Fdtxjr

$$R_C(x,y)=1over2int_0^inftyfracdt(t+y)sqrtt+x,quad(xge0)$$
$R_C$ has a singularity in the denominator when $y$ is negative. Wikipedia says in that case $R_C$ can be evaluated as follows
$$
PV;R_C(x,-y)=sqrtfracxx+yR_C(x+y,y),quad(y>0)
$$

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.

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 $x gt y$:

$$beginalignR_C(x,y) &= frac12 int_0^infty fracdt(t+y) sqrtt+x \ &=int_sqrtx^infty fracdtt^2-(x-y)\ &= frac1sqrtx-y operatornamearctanhleft (fracsqrtx-ysqrtx right )endalign $$

Then we want to show that, for $y lt 0$:

$$PV left [R_C(x,-y)right ] = sqrtfracxx+y R_C(x+y,y) = sqrtfracxx+y frac1sqrtx operatornamearctanhleft (fracsqrtxsqrtx+y right ) = frac1sqrtx+y operatornamearctanhleft (fracsqrtxsqrtx+y right )$$

2) We will now proceed along those lines. Consider the following contour integral in the complex plane for $x$ and $y gt 0$:

$$oint_C dz fraclogz(z-y)sqrtz+x $$

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:

$$PV int_0^infty dt fraclogt(t-y) sqrtt+x -PV int_0^infty dt fraclogt+i 2 pi(t-y) sqrtt+x - i pi fraclogysqrtx+y - i pi fraclogy+i 2 pisqrtx+y \ +i 2 int_x^infty dt fraclogt+i pi(t+y) sqrtt-x $$

By Cauchy's theorem, the contour integral is zero. Simplifying, we get that

$$-i 2 pi PV int_0^infty fracdt(t-y) sqrtt+x = i 2 pi fraclogysqrtx+y +i 2 pi fraci pisqrtx+y - i 2 int_x^infty dt fraclogt+i pi(t+y) sqrtt-x$$

or

$$beginalignPV left [R_C(x,-y) right ] &= -fraclogy2 sqrtx+y - fraci pi2 sqrtx+y + frac1pi int_0^infty dt fraclog(t^2+x)t^2+x+y + i int_0^infty fracdtt^2+x+y endalign $$

Note that the second and fourth terms cancel. Thus,

$$beginalignPV left [R_C(x,-y) right ] &= -fraclogy2 sqrtx+y + frac1pi int_0^infty dt fraclog(t^2+x)t^2+x+y endalign $$

The latter integral is

$$int_0^infty dt fraclog(t^2+x)t^2+x+y = fracpisqrtx+y operatornamearctanhleft (fracsqrtxsqrtx+y right ) + fracpi logy2 sqrtx+y $$

Thus, we have shown that

$$PV left [R_C(x,-y) right ] = frac1sqrtx+y operatornamearctanhleft (fracsqrtxsqrtx+y right ) = sqrtfracxx+y R_C(x+y,y) $$

$$newcommandPVoperatornamePV
beginalign
R(x,y)
&=frac12,PV!!int_0^inftyfracmathrmdz(z+y)sqrtz+xtag1\[6pt]
&=frac12sqrtx-y,PV!!int_sqrtfracxx-y^inftyfrac2,mathrmdzz^2-1tag2\[3pt]
&=frac12sqrtx-y,PV!!int_sqrtfracxx-y^inftyleft(frac1z-1-frac1z+1right)mathrmdztag3\
&=frac12sqrtx-y,PV!!int_sqrtfracxx-y,-1^sqrtfracxx-y,+1fracmathrmdzztag4\
&=frac12sqrtx-yint_left^sqrtfracxx-y,+1fracmathrmdzztag5\
&=bbox[5px,border:2px solid #C0A000]tag6\[6pt]
&=sqrtfracu-vufrac12sqrtu-vlogleft|fracsqrtu-v+sqrtusqrtu-v-sqrturight|tag7\[6pt]
&=sqrtfracu-vu,R(u,v)tag8\[9pt]
&=sqrtfracxx-y,R(x-y,-y)tag9
endalign
$$
Explanation:

$(2)$: substitute $zmapsto(x-y)z^2-x$

$(3)$: partial fractions

$(4)$: substitute $zmapsto z+1$ and $zmapsto z-1$ 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 $ymapsto-y$ in $(9)$ yields
$$
R(x,-y)=sqrtfracxx+y,R(x+y,y)tag10
$$