Complex Euler beta integral for a four-point sphere correlator

Set $\alpha =1-a$ and $\beta =1-b$. The hypotheses imply $\mathrm{Re}\alpha ,\mathrm{Re}\beta >0$ and $\mathrm{Re}(\alpha +\beta -1)>0$. To check absolute convergence, on $\mathrm{\mid }z\mathrm{\mid }\le \frac{1}{2}$ the factor involving $\mathrm{\mid }1-z\mathrm{\mid }$ is bounded and the integrand is $O(\mathrm{\mid }z{\mathrm{\mid }}^{2\mathrm{Re}a-2})$, whose planar integral is finite because $\mathrm{Re}a>0$. The same argument on $\mathrm{\mid }1-z\mathrm{\mid }\le \frac{1}{2}$ uses $\mathrm{Re}b>0$. For $\mathrm{\mid }z\mathrm{\mid }\ge 2$, $\mathrm{\mid }1-z\mathrm{\mid }$ and $\mathrm{\mid }z\mathrm{\mid }$ are comparable, so the integrand is $O(\mathrm{\mid }z{\mathrm{\mid }}^{2\mathrm{Re}(a+b)-4})$; its planar tail is finite because $\mathrm{Re}(a+b)<1$. It is bounded on the remaining compact region. For $z\mathrm{\ne }0$, the substitution $u=x\mathrm{\mid }z{\mathrm{\mid }}^2$ in [F1] gives $\mathrm{\mid }z{\mathrm{\mid }}^{2a-2}=\frac{1}{\mathrm{\Gamma }(1-a)}{\int }_0^{\mathrm{\infty }}{x}^{-a}{e}^{-x\mathrm{\mid }z{\mathrm{\mid }}^2}dx,$ and the same formula with $b$ and $\mathrm{\mid }1-z\mathrm{\mid }$ holds away from $z=1$. The two exceptional points have planar measure zero. The integral of the absolute double Schwinger kernel over $x,y,z$ is a finite constant times $I(\mathrm{Re}a,\mathrm{Re}b)$, since integrating first in $x$ and $y$ gives the displayed real power factors multiplied by $\mathrm{\Gamma }(1-\mathrm{Re}a)\mathrm{\Gamma }(1-\mathrm{Re}b)$. The bounds above make this finite, so [F4] permits inserting both representations and exchanging the $x,y,z$ integrations.

Completing the square gives $x\mathrm{\mid }z{\mathrm{\mid }}^2+y\mathrm{\mid }1-z{\mathrm{\mid }}^2=(x+y)\mathrm{\mid }z-y\mathrm{/}(x+y){\mathrm{\mid }}^2+xy\mathrm{/}(x+y)$. By [F3], integration over $z$ contributes $2\pi (x+y{)}^{-1}{e}^{-xy\mathrm{/}(x+y)}$ for the declared measure $d\mu$. Put $\lambda =x+y$ and $t=x\mathrm{/}(x+y)$; the Jacobian is $dxdy=\lambda d\lambda dt$. The remaining integral is $\frac{2\pi }{\mathrm{\Gamma }(1-a)\mathrm{\Gamma }(1-b)}{\int }_0^1{t}^{-a}(1-t{)}^{-b}{\int }_0^{\mathrm{\infty }}{\lambda }^{-a-b}{e}^{-\lambda t(1-t)}d\lambda dt\mathrm{.}$ Using [F1] on the inner integral yields $\mathrm{\Gamma }(1-a-b)[t(1-t){]}^{a+b-1}$.

The resulting $t$ integral is $B(b,a)$ because its integrand is $t^{b-1}(1-t{)}^{a-1}$. Applying [F2] gives the displayed Gamma quotient on the stated convergence domain.

Fix any $(a_0,b_0)$ in that domain and choose $\rho >0$ with $\mathrm{Re}{a}_0-\rho >0$, $\mathrm{Re}{b}_0-\rho >0$, and $\mathrm{Re}(a_0+b_0)+2\rho <1$. For $\mathrm{\mid }a-a_0\mathrm{\mid },\mathrm{\mid }b-b_0\mathrm{\mid }<\rho$, expand the integrand as its value at $(a_0,b_0)$ times $\exp (2(a-a_0)\log \mathrm{\mid }z\mathrm{\mid }+2(b-b_0)\log \mathrm{\mid }1-z\mathrm{\mid })$. The absolute sum of this double power series is bounded near $0$ by $C\mathrm{\mid }z{\mathrm{\mid }}^{2(\mathrm{Re}{a}_0-\rho )-2}$, near $1$ by $C\mathrm{\mid }1-z{\mathrm{\mid }}^{2(\mathrm{Re}{b}_0-\rho )-2}$, and at infinity by $C\mathrm{\mid }z{\mathrm{\mid }}^{2(\mathrm{Re}(a_0+b_0)+2\rho )-4}$; each is integrable in the respective planar region by the strict inequalities. Thus Tonelli/Fubini [F4] permits integrating the power series term by term, yielding a convergent local power series for $I(a,b)$, so $I$ is holomorphic throughout the convergence domain. By [F5] the displayed quotient is meromorphic on ${\mathbb{C}}^2$ and, by [F2]–[F3], agrees with $I$ on that domain; it is therefore a meromorphic continuation.