5
$\begingroup$

In my research I encountered the following integral $$J = \int_0^1 I_t^{-1}(a_1,b_1) \: I_t^{-1}(a_2,b_2) \: \mathrm{d}t$$ which I would like to evaluate as a closed form expression, that is, as a function of the parameters $a_{1}, b_{1}, a_{2}, b_{2} >0$. In words, this is the integral of the product of two inverse regularized incomplete beta functions.

This was posted in Math StackExchange about two months back but received no answer. That link shows my naive attempt via integration by parts. If a closed form (or lack of it) is well known, I would appreciate a reference. Any ideas or approximation for the integral are welcome.

$\endgroup$
2
$\begingroup$

for $a_1=a_2$ and $b_1=b_2$ there is a closed form expression in terms of the incomplete Beta function:

abbreviate $Q_t=I^{-1}_t(a,b)$ and then Mathematica tells me that

$$\int_0^1 Q_t^2 \,dt=\frac{1}{bB(a,b)}$$ $$\times\left[(1-Q_0){}^b Q_0{}^{a+1}-(1-Q_1){}^b Q_1{}^{a+1}+(a+1) B_{Q_1}(a+1,b+1)-(a+1)B_{Q_0}(a+1,b+1)\right]$$

I would be surprised if this can be generalized to $a_1\neq a_2$, $b_1\neq b_2$.

$\endgroup$
  • 1
    $\begingroup$ In fact, since $Q_{0} = 0$, $Q_{1} = 1$, hence in this special case the integral equals to a rational function of parameters: $\int_{0}^{1} Q_{t}^{2} \: dt = \frac{a(a+1)}{(a+b)(a+b+1)}$. $\endgroup$ – Abhishek Halder Dec 28 '16 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.