I need to evaluate this integral: $$Q=\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx.$$ I tried it in Mathematica, but it was not able to find a closed form. A numerical integration returned $$Q\approx0.670803371410017436741...$$

Is it possible to find a closed form for $Q$?

This integral is not from a book, it is part of some calculations related to theoretical physics. I do not have a particular reason to be sure that a closed form exists.

  • $\begingroup$ Start by using the fact that $\ln\frac a{bc}=\ln a-\ln b-\ln c$ to break it up into three simpler parts. $\endgroup$
    – Lucian
    Nov 14 '13 at 5:38



Looking through Table of Integrals, Series, and Products, $7^{th}$ Edition, I.S. Gradshteyn, I.M. Ryzhik, I noticed that the formula 3.255 could be taken as a starting point: $$\int_0^1\frac{x^{\mu+\frac12}(1-x)^{\mu-\frac12}}{\left(c+2\,b\,x-a\,x^2\right)^{\mu+1}}dx=\frac{\sqrt\pi}{\left(a+\left(\sqrt{c+2\,b-a}+\sqrt{c}\right)^2\right)^{\mu+\frac12}\sqrt{c+2\,b-a}}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}$$ Fixing the parameters $a=0,\ b=1,\ c=1$, we can make an observation that the integrand in $Q$ can be represented as a derivative of the integrand in 3.255: $$\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\ln\left(\frac{1+2\,x}{x\,(1-x)}\right)=\partial_\mu\left(\frac{x^{\mu+\frac12}(1-x)^{\mu-\frac12}}{(1+2\,x)^{\mu+1}}\right)_{\mu=\frac13}$$ Now we need to calculate the derivative of the right-hand side of 3.255: $$Q=\partial_\mu\left(\frac{\sqrt\pi}{\left(\sqrt3+1\right)^{2\mu+1}\sqrt3}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}\right)_{\mu=\frac13}\\=\sqrt{\frac\pi3}\left(\frac{\psi\left(\mu+\frac12\right)-\psi(\mu+1)-2\ln\left(\sqrt3+1\right)}{\left(\sqrt3+1\right)^{2\mu+1}}\cdot\frac{\Gamma\left(\mu+\frac12\right)}{\Gamma(\mu+1)}\right)_{\mu=\frac13}\\=\sqrt{\frac\pi3}\cdot\frac{\psi\left(\frac56\right)-\psi\left(\frac43\right)-2\ln\left(\sqrt3+1\right)}{\left(\sqrt3+1\right)^{\frac53}}\cdot\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac43\right)}$$ Here $\psi(z)=\partial_z\ln\Gamma(z)$ is the digamma function. Using Gauss digamma theorem we can expand the values of the digamma function that appear in this formula: $$\psi\left(\frac56\right)=\frac{\pi\,\sqrt3}2-\frac{3\ln3}2-2\ln2-\gamma$$ $$\psi\left(\frac43\right)=3-\frac{\pi\,\sqrt3}6-\frac{3\ln3}2-\gamma$$ Plugging these values back to the previous formula, we get the desired result.

Addendum (By editor): The formula $3.255$ in G&R that Vladimir quoted, can be proved by 6 substitutions, that is: $$x\to 1-t, t\to \frac1u, u\to v+1, v\to w^2, w\to y \sqrt[4]{\frac a{a+b-c}}$$ and the final one is the a V. Moll's variant of Cauchy-Schlomilch transform (i.e. dealing with integrals involving $\left(\frac{x^2}{x^4+2ax^2+1}\right)^c$), which can be find in Theorem $4.1$ of:

Amdeberhan, T. , et al. "The Cauchy-Schlomilch transformation." Mathematics (2010).

This justifies the correctness of $3.255$.

  • $\begingroup$ Vladimir: could we see a bit more? The result is far from trivial. $\endgroup$
    – Ron Gordon
    Nov 14 '13 at 14:46
  • $\begingroup$ @RonGordon Unfortunately, I do not yet have a proof of the formula 3.255. Would you like to see that part? Or how I manipulated it to get the result? $\endgroup$ Nov 14 '13 at 16:52
  • $\begingroup$ What you did: the derivative and the simplification. $\endgroup$
    – Ron Gordon
    Nov 14 '13 at 17:15

The same integral without the log term is given by $$ Q(a,b,c) = \int_0^1 \frac{x^a}{(1-x)^b(1+2x)^c}\,dx = \frac{\Gamma(1+a)\Gamma(1-b)}{\Gamma(2+a-b)} F\left(\begin{array}{c}1+a,c\\2+a-b\end{array}\middle|-2\right), $$ because this integral is one of the common integral representations of the hypergeometric function, such as DLMF 15.6.1.

At the point $(a,b,c)=(\frac56,\frac16,\frac43)$, the integral you want is given by a sum of partial derivatives of $Q$ at that point: $$ (-\partial_a+\partial_b-\partial_c)Q(a,b,c). $$ Derivatives of the hypergeometric function with respect to parameters generally aren't expected to have a closed form, except in terms of functions such as Appell functions or Kampe de Feriet functions if those can be considered closed form.


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.