I have the following integral,


where $B_x(c,d) = \int_{0}^xt^{c-1}(1-t)^{d-1}dt$ is the incomplete beta function, and $a,b,c,d>0$.

Question: Does this have a closed form?

My attempt:

  • First, playing around in Wolfram Alpha makes me think that there may be a (simple?) closed form: example 1 and example 2. The second example can also be written as $\int x^2(1-x)B_x(12,2)dx = \frac{1}{4}(B_x(16,2) - x^4B_x(12,2))+\frac{1}{3}(x^3B_x(12,2)-B_x(15,2))$.

  • It seems there is a reduction when $a=c$, $b=d$ as $$\int_0^1 x^{a-1}(1-x)^{b-1}B_x(a,b)dx=\frac{1}{2}\left(\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\right)^2$$ where $\Gamma(a)$ is the gamma function. However, for the case when $a\neq c$, $b\neq d$, things are not quite as clear for me.

  • I found a similar question here but it has not been answered.

  • I also found this but I'm not sure if it is useful.

  • 1
    $\begingroup$ do you have any reason to expect that it does? Also: what have you tried? $\endgroup$ – clathratus Mar 4 at 22:40
  • $\begingroup$ @clathratus I've updated my post to address your questions $\endgroup$ – jonem Mar 5 at 17:24
  • 1
    $\begingroup$ Thank you for your update. I have retracted my close vote and up-voted your question. $\endgroup$ – clathratus Mar 5 at 17:25
  • 1
    $\begingroup$ First note that $B_x(c,\,d)=\int_0^1x^cz^{c-1}(1-zx)^{d-1}dz$ so your integral is $$\int_{[0,\,1]^2}x^{a+c-1}(1-x)^{b-1}(1-zx)^{d-1}dxdz.$$It's mentioned in en.wikipedia.org/wiki/Hypergeometric_function#Euler_type that $$\Re c>\Re b>0\implies\int_0^1x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx=\operatorname{B}(b,\,c-b)\,{}_2F_1(a,\,b;\,c;\,z).$$Unfortunately the notation is a little different from your problem's, but at least you can now write the result as an $\int_0^1dz$ integral of a hypergeometric function. $\endgroup$ – J.G. Mar 5 at 22:17
  • $\begingroup$ @J.G. Thanks! Do you know of any simplification of $\int_0^1 {}_2F_1(a,b;c;z)dz$ when $a,b,c$ are positive integers? $\endgroup$ – jonem Mar 6 at 16:43

Using the hypergeometric representation of the incomplete Beta function \begin{equation} B_x\left( c,d \right)=\frac{x^c}{c}{}_2F_1\left( c,1-d;1+c;x \right) \end{equation} the integral can be written as \begin{align} I\left( a,b,c,d \right)&=\int_{0}^1x^{a-1}(1-x)^{b-1}B_x(c,d)\,dx\\ &=\frac{1}{c}\int_{0}^1x^{a+c-1}(1-x)^{b-1}{}_2F_1\left( c,1-d;1+c;x \right)\,dx\\ &=\frac{1}{c}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \int_0^1x^{a+c+k-1}(1-x)^{b-1}\,dx\\ &=\frac{1}{c}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \frac{\Gamma(b)\Gamma(a+c+k)}{\Gamma(a+b+c+k)}\\ &=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\sum_{k=0}^\infty \frac{(c)_k(1-d)_k}{(1+c)_kk!} \frac{(a+c)_k}{(a+b+c)_k}\\ &=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\,{}_3F_2\left( 1-d,a+c,c;1+c,a+b+c ;1\right) \end{align} Using this identity for the generalized hypergeometric function: \begin{equation} {}_3F_2\left( a_1,a_2,a_3;b_1,b_2;1 \right)=\frac{\Gamma(b_1)\Gamma(b_1+b_2-a_1-a_2-a_3)}{\Gamma(b_1-a_1)\Gamma(b_1+b_2-a_2-a_3)}{}_3F_2\left( a_1,b_2-a_2,b_2-a_3;b_2,b_1+b_2-a_2-a_3;1 \right) \end{equation} Here we choose $a_1=1-d,a_2=a+c,a_3=c,b_1=a+b+c,b_2=1+c$ to obtain \begin{align} I\left( a,b,c,d \right)&=\frac{1}{c}\frac{\Gamma(b)\Gamma(a+c)}{\Gamma(a+b+c)}\frac{\Gamma(a+b+c)\Gamma(b+d)}{\Gamma(a+b+c+d-1)\Gamma(b+1)}\,{}_3F_2\left( 1-d,1-a,1;1+c,1+b ;1\right)\\ &=\frac{1}{bc}\frac{\Gamma(a+c)\Gamma(b+d)}{\Gamma(a+b+c+d-1)}\,{}_3F_2\left( 1,1-a,1-d;1+b,1+c ;1\right)\\ &=\frac{a+b+c+d-1}{bc}B(a+c,b+d)\,{}_3F_2\left( 1,1-a,1-d;1+b,1+c ;1\right) \end{align} which gives simple results if $a$ or $d$ are positive integer, as the hypergeometric series is finite: as $1-a\le0$ or $1-b\le0$, in the series definition of the hypergeometric function, there are $\operatorname{min}(a,b)$ terms (numerator of the coefficients cancel after that). One can also check the given result when $a=c$ and $b=d$.

  • $\begingroup$ Thank you! I have a few questions: 1) is $z$ in the RHS of the first equation suppose to be $x$? 2) I think the identity was applied incorrectly, the second to last line I obtain is $I(a,b,c,d) = \frac{1}{bc}\frac{\Gamma(a+c)\Gamma(b+d)}{\Gamma(c+d)} {}_{3}F_2(1-d,b,a+b;a+b+c,b+1;1)$. 3) I can assume all numbers are integers greater than 0 (I should have included this in the problem statement); how does the hypergeometric series simplify (can you point me to a reference)? $\endgroup$ – jonem Mar 6 at 1:44
  • $\begingroup$ You're welcome. 1) sorry, typo corrected. 2) and 3) Details are added in the answer. A typo is also corrected in the last expression. $\endgroup$ – Paul Enta Mar 6 at 13:03

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