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!


$$ B\left( T+\tfrac{1}{2},\tfrac{3}{2}\right)\cdot {}_2 F_1\left(\tfrac{1}{2},T+\tfrac{1}{2};T+2,z\right) = \int_{0}^{1}\frac{x^T}{\sqrt{x(1-x)(1-xz)}}\,dx$$ 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 $$\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n(n+1)}\left[H_{n-1/2}-H_n\right] =\color{red}{2\pi-4}. $$

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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: $$F(z,k)=\int_{0}^{z}\dfrac{du}{\sqrt{1-u^2}\sqrt{1-k^2u^2}}.$$

  • $\begingroup$ You can get a tighter ub using $x(1-x) \leq \frac{1}{4}$ $\endgroup$ – Jonathan Simon Oct 29 '17 at 22:37
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.