Direct evaluation of complete elliptic integral

Question

In comments to this question, @RobertIsrael asserted that, for $-1<x<1$, $$ \int_0^{2\pi} \frac{1-x \cos(\phi)}{\left(1 - 2 x \cos(\phi) + x^2\right)^{3/2}} \mathrm{d} \phi = \frac{4}{1-x^2} \operatorname{E}(x^2) \tag{1} $$ where $E(m)$ is the complete elliptic integral of the second kind: $E(m) = \int_0^{\pi/2} \sqrt{1-m \sin^2(\theta)} \mathrm{d} \theta$.

It is easy to verify that the series expansion of the integrand, integrated term-wise, agrees with the series expansion of Robert's elegant answer.

I am very much interested if there is a way to directly establish $\text{eq. (1)}$ from the integral.

Answer 1

The first step towards the goal is to perform a rational transformation. As in the previous question, using the symmetry $\phi \to (2\pi - \phi)$ and the change of variables $t = \sin^2\left(\phi/2\right)$: $$ \mathcal{I} = \int_0^{2 \pi} \frac{1-x \cos(\phi)}{\left(1-2 x \cos(\phi) + x^2\right)^{3/2}} \mathrm{d} \phi = \frac{2}{(1-x)^2} \int_0^1 \frac{1+\frac{ 2 x}{1-x} t}{1+\frac{4x}{(1-x)^2} t} \frac{\mathrm{d} t}{\sqrt{ t(1-t)\left( 1+\frac{4x}{(1-x)^2} t\right)}} $$ Now, perform a rational substitution: $$ t = \frac{1-x}{2} \frac{y+1}{1- x y} \qquad \text{with} \qquad \mathrm{d} t = \frac{1-x^2}{( 1-x y)^2} \frac{\mathrm{d} y}{2} $$ which maps $0 <t<1$ into $-1<y<1$. With it: $$ t (1-t) \left( 1+\frac{4x}{(1-x)^2} t\right) = \frac{(1+x)^2}{4 (1-x y)^4} (1-y^2) (1- x^2 y^2) $$ and $$ \frac{1+\frac{ 2 x}{1-x} t}{1+\frac{4x}{(1-x)^2} t} = \frac{1-x}{1+ x y} $$ Combining, and using $1-x>0$ and $1-x y>0$: $$ \mathcal{I} = \frac{2}{1-x} \int_{-1}^1 \frac{(1-x)^2}{(1+ x y)} \frac{\mathrm{d} y}{\sqrt{(1-y^2)(1-x^2 y^2)}} = 2(1-x) \int_{-1}^1 \frac{1}{1 + x y} \frac{\mathrm{d} y}{\sqrt{(1-y^2)(1-x^2 y^2)}} $$ The integral above is reduced to the rational form with substitution $y = \operatorname{sn}(u| x^2)$, where $\operatorname{sn}(u|m)$ stands for the Jacobi elliptic sine function. Indeed: $$ (1-y^2)(1-x^2 y^2) = \left( 1- \operatorname{sn}^2(u|x^2)\right)\left( 1- x^2 \operatorname{sn}^2(u|x^2)\right) = \operatorname{cn}^2(u|x^2) \operatorname{dn}^2(u|x^2) $$ $$ \mathrm{d} y = \operatorname{cn}(u|x^2) \operatorname{dn}(u|x^2) \mathrm{d} u $$ The substitution maps $-1<y<1$ into $-K(x^2) < u < K(x^2)$, where $K(x^2)$ is the complete elliptic integral of the first kind, and both $\operatorname{cn}(u|x^2) > 0$ and $\operatorname{dn}(u|x^2) > \sqrt{1-x^2} > 0$ on this interval: $$ \mathcal{I} = \int_{-K(x^2)}^{K(x^2)} \frac{2(1-x) \mathrm{d u}}{1 + x \operatorname{sn}(u| x^2) } = \frac{2}{1-x^2} \left. \left(\operatorname{E}\left( \operatorname{am}(u|x^2), x^2\right) + x \frac{\operatorname{cn}(u|x^2) \operatorname{dn}(u|x^2) }{1+ x \operatorname{sn}(u|x^2)} \right) \right|_{-K(x^2)}^{K(x^2)} $$ Since $\operatorname{cn}\left( \pm K(x^2)| x^2\right) = 0$, and $\operatorname{am}(\pm K(x^2)| x^2) = \pm \frac{\pi}{2}$ we arrive at the desired result: $$ \mathcal{I} = \frac{4}{1-x^2} \operatorname{E}(x^2) $$

Answer 2

Following up on my comments,

$$ \begin{align} \int_0^{2\pi} \frac{1-x \cos\phi}{\left(1 - 2 x \cos\phi + x^2\right)^{3/2}} \mathrm{d} \phi &= \left[\int_0^{2\pi} \frac{d-x \cos\phi}{\left(d^2 - 2 dx \cos\phi + x^2\right)^{3/2}} \mathrm{d} \phi\right]_{d=1} \\ &= \left[-\frac{\partial}{\partial d}\int_0^{2\pi} \frac1{\left(d^2 - 2 dx \cos\phi + x^2\right)^{1/2}} \mathrm{d} \phi\right]_{d=1} \\ &= \left[-\frac{\partial}{\partial d}\int_0^{2\pi} \frac1{\left((d-x)^2 +4dx\sin^2\dfrac\phi2\right)^{1/2}} \mathrm{d} \phi\right]_{d=1} \\ &= \left[-4\frac{\partial}{\partial d}\int_0^{\pi/2} \frac1{\left((d-x)^2 +4dx\sin^2\theta\right)^{1/2}} \mathrm{d} \theta\right]_{d=1} \\ &= \left[-4\frac{\partial}{\partial d}\frac{K\left(\dfrac{2\mathrm i\sqrt{dx}}{|d-x|}\right)}{|d-x|} \right]_{d=1} \;. \end{align} $$

I'm not sure whether this gets us any closer to the result, but you could try using the derivative and differential equation given at Wikipedia, noting that $k=\dfrac{2\mathrm i\sqrt{dx}}{|d-x|}$ leads to $\sqrt{1-k^2}=\dfrac{d+x}{|d-x|}$.