Time ordered integral involving beta function:

Question

Any help on unpacking integrals of the following type, would be helpful: $$ \int_0^1 \int_0^r r^a (1-r)^b t^n (1-t)^m dr dt $$ where $a, b, n, m \in \mathbb{N}$ and $0 \le t \le 1$.

Edit/Update (2/9/19)

Some background: For the case ``$n=1$'', here is how the solution and proof goes... Label the integral $\beta_1(n,m):= \int_0^1 t^n (1-t)^m dt$. Then we first note that $\beta_1(n,0) = \frac{1}{n+1}$. Next we consider the expression, $$R: = \int_0^1 \frac{d}{dt} \left( t^n (1-t)^m \right) dt.$$ On the one hand, the fundamental theorem of calculus yields $$R = t^n (1-t)^m \Big\vert_0^1 = 0.$$ On the other hand, by using the product rule we get, $$R = n \beta_1(n-1,m) - m \beta_1(n, m-1).$$ Together we see that we can write, $$\beta_1(n,m) = \frac{m}{m+1}\beta_1(n+1, m-1).$$ Finally we put together our relations on $\beta_1(n,m)$ with our initial condition, $\beta_1(n,m) = \frac{1}{n+1}$, to obtain $$\beta_1(n,m) = \frac{n! m!}{(n+m + 1)!}$$

Answer 1

I presume with "the case $n=2$" you mean the integral $$\beta_2(n,m,p,q)=\int _0^1\int _0^t(1-t)^m t^n s^p (1-s)^qdsdt=\frac{m!(n+p+1)!}{(p+1) (m+n+p+2)!}\,\, _3F_2(p+1,n+p+2,-q;p+2,m+n+p+3;1)$$ which might well be further simplified, at least in special cases. For example, $$\beta_2(1,m,p,q)=\frac{[(m+1)!(p+1)!+q] p! (m+q+1)!}{(m+1) (m+2) (m+p+q+3)!},$$ $$\beta_2(n,1,p,q)=\frac{q!}{(n+1)(n+2)}\left(\frac{p!}{(p+q+1)!}-\frac{(n q+2n+p+2 q+4) (n+p+1)!}{(n+p+q+3)!}\right),$$ $$\beta_2(n,m,p,1)=\frac{m (m-1)! (mp+2m+n+2 p+4) (n+p+1)!}{\left(p^2+3 p+2\right) (m+n+p+3)!}.$$