About definite integral and Gauss hypergeometric function

Question

I feel very shameful asking questions on this forum, but while solving complex mathematical papers, sometimes I just cannot understand how the authors have proceeded. :(

Not sure if this is the case with everyone or is it just me

$$G(\theta) = 1+d\cos(n\theta)$$

$$\boxed{I = \int^{2\pi}_0 G(\theta)^{2/n} d\theta}$$

Authors claim that this integral can be simplified to:

$$= \pi\Big[ (1-d)^{2/n}{_2F_1}\Big(\frac{1}{2},-\frac{2}{n},1,\frac{2d}{d-1}\Big)+ (1+d)^{2/n}{_2F_1}\Big(\frac{1}{2},-\frac{2}{n},1,\frac{2d}{d+1}\Big)\Big] $$

$$= 2\pi - \frac{\pi(n-2)}{n}d^2 + \mathcal{O}(d^4)$$

where ${_2F_1}$ is the Gaussian hypergeometric function.

I have no idea where this come from. I looked at the book thoroughly : "Table of integrals, series, and products / I.S. Gradshteyn and I.M. Ryzhik ; Daniel Zwillinger" but could not find.

PS. There are some other details later, which I am unable to grasp fully. Also I can't find how $\mathcal{O}(.)$ comes from. Please find the figure attached.

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Answer 1

Assuming $|d|<1$, you can write $(1 + d \cos(n\theta))^{2/n}$ as a binomial series $$ (1+d \cos(n\theta)^{2/n} = \sum_{k=0}^\infty {2/n \choose k} (d\cos(n\theta))^k$$ and integrate term-by-term. Assuming $n$ is a nonzero integer,

$$ \int_0^{2\pi} \cos(n\theta)^k \; d\theta = \cases{0 & if $k$ is odd\cr {k \choose k/2} 2^{1-k} \pi & if $k$ is even}$$ and then your integral becomes $$ \sum_{j=0}^\infty {2/n \choose 2j} {2j \choose j} 2^{1-2j} \pi d^{2j} = 2\,\pi\, {\mbox{$_2$F$_1$}(-1/n,(n-2)/(2n);\,1;\,{d}^{2})} $$

I don't know where the other hypergeometric expression comes from, but I hope this helps.

Answer 2

I don't see why it would be shameful to ask for a clarification, especially since the series expansion in (13) appears to contain a mistake. I'll use the closed form given by Robert Israel. The coefficient at $d^2$ in the expansion around $d = 0$ is $$\frac {2 \pi \left( -\frac 1 \eta \right)_{\hspace{-2px}1} \left( \frac 1 2 - \frac 1 \eta\right)_{\hspace{-2px}1}} {(1)_1} = \frac {\pi (2 - \eta)} {\eta^2}.$$ For $d = 1$, Gauss's theorem gives $$2 \pi \,{_2F_1} \!\left( -\frac 1 \eta, \frac 1 2 - \frac 1 \eta; 1; 1 \right) = \frac {2 \pi \Gamma(1) \Gamma \!\left( \frac 1 2 + \frac 2 \eta \right)} {\Gamma \!\left( \frac 1 2 + \frac 1 \eta \right) \Gamma \!\left( 1 + \frac 1 \eta \right)},$$ and (14) is the expansion around $\eta = \infty$.