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Why does $|x-1| \; E\left(-\dfrac{4x}{(x-1)^2}\right) = |x+1| \; E\left(\dfrac{4x}{(x+1)^2}\right)$, where $E(m)$ is the complete elliptic integral of the second kind, with parameter $m$?

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1 Answer 1

up vote 4 down vote accepted

Note that the complete elliptic integral of the second kind satisfies the imaginary modulus identity (which I have specialized here to the complete case, $\phi=\pi/2$):

$$E(-m)=\sqrt{1+m}\,E\left(\frac{m}{1+m}\right)$$

Replacing $m$ with $\frac{4x}{(x-1)^2}$ in this identity gives

$$E\left(-\frac{4x}{(x-1)^2}\right)=\sqrt{1+\frac{4x}{(x-1)^2}}\;E\left(\frac{\frac{4x}{(x-1)^2}}{1+\frac{4x}{(x-1)^2}}\right)$$

which simplifies to

$$E\left(-\frac{4x}{(x-1)^2}\right)=\left|\frac{x+1}{x-1}\right|\;E\left(\frac{4x}{(x+1)^2}\right)$$

Multiplying both sides of the equation by $|x-1|$ turns this into what you have.

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