# Evaluating the elliptic integral $\int_{-\pi}^\pi\frac{dx}{\sqrt{(t-2\cos x)^2-4}}$

I have the following integral,

$$I(t)=\int_{-\pi}^\pi\frac{dx}{\sqrt{(t-2\cos x)^2-4}},$$

where $t>4$ is a real parameter. I know from messing around numerically and playing with Mathematica that

$$I(t)=\frac{4}{t}K\left(\frac{16}{t^2}\right),$$

where $K$ is the complete elliptic integral of the first kind with parameter $m=k^2=16/t^2$. However, I seek proof of that fact. I have tried a handful of changes of variables which didn't get the job done, and I've searched tables of integrals without finding this integrand or similar. Any suggestions or hints would be appreciated.

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I cannot believe that this integral yields an elliptic function. Maybe you have forgot the square root in the denominator? – Fabian Aug 13 '12 at 22:32
Err, you are right, Fabian. Thank you. – Jonathan Aug 13 '12 at 22:34

Here is a rather circuitous route. Maybe there is a more compact way to do this:

\require{cancel}\begin{align*} \int_{-\pi}^\pi \frac{\mathrm dx}{\sqrt{(t-2\cos x)^2-4}}&=2\int_0^\pi \frac{\mathrm dx}{\sqrt{(t-2\cos x)^2-4}}\\ &=4\int_0^\infty \frac{\mathrm du}{\sqrt{((t+4)u^2+t)(tu^2+t-4)}} \qquad \small{\left(u=\tan\frac{x}{2}\right)}\\ &=\frac4{\sqrt{t(t+4)}}\int_0^\infty \frac{\mathrm du}{\sqrt{\left(u^2+\frac{t}{t+4}\right)\left(u^2+\frac{t-4}{t}\right)}}\\ &=\frac{4\cancel{\sqrt{t(t+4)}}}{\cancel{\sqrt{t(t+4)}}}\int_0^{\pi/2} \frac{\mathrm dv}{\sqrt{t^2-16+16\sin^2 v}} \quad \small{\left(u=\sqrt{\frac{t}{t+4}}\tan\,v\right)}\\ &=4\int_0^{\pi/2} \frac{\mathrm dv}{\sqrt{t^2-16+16\sin^2 v}}\\ &=4\int_{-\pi/2}^0 \frac{\mathrm dv}{\sqrt{t^2-16+16\sin^2 v}}\qquad\text{(symmetry)}\\ &=4\int_0^{\pi/2} \frac{\mathrm dv}{\sqrt{t^2-16+16\cos^2 v}}=4\int_0^{\pi/2} \frac{\mathrm dv}{\sqrt{t^2-16\sin^2 v}}\\ &=\frac4{t}K\left(\frac{16}{t^2}\right) \end{align*}

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Excellent, thank you very much. I'll accept this soon if nobody else comes up with something slick. – Jonathan Aug 14 '12 at 4:00
No problem, take your time. If I conjure up something shorter, I'll edit this. – J. M. Aug 14 '12 at 4:13

Since $\cos(-x) = \cos(x)$ we can reduce the integration range to $(0,\pi)$, and then do the change of variable $u = \cos(x)$: $$\int\limits_{-\pi}^\pi \frac{\mathrm{d} x}{\sqrt{(t-2 \cos(x))^2-4}} = \int_0^\pi \frac{\mathrm{d} x}{\sqrt{\left(\frac{t}{2} - \cos(x)\right)^2 -1}} = \int_{-1}^1 \frac{2 \mathrm{d} u}{\sqrt{1-u^2} \sqrt{4 u^2 + 4 u t + t^2-4}}$$ Now perform a change of variables: $$u = \frac{v-z}{1- v z}, \qquad \text{where} \quad v = \frac{t - \sqrt{t^2-16}}{4}$$ that maps $-1<u<1$ into $-1<z<1$, that leads to $$\int_{-1}^1 \frac{ v \, \mathrm{d} z}{\sqrt{1-z^2} \sqrt{1-v^4 z^2}} = 2 v \cdot \mathrm{K}\left( v^4\right)$$ This is not exactly what the OP asked for, but quite elegant nonetheless.

Here is a confirmation of the equivalence in Mathematica:

The equivalence of the above answer to the one conjectured by OP and established by @J.M. is through the quadratic transformation: $$\mathrm{K}(z) = \frac{2}{1+\sqrt{1-z}} \mathrm{K} \left( \left(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}} \right)^2 \right)$$ where $z = \frac{16}{t^2}$. Indeed: $$\frac{2}{1+\sqrt{1-\frac{16}{t^2}}} = \frac{2t}{t + \sqrt{t^2-16}} = \frac{2t}{t + \sqrt{t^2-16}} \cdot \frac{t - \sqrt{t^2-16}}{t - \sqrt{t^2-16}} = \frac{t}{4} \cdot \frac{t-\sqrt{t^2-16}}{2}$$ and similarly: $$\left(\frac{1-\sqrt{1-\frac{16}{t^2}}}{1+\sqrt{1-\frac{16}{t^2}}} \right)^2 = \left(\frac{t-\sqrt{t^2-16}}{t+\sqrt{t^2-16}} \cdot \color\green{ \frac{t-\sqrt{t^2-16}}{t-\sqrt{t^2-16}} } \right)^2 = \left(\frac{\left(t-\sqrt{t^2-16}\right)^2}{16} \right)^2$$ Combining, we arrive at the equality: $$\frac{4}{t} \mathrm{K}\left(\frac{16}{t^2}\right) = \frac{t-\sqrt{t^2-16}}{2} \mathrm{K}\left(\left( \frac{t-\sqrt{t^2-16}}{4} \right)^4\right)$$

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Thank you for pointing this out. – Jonathan Aug 14 '12 at 16:28