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If $G$ is the principal complex inverse sine, all values reached by analytic continuation areThis follows from the monodromy reflections about the branch values $-\pi /2$ and $\pi /2$.

The complementary error function avoids subtracting a quantity close to one when describing a small Gaussian integral tail. For real $x>0$, $\mathrm{erfc}x=(2/\sqrt{\pi }){\int }_x^{\infty }{e}^{-t^2}dt$; analytic continuation defines it for complex arguments.

For $c=1-a-b$ with positive real parts $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c$, the integral converges at its two finite singularities and at infinity, and equalsTo prove it, use Schwinger parameterization with exponents $1-a$ and $1-b$. Integrating the resulting planar Gaussian integral gives $\pi /(\lambda +\rho )$ times $\exp [-\lambda \rho /(\lambda +\rho )]$. Set $q=\lambda +\rho$ and $x=\lambda /q$; the $q$ integral is a gamma function and the $x$ integral is $B(a,b)$, giving the stated ratio. This domain justifies the interchanges; elsewhere the answer is interpreted by analytic continuation. The string measure $d^{2}z=2dxdy$ multiplies the answer by two.

The dilogarithm is the order-two polylogarithm, defined for $|z|<1$ by ${\mathrm{Li}}_2(z)={\sum }_{n\ge 1}{z}^n/n^2$. Its derivative is $-\mathrm{Log}(1-z)/z$, with a removable value at zero; a chosen complex logarithm branch gives its analytic continuation.

For a parameter $k$, one form of the incomplete elliptic integral of the first kind isIts value depends on the choices of square-root branches and, under analytic continuation around the branch points, on the integration path.

For $\Re c>\Re b>0$ and a fixed branch away from the cut,Expand $(1-zs{)}^{-a}$ for $|z|<1$ and integrate termwise using the beta function to recover the Gauss hypergeometric function series. Analytic continuation extends the identity in $z$.

For $r=n+1/2$, $n\ge 0$, the convergent frequency zeta function, expressed through the Riemann zeta function, is ${\sum }_r{r}^{-s}=(2^s-1){\zeta }_R(s)$ when $\mathrm{Re}s>1$. Its analytic continuation at $s=-1$ gives $(1/2-1)(-1/12)=1/24$. A common exponential frequency cutoff gives ${\sum }_{r}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, so the same finite part follows by subtracting the leading divergence. In contrast, integer frequencies have regulated sum $-1/12$.

Put $d=c-a-b\notin \mathbb{Z}$ and choose a branch of $(1-z{)}^d$. The solution regular at zero isThe two characteristic exponents at a regular singular point are $0,d$. The Euler integral for the hypergeometric function determines the first coefficient by a finite endpoint limit and the second by scaling the singular endpoint; analytic continuation extends the parameter ranges.

Letwhere $P$ has a simple zero at $a$, and let $A$ be the limiting value of $W$ at $a$ on one sheet. Analytic continuation once around $a$ changes the sign of the square root and hence of $W^{\prime }$. The continued primitive agrees at $a$ with the original one, so it is