# Evaluating $-\int_0^1\frac{1-x}{(1-x+x^2)\log x}\,dx$

I was trying do variations of an integral representation for $\log\frac{\pi}{2}$ due to Jonathan Sondow, when I am wondering about if it is possible to evaluate

$$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx,\tag{1}$$ Wolfram Alpha online calculator provide me a closed-form with code

int -(1-x)/((1-x+x^2)log(x)) dx, from x=0 to x=1

Question. Please provide me hints to know how evaluate previous this definite integral as $$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx=\log \left(\frac{\Gamma(1/6)}{\Gamma(2/3)}\right)-\frac{\log \pi}{2}$$ as said Wolfram Alpha. Many thanks.

• Maybe this might be helpful: $x^2-x+1=\frac{x^3+1}{x+1}$ Jul 16, 2017 at 12:29
• i think this would make the question much better... Jul 16, 2017 at 12:34
• $$2I'_{a/3}=\sum_{n\geq0}\frac{(-1)^n}{n+a/3}=\psi(1-\frac a6)-\psi(\frac12-\frac a6)$$ and likewise $$I'_{a/3+2/3}=\sum_{n\geq0}\frac{(-1)^n}{n+a/3}=\psi(\frac 13-\frac a6)-\psi(\frac16-\frac a6)$$ where $\psi(z) =\log(\Gamma(z))'$ is the digamma function [the digamma function][1]. Integrating with respect to $a$ therefore yields $$6J_a=\log(\Gamma(1-\frac a6))-\log(\Gamma(\frac12-\frac a6))-\log(\Gamma(\frac13-\frac a6))+\log(\Gamma(\frac16-\frac a6))+C$$ Now your integral in question is just $J_1$... [1]: en.wikipedia.org/wiki/Digamma_function Jul 16, 2017 at 13:11
• @tired Or you just let $t=e^x$, and apply some complex analysis. Jul 16, 2017 at 13:11
• Please don't use int -(1-x)/((1-x+x^2)log(x)) dx, from x =0 to x=1 to replace $$\int_0^1 -\frac{(1-x)}{(1-x + x^2)\log(x)} = \int_0^1 \frac{x-1}{(x^2 -x+1)\log(x)}$$ Jul 16, 2017 at 16:13

Thanks to the power series representation of $1/(1+x^3)$ and the Dominated Convergence Theorem, the given integral is \begin{align*} -\int_0^1\frac{1-x^2}{(1+x^3)\log x}\,dx &=\int_0^1\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(1-x^2)x^{3k}}{\log x}\,dx\\ &=\int_0^1\sum_{k=0}^{\infty}\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\ &=\sum_{k=0}^{\infty}\int_0^1\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\&= \sum_{k=0}^{\infty}(\ln(6k+3)-\ln(6k+1)+\ln(6k+4)-\ln(6k+6))\\&= \ln\left(\prod_{k=0}^{\infty}\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right) =\ln\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right) \end{align*} where $\Gamma(x)$ is the Gamma function and we used the fact that $$\int_{0}^{1}\frac{x^n-1}{\log x}\,dx = \int_{0}^{+\infty}\frac{1-e^{-nt}}{t}\,e^{-t}dt = \log(n+1)$$ (apply Frullani's integral in the last step).

• @Zophikel Yes to interchange the sum and the integral I used Dominated Convergence Jul 16, 2017 at 12:59
• Although I am certain that your computation can be justified, I would like to mention that the Fubini's theorem does not apply directly since $$\int_{0}^{1} \sum_{k=0}^{\infty} \left| (-1)^{k+1} \frac{(1-x^2)x^{3k}}{\log x} \right| \, dx = \infty.$$ You may need more elaborated technique to justify that step, for instance regularizing the denominator and using a combination of both Fubini's theorem and the dominated convergence theorem. Jul 16, 2017 at 13:00
• Hmmmm,,, now looking back at the question maybe there's a simpler way via Asymptotics Jul 16, 2017 at 13:03
• I also tried a similar approach as @tired. I computed the integral $$I(s, a) := \int_{0}^{\infty} \frac{t^{s-1}e^{-at}}{1+e^{-t}} \, dt = \frac{\Gamma(s)}{2^s} \left[ \zeta\left(s, \frac{a}{2}\right) - \zeta\left(s, \frac{a+1}{2}\right) \right]$$ and then tried to use a functional equation for Hurwitz zeta function to compute the limit $$\int_{0}^{1} \frac{1-x}{(1-x+x^2)\log(1/x)} \, dx = \lim_{s\to 0} [ I(s, \tfrac{1}{3}) - I(s, 1) ].$$ But your approach is much neater and simpler. Jul 16, 2017 at 13:28
• @SangchulLee yeah the limiting procedure becomes quite daunting :( Jul 16, 2017 at 13:29

