Bounds on functions involving CDFs or Beta function

I have functions of the form \begin{align} I_i = \int_0^\infty F_0(x)^aF_1(x)^b(1-F_0(x))^c(1-F_1(x))^ddF_i(x)~~~~i = 0,1 \end{align}

$F_0(x)$ and $F_1(x)$ are CDFs corresponding to the random variables $X_0, X_1$, sith support $[0,\infty)$. The closed form expression for CDFs is extremely complicated - so substitution is out of the question.

I am trying to get a more tractable form of this in the form of upper or lower bounds. Does anybody have any suggestion for tricks that can be tried. This seems a simple enough form, but I am missing something trivial.

I was trying to relate this to the total variational distance between the distributions $TV(X_0,X_1)$. If we assume $$TV(X_0, X_1) < \epsilon,$$ then $$|F_0(x) - F_1(x)| < \epsilon, ~~~\forall x.$$ For example, simplify $I_0$, we use $F_1(x) < F_0(x) + \epsilon$, to obtain

\begin{align} I_0 \leq & \int_0^\infty F_0(x)^a(F_0(x)+\epsilon)^b(1-F_0(x))^c((1-F_0(x)) + \epsilon)^ddF_0(x) \\ = & \int_0^\infty F_0(x)^a\left(\sum_{j=0}^b {b \choose j}F_0(x)^j\epsilon^{b-j}\right)(1-F_0(x))^c\left(\sum_{k=0}^d {d \choose k}(1-F_0(x))^k\epsilon^{d-k}\right) dF_0(x) \\ = & \int_0^\infty \sum_{j=0}^b \sum_{k=0}^d {b \choose j} {d \choose k} \epsilon^{b+d-j-k}F_0(x)^{a+j}(1-F_0(x))^{c+k} dF_0(x) \\ = & \sum_{j=0}^b \sum_{k=0}^d {b \choose j} {d \choose k} \epsilon^{b+d-j-k}\int_0^\infty F_0(x)^{a+j}(1-F_0(x))^{c+k} dF_0(x) \\ = & \sum_{j=0}^b \sum_{k=0}^d {b \choose j} {d \choose k} \epsilon^{b+d-j-k}\mathcal B(a+j+1,c+k+1)\\ \end{align} Here $\mathcal B(m,n)$ is the Beta function. Since the arguments of the Beta function are not trivial, I am unable to simplify this further. In its current form, it is still not usable for me as I would be trying to bound the sum and difference of a few such terms.

My questions are

• Are there any bounds on the Beta function which can help simplify the expression above?

• Is there any better approach to bound the expression for $I_i$ so that we don't run into the Beta function issue?

Thanks!