Double integrals involving incomplete beta function

Question

I am trying to solve without success the following double integral $$I_1^{(p)}(N)\equiv\frac{1}{2^p}\int_0^1\text{d}x\int_0^1\text{d}y(1+y-x)^{N+p}(1+x-y)^{N-2}B\left(\frac{1}{1+y-x};N,p+1\right)\cdot\theta(y-x)\theta(1-x-y),$$ where $N\in\mathbb{N}$, $p>0$, $\theta(x)$ is the Heaviside step function and $B(x;a,b)$ is the incomplete beta function $$B(x;a,b)=\int_0^xt^{a-1}(1-t)^{b-1}\text{d}t.$$

The product of the two $\theta$ functions can be translated into one of the two following constraints

1) $\quad x\in\left(0,\frac{1}{2}\right)\longrightarrow x<y<1-x$

2) $\quad y\in\left(0,\frac{1}{2}\right)\longrightarrow x<y,\quad y\in\left(\frac{1}{2},1\right)\longrightarrow x<1-y$

so, e.g. in the first case, the integral becomes $$I_1^{(p)}(N)=\frac{1}{2^p}\int_0^{\frac{1}{2}}\text{d}x\int_x^{1-x}\text{d}y(1+y-x)^{N+p}(1+x-y)^{N-2}B\left(\frac{1}{1+y-x};N,p+1\right).$$ At this point I tried some substitutions, such as $t=\frac{y-x}{1-2x}$ in order to get $\int_x^{1-x}\text{d}y\rightarrow\int_0^1\text{d}t$, but the expression remained not tractable for me.

The same happened when I rewrote the incomplete beta function in terms of a hypergeometric one, e.g. by $$B(x;a,b)=\frac{x^a(1-x)^{b-1}}{a}{}_2F_1\left(1,1-b;a+1;\frac{x}{x-1}\right),$$ hoping to be able to use one of the relations that I found here.

Any help would be greatly appreciated.

Edit 1. The above integral is part of a larger expression, containing two other similar terms which can be obtained from $I_1^{(p)}(N)$ with the following substitutions respectively

$I_2^{(p)}(N):\quad B\left(\frac{1}{1+y-x};N,p+1\right)\rightarrow -B\left(\frac{1-y-x}{1+y-x};N,p+1\right)$

$I_3^{(p)}(N):\quad (1+y-x)^{N+p}B\left(\frac{1}{1+y-x};N,p+1\right)\rightarrow 2^p(1-x)^{N+p}B\left(\frac{1-y-x}{1-x};N,p+1\right)$

The structure does not change that much, so I thought that the solution procedure could be similar in the three cases. Nonetheless, probably it is better to report every detail, because the hypothesis that a simplication can occur between different terms cannot be discharged, even if I failed in doing that.

Edit 2. Proceeding as illustrated by @G Cab in his answer below, the result I obtained is \begin{equation}\begin{split} I_1^{(p)}(N)&=2^{2N-1}\left[B(N+p+1,N)B\left(\frac{1}{2};N,p+1\right)-B(N,p+1)B\left(\frac{1}{2};N+p+1,N\right)\right.\\[6pt] &\left.\quad+\int_{\frac{1}{2}}^1t^{N-1}(1-t)^pB\left(\frac{1}{2t};N+p+1,N\right)\text{d}t\right]. \end{split}\end{equation} I am pretty satisfied by the simplication with respect to the starting expression, but now I wonder whether the remaining single integral can be elaborated somehow.

Answer 1

a) the Incomplete Beta function is $$ B \left( {x\;;a,b} \right) = \int_{t\, = \,0}^{\;x} {t^{\,a - 1} \left( {1 - t} \right)^{\,b - 1} dt} $$ the integration variable is different from the upper bound

b) It might be useful to change the Step function with the Iverson bracket.

