# Closed form for $\int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx$

This is somewhat similar to my previous question: Closed form for $\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx$

Is it possible to find a closed form for this integral? $$Q=\int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx$$

• Break it up into three parts, using $\ln\frac a{bc}=\ln a-\ln b-\ln c$. – Lucian Nov 14 '13 at 8:09

First, we transform the integral into a more computable form by using some substitutions.

\begin{align*}\displaystyle Q &= \int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx\\ &=\int_0^1 \sqrt{\frac{1+u}{u(1-u)}}\log \left( \frac{(1+u)(1-u)}{u}\right)du \quad \color{blue}{\text{where }u=1-x}\\ &= \int_0^1 \frac{1+u}{\sqrt{u(1-u^2)}}\log \left( \frac{1-u^2}{u}\right)du \\ &= \frac{1}{2}\int_0^1 \frac{1+\sqrt{t}}{t^{\frac{3}{4}}\sqrt{1-t}}\log \left(\frac{1-t}{\sqrt{t}} \right)dt \quad \color{blue}{\text{where }t=u^2} \\ &= \frac{1}{2}\int_0^1 \frac{\log(1-t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt-\frac{1}{4}\int_0^1 \frac{\log(t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt \\ &\quad +\frac{1}{2}\int_0^1 \frac{\log(1-t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt-\frac{1}{4}\int_0^1 \frac{\log(t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt \tag{1} \end{align*}

These four integrals can be evaluated by calculating derivatives of beta function in terms of digamma function. For e.g.

\begin{align*} \int_0^1 \frac{\log(1-t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt &= \frac{d}{dz}\left\{ \int_0^1 t^{-\frac{3}{4}}(1-t)^{z-1} \; dt\right\}_{z=\frac{1}{2}}\\&= \frac{d}{dz}\left\{ \frac{\Gamma \left( \frac{1}{4}\right)\Gamma(z)}{\Gamma \left( \frac{1}{4}+z\right)} \right\}_{z=\frac{1}{2}}\\ &= \frac{\Gamma \left( \frac{1}{4}\right)\sqrt{\pi}}{\Gamma \left( \frac{3}{4}\right)}\left\{\psi_0 \left(\frac{1}{2} \right) -\psi_0 \left(\frac{3}{4} \right)\right\}\\ &= \pi^{3/2}\frac{\sqrt{2}}{\Gamma \left( \frac{3}{4}\right)^2}\left\{\log 2-\frac{\pi}{2} \right\} \tag{2} \\ \end{align*} To get the last expression, I used the special values \begin{align*} \psi_0 \left(\frac{3}{4}\right) &= -\gamma +\frac{\pi}{2}-3\log 2 \\ \psi_0 \left(\frac{1}{2}\right) &= -\gamma -2\log 2 \end{align*} Using the same technique, the other three integrals can be evaluated: \begin{align*} \int_0^1 \frac{\log(t)}{t^{\frac{3}{4}}\sqrt{1-t}}dt&= -\pi^{5/2}\frac{\sqrt2}{\Gamma \left( \frac{3}{4}\right)^2}\tag{3} \\ \int_0^1 \frac{\log(t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt&=\frac{(4\pi-16)\Gamma \left( \frac{3}{4}\right)^2}{\sqrt{2\pi}} \tag{4} \\ \int_0^1 \frac{\log(1-t)}{t^{\frac{1}{4}}\sqrt{1-t}}dt&=\frac{2(-8+\pi+2\log 2)\Gamma \left( \frac{3}{4}\right)^2}{\sqrt{2\pi}} \tag{5} \end{align*} Substituting the results of equations $(2),(3),(4)$ and $(5)$ in $(1)$ gives

$$Q=\frac{\Gamma\left(\frac34\right)^{-2}\pi^2\log2-\Gamma\left(\frac34\right)^2(4-2\log2)}{\sqrt{2\,\pi}}$$

$$Q=\frac{\Gamma\left(\frac34\right)^{-2}\pi^2\ln2-\Gamma\left(\frac34\right)^2(4-\ln4)}{\sqrt{2\,\pi}}$$

• Cleo, can you enlighten us as to how you got this result? – Ron Gordon Nov 14 '13 at 21:45
• (-1) downvoted because this only throw us an answer w/o giving us any hint/ref how to get it. – achille hui Nov 19 '13 at 8:27
• Do you downvote Fermat's post, which was made 350 years ago, too? – Math.StackExchange Nov 19 '13 at 8:40
• @AranKomatsuzaki I won't downvote Fermat's comment on a margin because it is a comment. Math.SE is an Q & A site, being an answer, in particular one with a big bonus attached, we will expect one with more details included. – achille hui Nov 19 '13 at 8:45
• @RonGordon It happens frequently over here. Sometimes ago, I make a lot of calculations to derive an answer and I got one upVote from the OP. However, somebody else wrote four lines of a 'philosophical speech' but never wrote any answer. He or she just said something like 'it's trivial' or so. It looks like there are 'circle of friends' over here. – Felix Marin Nov 24 '13 at 3:15