Real part of elliptic integral Suppose complete elliptic integral of the first kind:
$$
\tag 1 K(k) \equiv \int_{0}^{\frac{\pi}{2}}\frac{\mathrm{d}y}{\sqrt{1-k^2\sin^{2}(x)}}.
$$
I need to evaluate its real part on interval $0 < k < \infty$.
Since $K(k)$ has discontinuity at $k = 1$ and acquires imaginary part, we need to use analytical continuation:
$$
\tag 2 K(k) = \frac{1}{k}\left(K\left(\frac{1}{k}\right) \pm i K\left( \sqrt{1-\frac{1}{k^{2}}}\right)\right).
$$
The great surprise for me is that by using the relation
$$
K(x) = \frac{1}{1+x}K\left(\frac{2\sqrt{x}}{x+1}\right)
$$
for $K(k)$ on the interval $0<k<1$ and for $\dfrac{1}{k}K\left(\dfrac{1}{k}\right)$ on the interval $1<k<\infty$, we find that expressions for $\mathrm{Re}K(k), 0 < k<1$ and $\mathrm{Re}K(k), 1 < k < \infty$ coincide:
$$
\mathrm{Re}K(k) = \frac{1}{k+1}K\left(\frac{2\sqrt{k}}{k+1}\right). \quad 0 < k < \infty
$$
Why this is true? I didn't expect this because of singularity of $K(k)$ at $k = 1$... For example, imaginary part in the region $0 < k < 1$ is zero, while in the region $1 < z < \infty$ it is non-zero with non-trivial functional dependence, making a jump at $z = 1$. 
Is this somehow related to general properties of analytical continuation (which in our particular case is given by (2))?
 A: *

*First note: the function is not "analytic" at $k=1$. The real part has a logarithmic divergence.

*Second note: it is quite natural that "jump" across the branch cut, just occurs in imaginary component. A similar observation follows for functions that are defined (on some part of the real axis) as real functions due to Schwarz reflection principle. E.g., you observe a similar behaviour for the elementary function $\log z$ on the (standard) branch cut along the negative real line. 

*Note that you may make the real part more regular without removing the "jump" of the imaginary part across the branch cut. For that you can take a look at the family of functions
$$z^m \log z$$
with $m\geq 1$.

*Last note: in this context, the Hilbert transform might help to guide your intuition. Given any real function $u(x)$, the Hilbert transform $v= H(u)$ provides another function such that $f_1(x) = u(x) +i v(x)$ is analytic in the upper half plane (thus is is an analytic continuation in the upper half plane). By Schwarz reflection principle $f_2(x) = u(x)-i v(x)$ is an analytic function on the lower half plane. If $v(x)=0$ for some interval along the real axis (given by the function $f(x)=u(x)$), then $f_1$ and $f_2$ are the analytical continuation of this function in the upper/lower half plane. In the region, where $v\neq 0$, they obviously just differ by an imaginary number.
