Principal value of Carlson elliptic integral $R_C$ $$R_C(x,y)={1\over2}\int_0^\infty\frac{dt}{(t+y)\sqrt{t+x}},\quad(x\ge0)$$
$R_C$ has a singularity in the denominator when $y$ is negative. Wikipedia says in that case $R_C$ can be evaluated as follows
$$
PV\;R_C(x,-y)=\sqrt{\frac{x}{x+y}}R_C(x+y,y),\quad(y>0)
$$

Q: How can we prove the above equality?

The integrand has an easily computable anti-derivative but I don't know if it's useful in this context. 
 A: $$\newcommand{\PV}{\operatorname{PV}}
\begin{align}
R(x,y)
&=\frac12\,\PV\!\!\int_0^\infty\frac{\mathrm{d}z}{(z+y)\sqrt{z+x}}\tag{1}\\[6pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\frac{2\,\mathrm{d}z}{z^2-1}\tag{2}\\[3pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\left(\frac1{z-1}-\frac1{z+1}\right)\mathrm{d}z\tag{3}\\
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}\,-1}^{\sqrt{\frac{x}{x-y}}\,+1}\frac{\mathrm{d}z}z\tag{4}\\
&=\frac1{2\sqrt{x-y}}\int_{\left|\sqrt{\frac{x}{x-y}}\,-1\right|}^{\sqrt{\frac{x}{x-y}}\,+1}\frac{\mathrm{d}z}z\tag{5}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac1{2\sqrt{x-y}}\log\left|\frac{\sqrt{x}+\sqrt{x-y}}{\sqrt{x}-\sqrt{x-y}}\right|}\tag{6}\\[6pt]
&=\sqrt{\frac{u-v}{u}}\frac1{2\sqrt{u-v}}\log\left|\frac{\sqrt{u-v}+\sqrt{u}}{\sqrt{u-v}-\sqrt{u}}\right|\tag{7}\\[6pt]
&=\sqrt{\frac{u-v}{u}}\,R(u,v)\tag{8}\\[9pt]
&=\sqrt{\frac{x}{x-y}}\,R(x-y,-y)\tag{9}
\end{align}
$$
Explanation:
$(2)$: substitute $z\mapsto(x-y)z^2-x$
$(3)$: partial fractions
$(4)$: substitute $z\mapsto z+1$ and $z\mapsto z-1$ and subtract integrals
$(5)$: Principal Value is easy with an odd function
$(6)$: evaluate integral
$(7)$: $u=x-y$ and $v=-y$
$(8)$: apply $(6)$ to $(7)$
$(9)$: undo the substitution in $(7)$
Substituting $y\mapsto-y$ in $(9)$ yields
$$
R(x,-y)=\sqrt{\frac{x}{x+y}}\,R(x+y,y)\tag{10}
$$
A: One proves this identity by doing two things: 1) evaluating the integral for $y \gt 0$ and then 2) setting up the Cauchy PV integral in the complex plane.
1) The evaluation of the Carson integral is straightforward.  Let's consider the case $x \gt y$:
$$\begin{align}R_C(x,y) &= \frac12 \int_0^{\infty} \frac{dt}{(t+y) \sqrt{t+x}} \\ &=\int_{\sqrt{x}}^{\infty} \frac{dt}{t^2-(x-y)}\\ &= \frac1{\sqrt{x-y}} \operatorname{arctanh}{\left (\frac{\sqrt{x-y}}{\sqrt{x}} \right )}\end{align} $$
Then we want to show that, for $y \lt 0$:
$$PV \left [R_C(x,-y)\right ] = \sqrt{\frac{x}{x+y}} R_C(x+y,y) = \sqrt{\frac{x}{x+y}} \frac1{\sqrt{x}} \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )} = \frac1{\sqrt{x+y}}  \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )}$$
2) We will now proceed along those lines. Consider the following contour integral in the complex plane for $x$ and $y \gt 0$:
$$\oint_C dz \frac{\log{z}}{(z-y)\sqrt{z+x}} $$
where $C$ is the following contour:

Note that $C$ deals with the asymmetric integral over $[0,\infty)$ by introducing a log with a branch cut along the positive real axis.  The contour $C$ avoids the pole at $z=y$ by introducing semicircular detours about that pole of radii $\epsilon$.  $C$ also goes around the branch cut of the square root along the negative real axis.
I will take the limit as $\epsilon \to 0$ and the radius of the large circle $R \to \infty$.  The result is, for the contour integral in this limit:
$$PV \int_0^{\infty} dt \frac{\log{t}}{(t-y) \sqrt{t+x}} -PV \int_0^{\infty} dt \frac{\log{t}+i 2 \pi}{(t-y) \sqrt{t+x}} - i \pi \frac{\log{y}}{\sqrt{x+y}} - i \pi \frac{\log{y}+i 2 \pi}{\sqrt{x+y}}  \\ +i 2 \int_x^{\infty} dt \frac{\log{t}+i \pi}{(t+y) \sqrt{t-x}} $$
By Cauchy's theorem, the contour integral is zero.  Simplifying, we get that
$$-i 2 \pi PV \int_0^{\infty} \frac{dt}{(t-y) \sqrt{t+x}} = i 2 \pi \frac{\log{y}}{\sqrt{x+y}} +i 2 \pi \frac{i \pi}{\sqrt{x+y}} - i 2 \int_x^{\infty} dt \frac{\log{t}+i \pi}{(t+y) \sqrt{t-x}}$$
or
$$\begin{align}PV \left [R_C(x,-y) \right ] &= -\frac{\log{y}}{2 \sqrt{x+y}} - \frac{i \pi}{2 \sqrt{x+y}} + \frac{1}{\pi} \int_0^{\infty} dt \frac{\log{(t^2+x)}}{t^2+x+y} + i  \int_0^{\infty} \frac{dt}{t^2+x+y}  \end{align} $$
Note that the second and fourth terms cancel.  Thus,
$$\begin{align}PV \left [R_C(x,-y) \right ] &= -\frac{\log{y}}{2 \sqrt{x+y}} + \frac{1}{\pi} \int_0^{\infty} dt \frac{\log{(t^2+x)}}{t^2+x+y} \end{align} $$
The latter integral is
$$\int_0^{\infty} dt \frac{\log{(t^2+x)}}{t^2+x+y} = \frac{\pi}{\sqrt{x+y}} \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )} + \frac{\pi \log{y}}{2 \sqrt{x+y}} $$
Thus, we have shown that

$$PV \left [R_C(x,-y) \right ] = \frac{1}{\sqrt{x+y}} \operatorname{arctanh}{\left (\frac{\sqrt{x}}{\sqrt{x+y}} \right )} =  \sqrt{\frac{x}{x+y}} R_C(x+y,y) $$

