This question comes from related “sum of incomplete gamma function” problems. Also,there are many research papers trying to make the Hypergeometric Incomplete. Here is one confluent example from

Extension of Incomplete Gamma, Beta and Hypergeometric Functions formula (3)


M. A. Chaudhry, A. Qadir, H. M. Srivastava and R. B. Paris, Extended hypergeometric and confluent hypergeometric functions, Appl. Math. Comput. 159, 589–602 (2004):

$$\text F_p(a,b,c,z)=\sum_{m=0}^\infty \frac{(a)_m\text B_p(b+m,c-b)z^m}{\text B(b,c-b)m!}\implies \text F_p(a,b,c,z) \text B(b,c-b)= \sum_{m=0}^\infty \frac{(a)_m\text B_p(b+m,c-b) z^m}{m!} $$

with the (Incomplete) Beta function and Pochhammer symbol. Here is an attempt at a closed form with the well known First Appell Hypergeometric function using a series expansion for the Incomplete Beta function:

$$\text F_p(a,b,c,z) \text B(b,c-b) = \sum_{m=0}^\infty \frac{(a)_m\text B_p(b+m,c-b) z^m}{m!} = p^b\sum_{m=0}^\infty \frac{(a)_m z^m}{m!}\sum_{n=0}^\infty\frac{(1-c+b)_np^n p^m}{(b+m+n)n!} $$

Notice that $\frac1{b+m+n}=\frac{(b)_{m+n}}{b(b+1)_{m+n}}$:

$$p^b\sum_{m=0}^\infty \frac{(a)_m z^m}{m!}\sum_{n=0}^\infty\frac{(1-c+b)_np^n p^m}{(b+m+n)n!} =\frac{p^b}{b}\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(b)_{m+n}(a)_m(b-c+1)_n (zp)^m p^n}{(b+1)_{m+n}m!n!}$$

when the First Appell Hypergeometric function is defined as:

$$\text F_1(a;b_1,b_2;c;z_1,z_2)=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{(a)_{m+n}(b_1)_m (b_2)_n z_1^m z_2^n}{(c)_{m+n}m!n!}$$


$$\text F_p(a,b,c,z) = \sum_{m=0}^\infty \frac{(a)_m\text B_p(b+m,c-b)z^m}{\text B(b,c-b)m!} =\frac{p^b}{b\cdot \text B(b,c-b)}\text F_1(b;a,b-c+1;b+1;pz,p)$$

with some possible special cases of:

$$\text F_1(a,b,c,z)=\frac{\Gamma(b+1)\Gamma(c-b)\,_2\text F_1(a,b;c;z)}{b\Gamma(c)\text B(b,c-b)}$$

$$\text F_p(a,b,c,1)=\frac{p^b \,_2\text F_1(b,a+b-c+1;b+1;p)}{\text B(b,c-b)b}$$

$$\text F_a(a,b,a,a)=\frac{\,_2\text F_1(a,b,b+1,-a)}{(1-a)^b\text B(b,a-b)}$$

where appears the Gauss Hypergeometric function. The concern here is that it is not known if this final result has a typo with the following restriction:

$$\text B(b,c-b)\text F_p(a,b,c,z) = \sum_{m=0}^\infty \frac{(a)_m\text B_p(b+m,c-b)z^m}{m!} =\frac{p^b}{b}\text F_1(b;a,b-c+1;b+1;pz,p),|z|<1,-a\not\in\Bbb N$$

This gives closed forms for many Incomplete Beta series which may be encountered in integrals and more. Could you please check this derivation for any mistakes and give some special cases for this function? The special cases should not be based off of the cases where the Incomplete Beta function simplifies into simpler functions of $2$ arguments or less. For example:

$$\sum_{m=0}^\infty \frac{(a)_m\text B_p(b+m,c-b)z^m}{m!}\mathop=^{b=-1,c=0} \sum_{m=0}^\infty \frac{(a)_m p^{m-1} z^m}{(m-1)m!} $$

which is just the hypergeometric function again. Please correct me and give me feedback!



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