So $$ \begin{split} I^{\left( p \right)} (N) &= \frac{1}{2^p} \int\limits_{x = 0}^{1} \int\limits_{y = 0}^{1} ( {1 + y - x} )^{\,N + p}( {1 + {x - y} } )^{\,N - 2} \\ &\qquad\cdot B\left( {{1 \over {1 + y - x}}\;;N,p + 1} \right)[ {0 \le y - x} ][ {y + x \le 1} ]dx\;dy \\ \\ & = \frac{1}{2^p}\int\limits_{x = 0}^{1} \int\limits_{y = 0}^{1} \!\!\int\limits_{t = 0}^{\frac{1}{1 + y - x}} ( {1 + y - x})^{\,N + p}( {1 + {x - y} } )^{\,N - 2} t^{\,N - 1} \left( {1 - t} \right)^{\,p} \\ &\qquad\cdot[ {0 \le y - x} ][ {y + x \le 1} ]dx\;dy\,dt \\ \\ & = \frac{1}{2^p}\iiint\limits_{(x,y,t) \in V} ( {1 + y - x})^{N + p} ( {1 + {x - y} } )^{\,N - 2} t^{N - 1} ( {1 - t} )^{\,p} dx dy dt \end{split} $$ where $$ V = \left\{ {(x,y,t)} \right\}:\;\;\left\{ \matrix{ 0 \le x \le 1 \hfill \cr 0 \le y \le 1 \hfill \cr 0 \le y - x \hfill \cr y + x \le 1 \hfill \cr 0 \le t \le {1 \over {1 + y - x}} \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ 0 \le x \le 1/2 \hfill \cr 0 \le x \le y \le 1 - x \hfill \cr 0 \le t \le {1 \over {1 + y - x}} \hfill \cr} \right. $$

Now it remains to change the variables appropriately, so that we can integrate in $t$ at last, after the others, and to reformulate accordingly the bounds on $V$.
Proceeding further the change of variables is $$\begin{cases} v=1+y-x\\ u=1-y-x\end{cases}\quad\Longrightarrow\quad\begin{cases}x=1-\frac{v+u}{2}\\ y=\frac{v-u}{2}\end{cases}$$ and so one obtains $dxdy=\frac{1}{2}dvdu$. It is also easy to check that the domain $V$ splits into two different parts $$V_1=\{(u,v,t)\}:\begin{cases}0<t<\frac{1}{2}\\ 1<v<2\\ 0<u<2-v\end{cases}\qquad\quad V_2=\{(u,v,t)\}:\begin{cases}\frac{1}{2}<t<1\\ 1<v<\frac{1}{t}\\ 0<u<2-v\end{cases} $$ This leads us to $$ \begin{split} I^{\left( p \right)} (N) &= \frac{1}{2^{p+1}}\iiint\limits_{(u,v,t) \in V_1\cup V_2} v^{N + p} u^{N - 2} t^{N - 1} ( {1 - t} )^{p} du dv dt \\ & = \frac{1}{2^{p+1}}\left[\;\int\limits_{t = 0}^{\frac{1}{2}} t^{N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{2} v^{N + p} ( {2 - v} )^{N - 2} dv \int\limits_{u = 0}^{2 - v} {du} \right.\\ &\quad\left.+\int\limits_{t = \frac{1}{2}}^{1} t^{N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{\frac{1}{t}} v^{N + p} ( {2 - v} )^{N - 2} dv \int\limits_{u = 0}^{2 - v} {du} \right]\\ & = \frac{1}{2^{p+1}}\left[\;\int\limits_{t = 0}^{\,\frac{1}{2}} t^{\,N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{2} {v^{N + p} ( {2 - v} )^{N - 1} dv}\right.\\ &\left.\quad+\int\limits_{t = \frac{1}{2}}^{\,1} t^{\,N - 1} ( {1 - t} )^{p} dt\int\limits_{v = 1}^{\frac{1}{t}} {v^{N + p} ( {2 - v} )^{N - 1} dv} \right]\\ & = 2^{2N -1} \left[\;\int\limits_{t = 0}^{\,\frac{1}{2}} {t^{N - 1} ( {1 - t} )^{p} dt \int\limits_{\frac{v}{2} = \frac{1}{2}}^{1} {\left( {{v \over 2}} \right)^{N + p} \left( {1 - {v \over 2}} \right)^{N - 1} d\left( {{v \over 2}} \right)} }\right.\\ &\left.\quad+\int\limits_{t = \frac{1}{2}}^{\,1} {t^{N - 1} ( {1 - t} )^{p} dt \int\limits_{\frac{v}{2} = \frac{1}{2}}^{\frac{1}{2t}} {\left( {{v \over 2}} \right)^{N + p} \left( {1 - {v \over 2}} \right)^{N - 1} d\left( {{v \over 2}} \right)} } \right] = \\ & = 2^{2N -1}\left\{\;\int\limits_{t = 0}^{\frac{1}{2}} {t^{N - 1} \left( {1 - t} \right)^{p} {B\left( {\frac{1}{2};N,N+p+1} \right) } dt}\right.\\ &\left.\quad+\int\limits_{t = \frac{1}{2}}^{1} {t^{N - 1} \left( {1 - t} \right)^{p} \left[ {B\left( {\frac{1}{2t};N + p + 1,N} \right) - B\left( {\frac{1}{2};N + p + 1,N} \right)} \right]dt}\right\} \end{split} $$