Elementary closed form of $\exp\left(b\int_0^x \frac{t^{a-1}}{1-t^b}dt\right);a,b\in\Bbb N$

Question

Our goal is an elementary expression of $$\exp\left(b\int_0^x \frac{t^{a-1}}{1-t^b}dt\right)=e^{\text B_{x^b}\left(\frac ab,0\right)}=e^{x^a\Phi\left(x^b,1,\frac ab\right)};a,b\in\Bbb N,0\le x\le1 \tag1$$

with the Incomplete Beta $\text B_z(u.v)$ function and Lerch Transcendent $\Phi(z,s,r)$. It was straightforward to theorize:

$$e^{\text B_{x^n}\left(\frac1n+m,0\right)}=\prod_{k=0}^{n-1}\left(1-(-1)^{\left(\frac1n+1\right)k}x\right)^{-(-1)^{\left(1-\frac1n\right)k}}\prod_{k=0}^me^{-\frac{nx^{kn+1}}{kn+1}};n,m\in\Bbb N\tag2$$

where both products do not operate on one another’s arguments. Unfortunately, I did not find a formula for $e^{\text B_{x^b}\left(\frac ab,0\right)}$ on Wolfram functions, but if $a$ divides $b$, then $a,b$ both decrease proportionally simplifying $(1)$.

Also, $(1)$’s explicit formula is similar to $(2)$’s since the $a$ value changes the upper bound of $\displaystyle\prod_{k=0}^m$ and the ${-(-1)^{\left(1-\frac1n\right)k}}$ part to a different power of $-1$.

I used fullsimplify functionexpand[exp(Beta(x^b,a/b,0))],x=eulergamma to make WolframAlpha simplify into a product with particular $a,b$.

Please note the “closed form” and “elementary-functions” tags. What is $\displaystyle\exp\left(b\int_0^x \frac{t^{a-1}}{1-t^b}dt\right)$ only in terms of elementary functions and finite products?

Answer 1

If you use the geometric series, and exchange sum and integral, $$\begin{split} \exp\left(b\int_0^x \frac{t^{a-1}}{1-t^b}dt\right) &= \exp \left(b\int_0^x \sum_{n\geq 0}t^{a+bn-1}dt\right)\\ &=\exp \left(b\sum_{n\geq 0}\frac{x^{a+bn}}{a+bn}\right)\\ &= \prod_{n\geq 0}e^{\frac{b\cdot x^{a+bn}}{a+bn}} \end{split}$$

Answer 2

I remembered this paper:

A note on some reduction formulas for the incomplete beta function and the Lerch transcedent

which does not have the $\frac ab$ case, but does have $a+\frac pq;a,p,q\in\Bbb N,p<q$:

$$\text B_x\left(a+\frac pq,0\right)=-\sum_{n=0}^{q-1}e^{-\frac{2\pi i n p}q}\ln\left(1-\sqrt[q]xe^{\frac{2\pi i n}q}\right)-\sum_{n=0}^{a-1}\frac{x^{n+\frac pq}}{n+\frac pq}$$

We use the principal root of $-1$ assuming $0\le x\le 1$:

$$\begin{align}e^{\text B_{x^q}\left(a+\frac pq,0\right)}=\exp\left(-\sum_{n=0}^{q-1}(-1)^{-\frac{2 n p}q}\ln\left(1-x(-1)^{\frac{2 n}q}\right)-\sum_{n=0}^{a-1}\frac{x^{nq+p}}{n+\frac pq}\right)= \prod_{n=0}^{q-1}\left(1-(-1)^\frac{2n}qx\right)^{-(-1)^{-\frac{2np}q}}\prod_{n=0}^{a-1}e^{-\frac{x^{nq+p}}{n+\frac pq}}\end{align}=\prod_{n=0}^{q-1}(1+a_n x)^{b_n}\prod_{n=0}^{a-1}e^\frac{x^{c_n}}{d_n}$$