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A derivative of the incomplete beta function $\text B_x(a,b)$ uses hypergeometric $_3\text F_2$

$$\frac{d\text B_x(a,b)}{da}=\ln(x)\text B_x(a,b)-\frac{x^a}{a^2}\,_3\text F_2(a,a,1-b;a+1,a+1;x)$$

Now reduce to a Gauss hypergeometric function $_2\text F_1(a,b;c;x)$:

$$\begin{align}_3\text F_2(a,a,1-b;a+1,a+1;x)=\,_2\text F_1(1-b,a;a+1;x)-a\,_2\text F_1^{(0,1,0,0)}(1-b,a,a+1;x)-a\,_2\text F_1^{(0,0,1,0)}(1-b,a,a+1;x)=\lim_{c\to a}\frac{c\,_2\text F_1(1-b,a;a+1;x)-a\,_2\text F_1(1-b,c;c+1;x)}{c-a}=\underbrace{{\lim_{c\to a}\frac{ac}{c-a}\left(\frac{\text B_x(a,b)}{x^a}-\frac{\text B_x(c,b)}{x^c}\right)= \lim_{c\to b}\frac{bc}{b-c}\left(\frac{\text B_x(a,c)}{x^c}-\frac{\text B_x(a,b)}{x^b}\right)}}_{(1)} \end{align}$$

Shown here is the beta result and $\frac {df(u,v)}{du}=f^{(1,0)}(u,v)$. However, one is unsure of the next step.

Alternatively, we prove a simpler form of $(1)$ using these beta integrals

$$\,_3\text F_2(a,a,1-b;a+1,a+1;x)=\underbrace{\frac{a^2}{x^a}\int\frac{\text B_x(a,b)}x dx}_{(2)}=\underbrace{\frac{a^2}{x^a}\lim_{c\to 0}\frac{x^c\text B_x(a,b)-\text B_x(a+c,b)}c}_{(3)}$$

All three attempts, $(1),(2),(3)$ use simpler functions, but are not closed forms.

Is there a closed form of $_3\text F_2(a,a,1-b;a+1,a+1;x)$ in terms of simpler functions?

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  • $\begingroup$ What is $(1)$? The relation $(2)$ has on the L.H.S. a generalized hypergeometric ${}_3F_2$, should we replace this side by a "simpler" expression? The places marked by $(1)$, $(2)$, $(3)$ are equalities, which expression should be replaced exactly by a function simpler than a ${}_3F_2$? Having one (and only one) clear question would help potential answerers to focus. $\endgroup$
    – dan_fulea
    Commented Dec 24, 2022 at 14:19
  • $\begingroup$ @dan_fulea Fixed tags. All of the equalities, $(1),(2),(3)$, are not closed forms since the use integrals, limits, or derivatives of simpler functions. All are three attempts,$(1),(2),(3)$, at a simpler form of $_3 F_2$, not multiple questions. Right? $\endgroup$
    – Тyma Gaidash
    Commented Dec 24, 2022 at 14:20

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