It is well known that the Beta function is defined as
$$B(x,y) =\int\limits_0^\infty \frac{t^{x-1}}{{(1+t)}^{x+y}}\,\mathrm{d}t$$
What if I have the exact same integral but with a non-zero lower bound, i.e.
$$\int\limits_a^\infty \frac{t^{x-1}}{{(1+t)}^{x+y}}\,\mathrm{d}t$$
where a>0. Can this integral be expressed in terms of Special functions (Beta, Hypergeometric, etc.)?
Thanks
Things turn out a bit nicer if we use a fixed lower limit rather than a fixed upper limit. We have $$ \int_0^a\frac{t^{x-1}}{(1+t)^{x+y}}dt = a^x \int_0^1 u^{x-1}(1+a u)^{-x-y}du = a^xB(1,x)\,_2F_1\left({x+y,\,x\atop x+1};-a\right), $$ found from this form of the hypergeometric integral. Since $B(1,x) = 1/x$, we have for the original integral $$ \int_a^\infty\frac{t^{x-1}}{(1+t)^{x+y}}dt =B(x,y) - \frac{a^x}{x}\,_2F_1\left({x+y,\,x\atop x+1};-a\right). $$