The integral representation of the Beta function $$ \mathrm{B}(\alpha, \beta) = \int_0^1 t^{\alpha - 1} (1-t)^{\beta - 1} \; \mathrm{d} t $$ is often generalised to give the incomplete Beta function, $$ \mathrm{B}_x(\alpha, \beta) = \int_0^x t^{\alpha - 1} (1-t)^{\beta - 1} \; \mathrm{d} t. $$

The integral representation of the hypergeometric function $$ {}_2 F_1 (\alpha, \beta, \gamma; z) = \frac{1}{\mathrm{B}(\beta, \gamma-\beta)} \int_0^1 t^{\beta-1} (1-t)^{\gamma-\beta-1} (1-zt)^{-\alpha} \; \mathrm{d}t $$ seems like it could be generalised in a similar way, by restricting the upper limit of the integral.

Has this been used and given a name somewhere? It would be especially helpful if anyone is aware of identities relating this to a hypergeometric function with different arguments, in the way the incomplete Beta function often is.

  • $\begingroup$ Can you do this special case? $\alpha=2/3,\beta=1,c=3/2,z=2$ $$\frac{1}{2}\int_{0}^{1/3}\!{\frac {1}{\sqrt {1-t} \left( -2\,t+1 \right) ^{2/3}}}\,{\rm d}t$$ $\endgroup$
    – GEdgar
    Dec 6, 2021 at 18:26
  • $\begingroup$ @GEdgar Quite frankly, probably not, at least judging by my attempts so far. Are there any special tricks I should have in mind for the manipulation? $\endgroup$
    – JCW
    Dec 6, 2021 at 20:51
  • $\begingroup$ It looks like an Appell function based off of software. $\endgroup$ Dec 6, 2021 at 22:50

1 Answer 1


Let’s see if there is a closed form in terms of some more general functions which appear in literature. Your problem is:

$$\frac{1}{\mathrm{B}(\beta, \gamma-\beta)} \int_0^1 t^{\beta-1} (1-t)^{\gamma-\beta-1} (1-zt)^{-\alpha} \; \mathrm{d}t$$

First let’s use the Binomial series to get a sum representation for $|zt|<1$ with easier variables:

$$ \int t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a} dt= \int t^{b-1} (1-t)^{c-b-1} \sum_{m=0}^\infty\binom{-a}{m} (-1)^m t^m dt $$

Remember the Incomplete Beta function:

$$\text B_p(a,b)=\int_0^p t^{a-1} (1-t)^{b-1} dt$$

Therefore we can integrate termwise:

$$\int t^{b-1} (1-t)^{c-b-1} \sum_{m=0}^\infty\binom{-a}{m} (-1)^m z^mt^m dt = \sum_{m=0}^\infty\binom{-a}{m} (-1)^m z^m \int_0^p t^{m+b-1} (1-t)^{c-b-1}dt = С+\sum_{m=0}^\infty\binom{-a}{m} (-1)^m z^m \text B_t(m+b,c-b)= \sum_{m=0}^\infty\frac{(-a)!}{m!(-m-a)!} (-1)^m z^m \text B_t(m+b,c-b)$$

From here, the goal would be to get a double hypergeometric series and use the First Appell function which is almost the same as this question where I derived a very similar problem’s closed form in terms of the First Appell function, but I am not sure what to do about the $-m$ factorial in the denominator as it cannot be converted into a factorial with a natural number coefficient, but maybe I will come up with something. Here is the answer anyways with the Pochhammer Symbol function $(a)_n$:

$$\int t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a}dt=\frac{t^b}{b}\text F_1(b;b-c,a;b+1;t,tz)+\frac{t^{b+1}}{b+1}\text F_1(b+1;b-c+1,a;b+2,t,tz)+C,\text F_1(a;b_1,b_2,c;z_1,z_2)=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{(a)_{m+n}(b_1)_m(b_2)_n z_1^m z_2^n}{(c)_mm! n!}$$

Let me try to derive it another way. Please correct me and give me feedback!


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