I was confused by the implicit use of the Frullani integral, so I think it bears mention: \begin{align} \int_0^1\frac{x^n-x^m}{\log(x)}\,\mathrm{d}x &=\int_0^\infty\frac{e^{-mu}-e^{-nu}}{u}e^{-u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-(m+1)u}-e^{-(n+1)u}}{u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_{(m+1)\epsilon}^{(n+1)\epsilon}\frac{e^{-u}}{u}\,\mathrm{d}u\\[3pt] &=\log\left(\frac{n+1}{m+1}\right) \end{align} \begin{align} -\int_0^1\frac{1-x}{\left(1-x+x^2\right)\log(x)}\,\mathrm{d}x &=-\int_0^1\frac{1-x^2}{\left(1+x^3\right)\log(x)}\,\mathrm{d}x\\ &=-\sum_{k=0}^\infty(-1)^k\frac{x^{3k}-x^{3k+2}}{\log(x)}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\log\left(\frac{3k+3}{3k+1}\right)\\ &=\sum_{k=0}^\infty\log\left(\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right)\\ &=\log\left(\prod_{k=0}^\infty\frac{\left(k+\frac12\right)\left(k+\frac23\right)}{\left(k+\frac16\right)(k+1)}\right)\\ &=\lim_{n\to\infty}\log\left(\prod_{k=0}^{n-1}\frac{\color{#C00}{\left(k+\frac12\right)}\color{#090}{\left(k+\frac23\right)}}{\color{#00F}{\left(k+\frac16\right)}(k+1)}\right)\\ &=\log\left(\lim_{n\to\infty}\color{#C00}{\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)}}\color{#090}{\frac{\Gamma\left(n+\frac23\right)}{\Gamma\left(\frac23\right)}}\color{#00F}{\frac{\Gamma\left(\frac16\right)}{\Gamma\left(n+\frac16\right)}}\frac{\Gamma(1)}{\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right)+\log\left(\lim_{n\to\infty}\frac{\Gamma\left(n+\frac12\right)\Gamma\left(n+\frac23\right)}{\Gamma\left(n+\frac16\right)\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right) \end{align} The last step is by Gautschi's Inequality.


Note that $\ds{\left.\vphantom{\Large A}\Psi\pars{z}\right\vert_{\ \Re\pars{z}\ >\ 0} = -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant. See $\ds{\color{#000}{\mathbf{6.3.22}}}$ in A & S Table.

I thought it might be instructive to present an approach that does not rely on Frullani's integral.

To that end, we first note that $\frac{x-1}{\log(x)}=\int_0^1 x^s\,ds$.

Therefore, we can write

\begin{align} \int_0^1 \frac{1+x}{1+x^3}\,\frac{x-1}{\log(x)}\,dx&=\int_0^1 \left(\int_0^1 \frac{x^s+x^{s+1}}{1+x^3} \right)\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n\int_0^1 (x^{s+3n}+x^{s+3n+1})\,dx\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n \left(\frac{1}{s+3n+1}+\frac{1}{s+3n+2}\right)\,ds\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{3n+3}{3n+1}\right)\\\\ &=\sum_{n=0}^\infty \log\left(\frac{(6n+3)(6n+4)}{(6n+1)(6n+6)}\right)\tag 1\\\\ &=\log\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right)\tag2 \end{align}

where in going from $(1)$ to $(2)$ we relied on the analysis posted in Rob's solution herein.

• The last = sign alone is a huge jump to anybody asking this, proposed with no justification.
– Did
Jul 17, 2017 at 16:25
• @Did There were already two posted solutions with a way forward on that huge jump. I thought it redundant to duplicate. Would you suggest that I duplicate those or just reference them? Jul 17, 2017 at 20:00
• The sentence you added, at least makes clear that there is a gap.
– Did
Jul 17, 2017 at 21:11
• @Did I noticed that the accepted answer had the same gap. Jul 17, 2017 at 21:15
• This is true. No idea why the value of the infinite product should be so obvious to any user asking this question.
– Did
Jul 17, 2017 at 21:26

Let

$$I=\int_0^1\frac{1+x}{1+x^3}\frac{x-1}{\ln(x)}dx$$ and

$$I(a)=\int_0^1\frac{1+x}{1+x^3}\frac{x^a-1}{\ln(x)}dx$$

and notice that $$I(1)=I$$ and $$I(0)=0.$$

$$I'(a)=\int_0^1\frac{x^a+x^{a+1}}{1+x^3}dx\overset{x=t^{1/3}}{=}\frac13\int_0^1\frac{t^{\frac{a-2}{3}}+t^{\frac{a-1}{3}}}{1+t}dt$$

By using

$$\int_0^1\frac{x^n}{1+x}dx=\frac12\psi\left(\frac{n+2}{2}\right)-\frac12\psi\left(\frac{n+1}{2}\right),$$

we have

$$I'(a)=\frac16\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)+\psi\left(\frac{a+5}{6}\right)-\psi\left(\frac{a+2}{6}\right)\right].$$

Integrate both sides from $$a=0$$ to $$a=1$$

$$\int_0^1 I'(a)da=I(a)|_0^1=I(1)-I(0)=I-0$$

$$=\int_0^1 \frac16\left[\psi\left(\frac{a+4}{6}\right)-\psi\left(\frac{a+1}{6}\right)+\psi\left(\frac{a+5}{6}\right)-\psi\left(\frac{a+2}{6}\right)\right]da$$

$$=\left[\ln\Gamma\left(\frac{a+4}{6}\right)-\ln\Gamma\left(\frac{a+1}{6}\right)+\ln\Gamma\left(\frac{a+5}{6}\right)-\ln\Gamma\left(\frac{a+2}{6}\right)\right]_0^1$$

$$=\left[\ln\Gamma\left(\frac{5}{6}\right)-\ln\Gamma\left(\frac{1}{3}\right)+\ln\Gamma\left(1\right)-\ln\Gamma\left(\frac{1}{2}\right)\right]$$

$$-\left[\ln\Gamma\left(\frac{2}{3}\right)-\ln\Gamma\left(\frac{1}{6}\right)+\ln\Gamma\left(\frac{5}{6}\right)-\ln\Gamma\left(\frac{1}{3}\right)\right]$$

$$=-\ln\Gamma\left(\frac{1}{2}\right)-\ln\Gamma\left(\frac{2}{3}\right)+\ln\Gamma\left(\frac{1}{6}\right)$$

$$=\ln\left(\frac{\Gamma\left(\frac16\right)}{\sqrt{\pi}\, \Gamma\left(\frac23\right)}\right).$$