${\displaystyle \mathrm {B} _{z}(a,b)=\int _{0}^{z}t^{a-1}\,(1-t)^{b-1}\,\mathrm {d} t\!}$

${\displaystyle I_{z}(a,b)={\frac {\mathrm {B} _{z}(a,b)}{\mathrm {B} (a,b)}}\!}$

${\displaystyle I_{0}(a,b)=0\,}$
${\displaystyle I_{1}(a,b)=1\,}$
${\displaystyle I_{x}(a,1)=x^{a}\,}$
${\displaystyle I_{x}(1,b)=1-(1-x)^{b}\,}$
${\displaystyle I_{x}(a,b)=1-I_{1-x}(b,a)\,}$
${\displaystyle I_{x}(a+1,b)=I_{x}(a,b)-{\frac {x^{a}(1-x)^{b}}{a\mathrm {B} (a,b)}}\,}$
${\displaystyle I_{x}(a,b+1)=I_{x}(a,b)+{\frac {x^{a}(1-x)^{b}}{b\mathrm {B} (a,b)}}\,}$
${\displaystyle \mathrm {B} (x;a,b)=(-1)^{a}\mathrm {B} \left({\frac {x}{x-1}};a,1-a-b\right)}$