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I've come across the following integral:

$$\int_{-\pi}^{\pi}\left[\frac{1}{A-R \cos(2\theta-\phi)}\right]^{\frac{N-1}{2}}d\theta$$

I know how to approximate this integral using the Laplace method, just wondering if:

a) Does this integral have an exact answer?

b) Is there a better approximation than Laplace method for this integral? If so, under what conditions will it be better?

My thinking is that it will be a hypergeometric function (mainly because every hard integral I've come across turns out to be one of these). Conditions (if needed) are $A>R>0$, and $N$ is an integer.

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You can drop the $\phi$, since you're integrating over two periods. Also the integral is twice the integral over a single period, say, $[0,\pi]$. – joriki Jun 14 '11 at 8:52
And the result is (a trivial transform of) a function of a single parameter, for example the parameter $c=(A-R)/R$ whose domain is $c>0$. // The answer to (a) is probably no if to have an exact answer means to be given by an explicit formula involving usual functions only. (But yes, this might depend on the extent of your repertoire of usual functions.) – Did Jun 14 '11 at 9:06
up vote 1 down vote accepted

As joriki points out, you can drop the $\phi$. Further assuming that $A > R$ so that the denominator never vanishes, the integral can be evaluated by resorting to complex analysis. Let $N = 2m + 1$ $$I = \int_{-\pi}^{\pi} \left(\frac1{A - R \cos(2 \theta)} \right)^{m} d \theta$$ Let $z = e^{i \theta}$, then $dz = i z d \theta$. Hence, the integral becomes $$I = \oint_{|z| = 1} \frac{dz}{iz \left(A - \frac{R}{2} \left(z^2 + \frac1{z^2} \right) \right)^m}$$ $$I = \oint_{|z| = 1} \frac{2^m z^{2m} dz}{iz \left(2Az^2 - R \left(z^4 + 1 \right) \right)^m}$$ $$I = -i \left(- \frac{2}{R} \right)^m \oint_{|z| = 1} \frac{z^{2m-1} dz}{\left(\left(z^4 + 1 \right) - 2 \frac{A}{R} z^2 \right)^m}$$ And evaluate the integral using residue theorem.

The poles of the integrand are at $$z= \pm \sqrt{\frac{A \pm \sqrt{A^2 - R^2}}{R}}$$ Since $A>R$, the poles within the contour lie at $$z = \pm \sqrt{\frac{A - \sqrt{A^2 - R^2}}{R}}$$

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From Maple... $K$ and $E$ are elliptic integrals. $$ \int_{0}^{\pi} \sqrt{\frac{1}{2 - \operatorname{cos} (2 t)}} d t = \frac{2 \sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{3} $$ $$ \int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)} d t = \frac{\pi \sqrt{3}}{3} $$ $$ \int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{3}{2}} d t = \frac{2 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{3} $$ $$ \int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-2)} d t = \frac{2 \pi \sqrt{3}}{9} $$ $$ \int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{5}{2}} d t = \frac{-2\sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{27} + \frac{16 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{27} $$ $$\int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-3)} d t = \frac{\pi \sqrt{3}}{6} $$ $$ \int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{7}{2}} d t = \frac{-32\sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{405} + \frac{202 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{405} $$ $$ \int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-4)} d t = \frac{11 \pi \sqrt{3}}{81} $$

This suggests a reduction formula, together with the first two.

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If $N$ is odd, then the exponent is even, so you've got a rational function of sine and cosine, so the Weierstrass substitution should work:

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