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I am seeking (preferably) a closed form for the integral

\begin{equation} \int_{0}^{1} \frac{\ln\left[x(1-x)\right]}{\sqrt{x(1-x)(1-zx)}} dx, \;\;\; z<1 \end{equation} I know from previous work that

\begin{equation} B(b,c-b)\,_2F_1(a,b;c;z) = \int_0^1 x^{b-1} (1-x)^{c-b-1}(1-zx)^{-a} \, dx \end{equation} so I am hoping that my integral will be some other form of a hypergeometric function. I have tried expressing the logarithm as a hypergeometric function using the relationship $\ln(1-z)=-zF(1,1;2;z)$ and using properties in the book "Higher Transcendental Functions." I have also scoured nist.gov and "Table of integrals, series, and products." for some kind of useful identity but I cant find anything. The closest thing I found is equation 7.512.9 from Table of Integrals Series and Products, which is a result for

\begin{equation} \int_0^{1}x^{\gamma-1}(1-x)^{\rho-1}(1-zx)^{-\sigma}F(\alpha,\beta;\gamma;x)dx \end{equation}

which I wont type because my integral doesn't conform.

At this point, an approximation would be fine. But if someone can see a way to solve this thing analytically, that would be amazing. Either way, I appreciate the help!

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$$ B\left( T+\tfrac{1}{2},\tfrac{3}{2}\right)\cdot {}_2 F_1\left(\tfrac{1}{2},T+\tfrac{1}{2};T+2,z\right) = \int_{0}^{1}\frac{x^T}{\sqrt{x(1-x)(1-xz)}}\,dx$$ hence by differentiating both sides with respect to $T$, then evaluating at $T=0$ we get that $$ -\tfrac{\pi}{2}\left(1+2\log 2\right)\cdot {}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};2,z\right)+\tfrac{1}{2}\sum_{n\geq 0}\frac{\Gamma\left(n+\tfrac{1}{2}\right)^2\left[1+2\log 2+H_{n-1/2}-H_n\right]}{n!\Gamma(n+2)}\, z^n $$ exactly equals $\int_{0}^{1}\frac{\log x}{\sqrt{x(1-x)(1-xz)}}\,dx$, and $\int_{0}^{1}\frac{\log(1-x)}{\sqrt{x(1-x)(1-xz)}}\,dx$ has an analogous closed form.
The previous line can be simplified as follows: $$\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n(n+1)}\left[H_{n-1/2}-H_n\right] z^n $$ and this is not, strictly speaking, a hypergeometric function, but it is pretty simple to approximate such object with hypergeometric functions by considering the asymptotic expansion of $H_{n-1/2}-H_n$. Campbell has shown that similar series has closed forms for many specific values of $z$, and together with Sondow we proved that Fourier-Legendre series expansions provide a very effective technique for evaluating $\int K(x)g(x)\,dx$. The above series is an instance: at $z=1$ we have $$\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n(n+1)}\left[H_{n-1/2}-H_n\right] =\color{red}{2\pi-4}. $$

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  • $\begingroup$ Thank you for this very elegant solution. I am working through the details(its not trivial to me as I am not a mathematician). I am wondering: 1) Is the equation you began with written somewhere? I did not see it referenced anywhere I looked. 2) I am also having a hard time with the second term in the derivative of the LHS of the provided equation. Additionally, I browsed your arXiv page and it seems like you do work with this function. Would you suggest a reference or two which would have a comprehensive list of identities and relationships? Thank you! $\endgroup$ – John Snyder Oct 29 '17 at 19:44
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    $\begingroup$ 1) The initial equation of this answer is just the second equation of your question, with a suitable choice of parameters 2) the second equation in my answer depends on the derivatives of Pochhammer symbols / Beta functions, i.e. on the $\psi$ function. For some insights on similar series, which turn to be natural generalization of Euler Beta function, have a look at the work of John Campbell. $\endgroup$ – Jack D'Aurizio Oct 29 '17 at 20:14
  • $\begingroup$ Thank you on 2), I will look into that. re: 1), $b=T+1/2$ but $c-b-1=-1/2$ would mean $c=T+1$ not $T+2$ though right? $\endgroup$ – John Snyder Oct 29 '17 at 20:30
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    $\begingroup$ @JohnSnyder: it looks like I manipulated something more complex than the wanted integral, you are right. Indeed $\int_{0}^{1}\frac{\log x}{\sqrt{x}(1-x)}\,dx = -\frac{\pi^2}{2}\neq 2\pi-4$. Luckily, the shown techniques applies also if $T+2$ is replaced by $T+1$, and they lead to a hypergeometric series with a "harmonic twist" anyway. So the provided references still applies. $\endgroup$ – Jack D'Aurizio Oct 29 '17 at 21:10
  • $\begingroup$ Thanks for your help, I posted a followup question to this one here $\endgroup$ – John Snyder Oct 31 '17 at 15:44
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Not an approximation but an upper bound for the integral

$$ \int_{0}^{1} \frac{\ln\left[x(1-x)\right]}{\sqrt{x(1-x)(1-zx)}} dx\leq \int_{0}^{1} \frac{\ln\left[1/4\right]}{\sqrt{x(1-x)(1-zx)}}$$ $$=\ln 8\dfrac{F\left[\sqrt{1-z},\sqrt{\dfrac{1}{1-z}}\right]-F\left[1,\sqrt{\dfrac{1}{1-z}}\right]}{\sqrt{z-1}} $$

In which the following elliptic function was used: $$F(z,k)=\int_{0}^{z}\dfrac{du}{\sqrt{1-u^2}\sqrt{1-k^2u^2}}.$$

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  • $\begingroup$ You can get a tighter ub using $x(1-x) \leq \frac{1}{4}$ $\endgroup$ – Jonathan Simon Oct 29 '17 at 22:37
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    $\begingroup$ @Jon_simon: Don't know how I came up with $1/2$ I actually wanted to have $1/4$ :D. $\endgroup$ – MrYouMath Oct 30 '17 at 8:53

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