I am a grad student working on some independent research trying to derive some exact formulas for a particular class of power series. During my study I came across the following integral which would lead to an exact solution of my power series if I can find an exact anti-derivative.

$\int^x_0 \frac{u^{r-1}}{1-u}\, du$

I am only really interested in values of $r$ between $0$ and $1$ as larger values could essentially be handled by a reindexing of the series. So far the only value of $r$ which I have found an exact anti-derivative for is $r = 1/2$ in which case the integral evaluates to $2 \mathrm{arctanh}(\sqrt{x})$. Can anyone think of any other values of $r$ for which the integral can be evaluated exactly? Or prove that there are no others?

If it is not possible to get any other exact solutions would approximating the integral via Simpson's rule or something similar generally be quicker then approximating a power series?

Any advice on the problem would be greatly appreciated, thanks in advance.

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Hi, Your integral is the incomplete beta function, and has elementary expressions when $r$ is equal to an integer or half-integer. For example, you found the value for $r=1/2$, and when $r=3$ it is

$\frac{1}{2} \left(-2 x-x^2-2 \log(1-x)\right) $.

I haven't found a good reference for this yet; a quick check in Mathematica gave me the evaluations for several integral and half-integral $r$ values. I hope this helps you get started,


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  • $\begingroup$ Thank you. I had heard of the beta function but have never used it before. I guess this kind of ruins my hope of finding some exact values but I may still be able to produce something useful from this. $\endgroup$ – Elliot Mar 15 '13 at 22:12
  • $\begingroup$ sorry but I did not immediately notice that you wanted r < 1; my example for r=3 did you no good... $\endgroup$ – Tom Dickens Mar 15 '13 at 22:37
  • $\begingroup$ That is fine, I had already dealt with the integer solutions as well, I forgot to mention this. But essentially the $-2x - x^2$ piece of the solution amounts to a correction of the index on my power series. $\endgroup$ – Elliot Mar 16 '13 at 0:07

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