Examining convergence of integral (Gamma/Beta function) Consider the following integral:
\begin{equation}
J=\int_0^{\infty}\frac{t^{2x-1}}{(1+t)^{x+y}}\ \textrm{d}t
\end{equation}

(1) Find the range of values of $x,y$ for which $J$ converges.


(2) Write $J$ in terms of a Gamma or Beta integral

My attempts:
(1) As $t\to+\infty$, convergence is governed by the $t^{2x-1}$ term. For convergence we thus need $2x-1<1$, i.e. $x<\frac{1}{2}$. As $t\to 0+$, convergence is governed by the $(1+t)^{-x-y}$ term, for which we need $x+y>0$, since $1+t>0$. Therefore $y>-1$, as we incorporate the bound for $x$ in this constraint. So overall we have $x<1, y>-1$ for convergence.
(2) is where I struggle a bit. This is what I think we can do
\begin{equation}
J=\int_0^{\infty}t^{(2x)-1}(1-(-t))^{-x-y}dt
\end{equation}
So if we have $p=2x$ and change $t\to -t$ we could get $-B(2x,1-x-y)$? But this seems off to me.
Can someone help with formalising my argument for (1) and giving hints for how to do (2)?
 A: As for the first part of the question:
You can write $J$ as a the sum of corresponding integrals on $[0,1]$ and $[1,\infty)$, i.e. $J=A+B$, where
\begin{equation}
A=\int_0^{1}\frac{t^{2x-1}}{(1+t)^{x+y}}\ \textrm{d}t
\quad \text{and} \quad
B = \int_1^{\infty}\frac{t^{2x-1}}{(1+t)^{x+y}}\ \textrm{d}t.
\end{equation}
We can estimate each of the integrals $A$ and $B$ separately:
$$
  \frac 1 2 \int_0^{1} t^{2x-1} \textrm{d}t
< A < \int_0^{1} t^{2x-1} \textrm{d}t
= \frac 1 {2x}\big[t^{2x}\big]_0^1, 
$$
so $A$ is finite if and only if $x>0$; and
$$
 \int_1^{\infty}\frac{t^{2x-1}}{t^{x+y}}\ \textrm{d}t
< B < \int_1^{\infty}\frac{(1+t)^{2x-1}}{(1+t)^{x+y}}\ \textrm{d}t
=\int_2^{\infty}\frac{t^{2x-1}}{t^{x+y}}\ \textrm{d}t,
$$
so that $B$ is finite if and only if
$$
\int_1^{\infty}\frac{t^{2x-1}}{t^{x+y}}\ \textrm{d}t
=\int_1^{\infty}t^{x-y-1}\ \textrm{d}t
= \frac{1}{x-y} \big[t^{x-y}\big]_1^\infty
$$
is finite, i.e. when $x-y<0$.
In conclusion, the original integral $J$ is finite if and only if $0<x<y$.
