4

Can either of the following integrals be expressed in explicit forms in terms of known constants? They can certainly be expressed in terms of special values of hypergeometric functions with many parameters, but I am interested in expressions in terms of some combination of simpler (multiple)-zeta values and L-values if such expressions exist.

Here I am interested in

$$I_2 = \int_{0}^{1/4} \frac{(\log(1-x))^2}{x \sqrt{1-4x}}dx,$$

$$J_2 = \int_{0}^{1/4} \frac{\log((1-x))^2}{(1-x) \sqrt{1-4x}}dx.$$

Some added clarification in light of a comment, this is the square of $\log(1-x)$, not the log of $(1-x)^2$ which simplifies. It turns out for comparison that

$$I_1 = \int_{0}^{1/4} \frac{(\log(1-x))}{x \sqrt{1-4x}}dx = - \frac{\pi^2}{18},$$

$$J_1 = \int_{0}^{1/4} \frac{\log((1-x))}{(1-x) \sqrt{1-4x}}dx = \frac{\pi \sqrt{3} \log 3}{9} - L(2,\chi_{-3})$$

where $L(2,\chi_{-3})$ is the Dirichlet L-value $\displaystyle{ \sum \frac{1}{(3n+1)^2} - \frac{1}{(3n+2)^2}}$.

Mathematica can "evaluate" the second integral (even as an indefinite integral) in terms of hypergeometric functions and so reduces the calculation in that case to ${}_{4}F_{3}(1/2,1/2,1/2,1/2;3/2,3/2,3/2;3/4)$, but I have no reason to expect either way whether this has a simpler form or not.

One reason to possibly imagine there might be an explicit answer is because the simpler versions $I_1$ and $J_1$ can be evaluated. One reason to imagine there might not be is that they are related to special values of Hypergeometric functions not at $z=1$ but at $z=3/4$ or $z=1/4$, and there seem to be fewer examples known of special values at this point.

SearchingForPeriods
  • 41
  • 1
    The square root term has a rather convenient representation as a series $$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty {2n \choose n} x^n$$ which would make calculation a bit easier. – Ninad Munshi Dec 19 '23 at 00:07
  • 1
    @NinadMunshi. i agree but we are left with a summation of other hypergeometric functions. Cheers :-) – Claude Leibovici Dec 19 '23 at 03:05
  • In the $2nd$ integral you can write $\log(1-x)^{2}=2\log(1-x)$. Therefore now substitute $\log(1-x)=t$ and you will get $\frac{-1}{1-x}dx=dt$. – Syamaprasad Chakrabarti Dec 19 '23 at 20:11
  • But still I don't know whether this substitution will make your question easier or not. But still try once. Write $x$ in terms of $e^{x}$. – Syamaprasad Chakrabarti Dec 19 '23 at 20:14
  • @SyamaprasadChakrabarti the $\log(1-x)^2$ does not simplify, as clarified, $\log(1-x)^2$ is meant as the square of $\log(1-x)$, not the logarithm of the square of $x-1$. – GTBailey Dec 19 '23 at 21:42
  • But then the OP should have written $\log^{2}(1-x)$ which is equal to $(\log(1-x))^{2}$. Genrally $\log(x^{2})=2\log(x)$. – Syamaprasad Chakrabarti Dec 20 '23 at 04:25

0 Answers0