This post is in reference to this question which asks about the general antiderivative of the following function with the arbitrary constants $a_i,b_i\in \mathbb{R}\forall i\in[1,n]\cap\mathbb{Z^{+}}$.
$$\int\prod_{i=1}^{n}(x-a_i)^{b_1}\mathrm dx$$
It seems easy at first to generalise because this is practically a polynomial, but the source of ambiguity are the real exponents, because they seem to make applying the binomial theorem invalid.
I am thinking about using Taylor series for the $i$th term of the product, but again would we then center the series about the $a_i$? Also how to make sense of something like $(1+x)^{e}$? How to get a grip on it? Any hints are appreciated. Thanks.
Update $1$:
Here's what I've done with the case when $n=2$, using the generalised binomial theorem for complex valued index.
$$\begin{aligned}\int\prod_{i=1}^{2}(x-a_i)^{b_i}\mathrm dx&=\int \prod_{i=1}^{2}\sum_{k=0}^{\infty}(-1)^{k}{b_i\choose k}x^{b_i-k}a_i^{k}\mathrm dx\\ &=\int\left(\sum_{k=0}^{\infty}(-1)^{k}{b_1\choose k}x^{b_1-k}a_1^k\right)\left(\sum_{j=0}^{\infty}(-1)^{j}{b_2\choose j}x^{b_2-j}a_2^{j}\right)\mathrm dx\end{aligned}$$
Considering the $k$th term of the series with parameters $a_1,b_1$ with the corresponding series with parameters $a_2, b_2$ and integrating term by term.
$$\begin{aligned}\int{b_1\choose k}\frac{(-1)^{k}a_1^k}{x^k}\sum_{j=0}^{\infty}{b_2\choose j}\frac{(-1)^{j}a_2^j}{x^{j-(b_1+b_2)}}\mathrm dx&= \sum_{j=0}^{\infty}(-1)^{j+k}{b_1\choose k}{b_2\choose j}a_1^ka_2^j\int\frac{\mathrm dx}{x^{k+j-(b_1+b_2)}}\\&=\sum_{j=0}^{\infty}(-1)^{j+k}{b_1\choose k}{b_2\choose j}a_1^ka_2^j\left(\frac{x^{(b_1+b_2)-(j+k)+1}}{(b_1+b_2)-(j+k)+1}\right)\end{aligned}$$
The integral we started with is now simply the summation of this $k$th term with $k$ varying discretely from $k=0$ to $k=\infty$.
$$\bbox[5px,border:2px solid #C0A000]{\int\prod_{i=1}^{2}(x-a_i)^{b_i}\mathrm dx=\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}(-1)^{j+k}{b_1\choose k}{b_2\choose j}a_1^ka_2^j\left(\frac{x^{(b_1+b_2)-(j+k)+1}}{(b_1+b_2)-(j+k)+1}\right)}$$
Similarly for the case when $n=3$
$$\begin{aligned}\int\left(\sum_{k=0}^{\infty}(-1)^{k}{b_1\choose k}x^{b_1-k}a_1^k\right)\left(\sum_{j=0}^{\infty}(-1)^{j}{b_2\choose j}x^{b_2-j}a_2^j\right)\left(\sum_{i=0}^{\infty}(-1)^{i}{b_3\choose i}x^{b_3-i}a_3^i\right)\mathrm dx\end{aligned}$$
Considering the $k$the term of the first sequence and the $j$th term of the second sequence with the third series.
$$\begin{aligned}(-1)^{j+k}{b_1\choose k}{b_2\choose j}a_1^ka_2^j\sum_{i=0}^{\infty}(-1)^{i}{b_3\choose i}a_3^i\int\frac{\mathrm dx}{x^{i+j+k-(b_1+b_2+b_3)}}\end{aligned}$$
Now how can this summation be written out as both $j$ and $k$ are varying parameters? Also is the approach correct, and whether it can be said that the general integral would be the sum with proper summation notation of the following expression $$\left((-1)^{\sum k_i}\prod_{i=1}^{n}{b_i\choose k_i}\prod_{i=1}^{n}a_i^{k_i}\right)\left(\frac{x^{\sum {b_i}-\sum k_i}+1}{\sum b_i-\sum {k_i}+1}\right)$$ where $k_i$ represents the value being taken on by the respective parameters or indexes of summation like the $i,j,k$ in the above cases?
Update $2$:
I think that the correct way to write this summation, since it is summing over all possible combinations of the indexes $k_i$, is as follows.
$$\bbox[5px,border:2px solid #C0A000]{\sum_{\sigma}\left((-1)^{\sum k_i}\prod_{i=1}^{n}{b_i\choose k_i}\prod_{i=1}^{n}a_i^{k_i}\right)\left(\frac{x^{\sum {b_i}-\sum k_i}+1}{\sum b_i-\sum {k_i}+1}\right), \sigma:=0\le k_i\lt\infty\forall i\in[1,n]\cap\mathbb{Z^{+}}}$$
Is this correct, could someone check my work? Thanks