Legendre relations for elliptic integrals with null imaginary part

I am trying to compute the transformation of $K(k)$ as $k \to 1/k$, where $K(k)$ is just the Legendre integral

$$K(k) := F\left(\frac{\pi}{2}, k\right).$$

The Digital Library of Mathematical Functions [1] gives a relation for this transformation that relies on the imaginary part of $k$:

$$K\left(1 / k^{\prime}\right)=k^{\prime}\left(K\left(k^{\prime}\right) \mp \mathrm{i} K(k)\right),$$

where $k^{\prime}$ is the complementary modulus of $k$, and the $\pm$ sign is for $\Im(k) \lessgtr 0$. My question is, how do I find the transformation in the case where $\Im(k)=0$? I understand that there is a branch cut for this formula for $\Im(k)=0$, but what do I do to get around that?

Solution 1:

Let $0<k<1$, then

$$ K\left(\frac{1}{k}\right) = \int_{0}^{1}\frac{dt}{\sqrt{\left(1-t^2\right)\left(1-\frac{1}{k^2}t^2\right)}}dt $$

Make the following change of variable: $\displaystyle w^2 \mapsto \frac{1}{k^2}t^2$

$$K\left(\frac{1}{k}\right) = k\int_{0}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw$$

Note that, since $0<k<1$ then $\displaystyle \frac{1}{k}>1$. Hence,

$$K\left(\frac{1}{k}\right) = k\int_{0}^{1}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw + k\int_{1}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw$$

The firts integral is just $K(k)$

Then

$$K\left(\frac{1}{k}\right) = kK(k) + k\int_{1}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw$$

For the second, make the substitution:

$$ w = \frac{1}{\sqrt{1-k'^2z^2}}$$

where $k'$ is the complementary modulus and $k' = \sqrt{1-k^2}$

Hence

$$ dw = \frac{k'^2zdz}{\sqrt{1-k'^2z^2}}$$ $$\frac{1}{\sqrt{1-k^2w^2}} = \frac{\sqrt{1-k'^2z^2}}{k'\sqrt{1-z^2}}$$ $$ \frac{1}{\sqrt{1-w^2}} = \frac{i\sqrt{1-k'^2z^2}}{k'z}$$ $$ z \to 0 \textrm{ if } x \to 1$$ $$ z \to 1 \textrm{ if } x \to \frac{1}{k}$$ Therfore

$$\int_{1}^{\frac{1}{k}}\frac{dt}{\sqrt{\left(1-k^2w^2\right)\left(1-w^2\right)}}dw = i\int_{0}^{1}\frac{dt}{\sqrt{\left(1-z^2\right)\left(1-k'^2z^2\right)}}dz= iK(k') $$

Hence, we can conclude

$$ \boxed{K\left(\frac{1}{k}\right) = k\left[K(k)+iK(k')\right] \quad 0<k<1 }$$