{4}N^{2}},\quad b=\frac{2-K^{2}(N^{2}+1)}{K^{3}N^{2}},\quad c=\frac1{N^2},\quad d=-\frac1{K^2}.$$ Thus $$\frac{K^2}{MN}I=fK\left(\frac1N\right)+gE\left(\frac1N\right)+h\int_{-1}^1\frac1{(v+1/K)\sqrt{(1-v^2/N^2)(1-v^2)}}\,dv$$ where \begin{align}f&=-\frac2{N^2}-\frac{2c(N^2+d)}a\left(1-\frac1{K^2N^2}\right)\\g&=\frac{2cN^2}a\left(1-\frac1{K^2N^2}\right)\\h&=\frac2{KN^2}-\frac ba\left(1-\frac1{K^2N^2}\right).\end{align}

Answer 2

Consider $$f\left( k \right)=2\int\limits_{0}^{\pi }{\sqrt{{{k}^{2}}{{\cos }^{2}}\left( t \right)-\cos \left( t \right)+1}dt}$$ Note that the derivative of ${{\left( {{k}^{2}}{{\cos }^{2}}\left( t \right)-\cos \left( t \right)+1 \right)}^{1/2}}$at $t=0$ doesn’t exist when $k=0$, and this is causing problems. So perhaps “zoom in” to the lower bound by perturbing it by say k, as a start, and then take the limit. For example suppose $$f\left( k \right)\simeq 2\int\limits_{k}^{\pi }{\sqrt{{{k}^{2}}{{\cos }^{2}}\left( t \right)+1-\cos \left( t \right)}dt}=\sqrt{\pi }\sum\limits_{m=0}^{\infty }{\frac{{{k}^{2m}}}{m!\Gamma \left( \tfrac{3}{2}-m \right)}}\int\limits_{k}^{\pi }{\frac{{{\cos }^{2m}}\left( t \right)}{{{\left( 1-\cos \left( t \right) \right)}^{m-\tfrac{1}{2}}}}dt}$$ So for $m=0,1$ say, we get $$f\left( k \right)\simeq 2\int\limits_{k}^{\pi }{\sqrt{1-\cos \left( t \right)}+\frac{{{k}^{2}}{{\cos }^{2}}\left( t \right)}{2\sqrt{1-\cos \left( t \right)}}dt}$$ Mathematica says this is $$f\left( k \right)\simeq -\frac{1}{3\sqrt{1-\cos \left( k \right)}}\left( 6{{k}^{2}}\sin\left( \frac{k}{2} \right)\log\left( \tan \left( \frac{k}{4} \right) \right)+2\sin \left( k \right)\left( {{k}^{2}}+{{k}^{2}}\cos \left( k \right)-6 \right) \right)$$ A series expansion about $k=0$ yields for small $k$ $$f\left( k \right)\simeq 4\sqrt{2}+\frac{12\log \left( 2 \right)-11}{3\sqrt{2}}{{k}^{2}}-\sqrt{2}{{k}^{2}}\log \left( k \right)+O\left( {{k}^{4}} \right)$$

This seems promising but we need to fix this up, so consider $$2\int\limits_{0}^{\pi }{\sqrt{{{k}^{2}}{{\cos }^{2}}\left( t \right)+1-\cos \left( t \right)}dt}\\=2\int\limits_{0}^{\phi }{k\cos \left( t \right)\sqrt{1+\frac{1-\cos \left( t \right)}{{{k}^{2}}{{\cos }^{2}}\left( t \right)}}dt}+2\int\limits_{\phi }^{\pi }{\sqrt{1-\cos \left( t \right)}\sqrt{\frac{{{k}^{2}}{{\cos }^{2}}\left( t \right)}{1-\cos \left( t \right)}+1}dt}$$ Where $\phi =\arccos \left( \frac{-1+\sqrt{1+4{{k}^{2}}}}{2{{k}^{2}}} \right)$ . A change of variables in the first integral and then expansion in $k$ yields $$\begin{aligned} 2\int\limits_{0}^{\phi }{k\cos \left( t \right)\sqrt{1+\frac{1-\cos \left( t \right)}{{{k}^{2}}{{\cos }^{2}}\left( t \right)}}dt}&=\int\limits_{0}^{1}{\cos \left( \phi t \right)\sqrt{1+\frac{1-\cos \left( \phi t \right)}{{{k}^{2}}{{\cos }^{2}}\left( \phi t \right)}}dt}\\& \simeq 2\sqrt{2}{{k}^{2}}\int\limits_{0}^{1}{\sqrt{1+{{t}^{2}}}dt}+O\left( {{k}^{4}} \right) \\ & =\left( 2+\sqrt{2}\arcsin h\left( 1 \right) \right){{k}^{2}}+O\left( {{k}^{4}} \right) \\ \end{aligned}$$ The second integral is tricky since it seems quite difficult to remove k from the terminals. So for example: $$\begin{aligned} 2\int\limits_{\sqrt{2}k}^{\pi }{\sqrt{1-\cos \left( t \right)}\sqrt{\frac{{{k}^{2}}{{\cos }^{2}}\left( t \right)}{1-\cos \left( t \right)}+1}dt}=2\sum\limits_{n=0}^{\infty }{\left( \begin{matrix} \tfrac{1}{2} \\ n \\ \end{matrix} \right){{k}^{2n}}}\int\limits_{\sqrt{2}k}^{\pi }{\frac{{{\cos }^{2n}}\left( t \right)}{{{\left( 1-\cos \left( t \right) \right)}^{n-1/2}}}dt}\end{aligned}$$

And taking $n=0,1$ we obtain for small $k$

$$\begin{aligned} 2\int\limits_{\sqrt{2}k}^{\pi }{\sqrt{1-\cos \left( t \right)}\sqrt{\frac{{{k}^{2}}{{\cos }^{2}}\left( t \right)}{1-\cos \left( t \right)}+1}dt}\\\simeq \int\limits_{\sqrt{2}k}^{\pi }{2\sqrt{1-\cos \left( t \right)}+\frac{{{k}^{2}}{{\cos }^{2}}\left( t \right)}{\sqrt{1-{{\cos }^{2}}\left( t \right)}}dt}\\\simeq 4\sqrt{2}+\frac{\left( 9\log \left( 2 \right)-14-6\log \left( k \right) \right){{k}^{2}}}{3\sqrt{2}}\end{aligned}$$

While higher terms of n contribute to say terms of $O\left( {{k}^{2}} \right)$ this is not the case for the $\log \left( k \right)$ term. So we have then $$2\int\limits_{0}^{\pi }{\sqrt{{{k}^{2}}{{\cos }^{2}}\left( t \right)+1-\cos \left( t \right)}dt}=4\sqrt{2}-\sqrt{2}\log \left( k \right){{k}^{2}}+O\left( {{k}^{2}} \right)$$