I know that $${\left\lvert \int_\alpha ^\beta (x-\alpha)^m(x-\beta)^ndx \right\rvert} $$ can be integrated to $${{n!m!(\beta-\alpha)^{m+n+1}}\over {(n+m+1)!} }$$ using recurrence relation.

Then, can the generalized form: $$\left\lvert \int_{a_1} ^{a_n} (x-a_1)^{b_1}(x-a_2)^{b_2}\cdots(x-a_n)^{b_n} dx \right\rvert $$ be integrated?

Let's assume that $ a_p<a_q $ when $p<q$.

$b_i$ values are non-negative integers.

  • $\begingroup$ Are the $b_i$ values non-negative integers? $\endgroup$ – JimB Aug 1 '19 at 16:39
  • $\begingroup$ @JimB yes, they are $\endgroup$ – Verthele Aug 1 '19 at 16:40
  • 1
    $\begingroup$ Doesn't look promising for a simple form. For $n=3$ Mathematica gives $b_1! (a_3-a_1)^{b_3} \left(e^{i \pi b_3} b_3! \left(\frac{1}{a_1-a_2}\right)^{-b_2} (a_3-a_1)^{b_1+1} \, _2\tilde{F}_1\left(b_1+1,-b_2;b_1+b_3+2;\frac{a_1-a_3}{a_1-a_2}\right)+\left(-1+e^{2 i \pi b_2}\right) b_2! e^{-i \pi (b_2-b_3)} (a_2-a_1)^{b_1+b_2+1} \, _2\tilde{F}_1\left(b_1+1,-b_3;b_1+b_2+2;\frac{a_1-a_2}{a_1-a_3}\right)\right)$. $\endgroup$ – JimB Aug 1 '19 at 16:58

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