二項係数付きゼータ関数の値
定理
$m\ge1$
$\beginend{align}{
f_m(x,y) \acoloneqq \sum_{n=0}^\infty \frac{(x)_n}{n!(y+n)^m} \\
F_m(x,y) \acoloneqq (m-1)!f_m(x,y)
}$
超幾何関数
$\displaystyle{
f_m(x,y) = y^{-m}\hygeo{m+1}{F}{m}{x,y,\cdots,y}{y+1,\cdots,y+1}{1}
}$
$\zeta^{\langle a\rangle}(s,q)$
$\displaystyle{
f_m(x,y) = \zeta^{\langle -x\rangle}(m,y)
}$
$F_m(x,y) = \lr\{{\beginend{alignat}{2 &{\rm B}(1-x,y) &&{\rm if}\ m=1 \\ &\sum_{k=0}^{m-2}(-1)^k \binom{m-2}{k}\lr({\psi^{(k)}(y-x+1)-\psi^{(k)}(y)})F_{m-1-k}(x,y) \quad &&{\rm otherwise} }}.$
$m=1$の時、
$\beginend{align}{
F_1(x,y) &=
f_1(x,y) =
\hygeo2F1{x,y}{y+1}1 =
\frac1y\frac{\Gamma(y+1)\Gamma(1-x)}
{\Gamma(y-x+1)\Gamma(1)} \\&=
{\rm B}(1-x,y)
}$
$m>1$の時、
$\beginend{align}{
g(x,y,z) \acoloneqq
\sum_{m=1}^\infty f_m(x,y)z^m =
\sum_{n=0}^\infty
\frac{(x)_n}{n!}
\sum_{m=1}^\infty
\lr({\frac z{y+n}})^m \\&=
\sum_{n=0}^\infty
\frac{(x)_n}{n!}
\frac{\frac z{y+n}}{1-\frac z{y+n}} =
z\sum_{n=0}^\infty
\frac{(x)_n}{n!}\frac1{y-z+n} \\&=
\frac z{y-z}\hygeo2F1{x,y-z}{y-z+1}1 \\&=
\frac z{y-z}
\frac{\Gamma(y-z+1)\Gamma(1-x)}
{\Gamma(y-x-z+1)\Gamma(1)} \\&=
z\frac{\Gamma(y-z)\Gamma(1-x)}
{\Gamma(y-x-z+1)}
}$
$\beginend{align}{
\frac{g^{(0,0,1)}(x,y,z)}{g(x,y,z)} &=
\frac\partial{\partial z}\ln g(x,y,z) \\&=
\frac1z+\psi(y-x-z+1)-\psi(y-z) \\
g^{(0,0,1)}(x,y,z) &=
g(x,y,z)\lr({\frac1z+\psi(y-x-z+1)-\psi(y-z)}) \\
\asupplement{-70pt}{$z=0$でのテイラー展開の係数を比較する。} \\
mf_m(x,y) &=
f_m(x,y)+
\sum_{k=0}^{m-2}
\frac{(-1)^k\lr({\psi^{(k)}(y-x+1)-\psi^{(k)}(y)})}
{k!}f_{m-1-k}(x,y) \\
(m-1)f_m(x,y) &=
\sum_{k=0}^{m-2}
\frac{(-1)^k\lr({\psi^{(k)}(y-x+1)-\psi^{(k)}(y)})}
{k!}f_{m-1-k}(x,y) \\
\frac{F_m(x,y)}{(m-2)!} &=
\sum_{k=0}^{m-2}
\frac{(-1)^k\lr({\psi^{(k)}(y-x+1)-\psi^{(k)}(y)})}
{k!(m-2-k)!}F_{m-1-k}(x,y) \\
F_m(x,y) &=
\sum_{k=0}^{m-2}(-1)^k
\binom{m-2}{k}\lr({\psi^{(k)}(y-x+1)-\psi^{(k)}(y)})F_{m-1-k}(x,y)
}$
特殊値
$m=2$
$\displaystyle\large \sum_{n=0}^\infty \frac{(x)_n}{n!(y+n)^2} = {\rm B}(1-x,y)(\psi(y-x+1)-\psi(y))$
$x=y=\dfrac12$
$\displaystyle\large \sum_{n=0}^\infty \frac{\binom{2n}n}{(2n+1)^24^n} = \frac{\pi\ln2}2$
$m=3$
$\displaystyle\large \sum_{n=0}^\infty \frac{(x)_n}{n!(y+n)^3} = \frac{{\rm B}(1-x,y)}2\lr[{(\psi(y-x+1)-\psi(y))^2-\lr({\psi'(y-x+1)-\psi'(y)})}]$
$x=y=\dfrac12$
$\displaystyle\large \sum_{n=0}^\infty \frac{\binom{2n}n}{(2n+1)^34^n} = \frac{\pi\ln^22}4+\frac{\pi^3}{48}$