I met this integral

$$ \int_{0}^{1} \left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} \qquad (*) $$

while evaluating this log-cosine integral. I made several attemps before being successful.

Find a closed form for

$$ I(s): = \int_{0}^{1} \left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{x^s }{1-x}\mathrm{d}x, \quad \Re (s)>-1. \qquad (**) $$

  • $\begingroup$ The first thing I would do is to consider $I'(s)$ instead of $I(s)$ in order to remove $\log x$ from the denominator of the first term of the integral, and deal with a more treatable integral. $\endgroup$ – Jack D'Aurizio Jul 21 '14 at 21:22

As stated in the comments, by following the Feynman's approach to integrals we have: $$\begin{eqnarray*} I'(s) &=& \int_{0}^{1}\left(1+\frac{\log x}{1-x}-\frac{\log x}{2}\right)\frac{x^s}{1-x}dx\\ &=& \frac{1}{2}\zeta(2,s+1)+\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{1-x}dx\\&=&\frac{1}{2}\zeta(2,s+1)-\frac{1}{2s+2}\phantom{}_3 F_2\left(1,1,2;3,s+2;1\right),\end{eqnarray*}$$ but probably there is a nicer form for the second term.

As a matter of fact, by treating the second term this way: $$\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{1-x}dx=\lim_{a\to 0^+}\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{(1-x)^{1-a}}$$ we get: $$ I'(s)=\frac{1}{2}\psi'(s+1)+s\,\psi'(s)+\psi(s)-\psi(s+1)-1\tag{1}$$ where $\psi$ is the digamma function $\psi(s)=\frac{\Gamma'(s)}{\Gamma(s)}$. By integrating $(1)$ we have: $$ I(s) = I(1)+\frac{3\gamma+1}{2}-s+\frac{\psi(s+1)}{2}-\log\Gamma(s+1)+s\,\psi(s).\tag{2}$$

All we need to find a closed expression for $I(s)$ is now to evaluate $I(1)$ or $I(0)$, since by $(2)$: $$ I(0)=I(1)+\gamma-\frac{1}{2}$$ holds. In virtue of $(2)$ we also have: $$\lim_{s\to +\infty} I(s) = I(1) + \frac{3\gamma-\log(2\pi)}{2},$$ but since $$ f(x)=\left(\frac{1}{\log x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{x}{1-x}$$ is a continuous, positive, increasing and bounded function ($f(x)\leq\frac{1}{12}$) on $(0,1)$, as long as $s\to+\infty$ we have $I(s)\leq\frac{1}{12s}$, so $\lim_{s\to +\infty}I(s)=0$. Hence we have:

$$ I(1) = \frac{\log(2\pi)-3\gamma}{2},\qquad I(0) = \frac{\log(2\pi)-\gamma-1}{2} \tag{3}$$


$$ I(s) = \frac{1+\log(2\pi)}{2}-s+\frac{\psi(s+1)}{2}-\log\Gamma(s+1)+s\,\psi(s).\tag{4}$$

As a bonus, by expanding $(4)$ in a neighbourhood of $+\infty$ and exploiting convexity, we can also show that: $$\frac{1}{12s+8}\leq I(s)\leq\frac{1}{12s+6}$$ holds for any $s\in\mathbb{R}^+$.

  • $\begingroup$ Thanks for your interest! Please, there might be a mistake, you have $I(s) \sim -\ln s$ as $s$ tends to $0^+$ in (2). $\endgroup$ – Olivier Oloa Jul 21 '14 at 23:18
  • $\begingroup$ You also have two extra terms in your line (1), namely:$\psi(s)$ and $-\psi(s+1)$, the rest is correct. $\endgroup$ – Olivier Oloa Jul 22 '14 at 0:45
  • $\begingroup$ @OlivierOloa: yes, there were two mistakes. The first one was a forgotten half when bringing back the Hurwitz zeta into its $\psi'$ form, the second one was a typo for which a $\psi'(s)$ became a $\psi'(s+1)$. Now the formula is right, I checked it numerically, too. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 0:46
  • $\begingroup$ The term $\psi(s)-\psi(s+1)=-\frac{1}{s}$ is right. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 0:48
  • $\begingroup$ OK. So what is the value of I:=I(0)? $(*)$ $\endgroup$ – Olivier Oloa Jul 22 '14 at 0:49

Recall Binet's formula $$ \log \Gamma(z)= \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) + \int_0^{\infty} \! \left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{e^{-zx}}{x} \mathrm{d}x,\quad \Re z >0 $$ which, upon making $x = - \log v$, can be written as

$$ \log \Gamma(z)= \! \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) - \!\! \int_0^{1} \! \left(\frac{1}{\log v}+\frac{1}{1-v}-\frac{1}{2}\right)\frac{v^{z-1}}{\log v}\mathrm{d}v, \, \Re z >0 $$

and Gauss' formula

$$ -\psi(z)+\log z = \int_0^{1} \left(\frac{1}{\log v}+\frac{1}{1-v}\right)v^{z-1} \mathrm{d}v. $$

One may check that \begin{multline} \displaystyle \left(\frac{1}{\log u} + \frac{1}{1-u} - \frac{1}{2} \right) \frac{u^s}{1-u} = u^{s} \frac{d}{du}\left\{u \left(\frac{1}{\log u} + \frac{1}{1-u}\right)\right\} \, \\ + \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u} - \frac{1}{2} \right) \frac{u^s}{\log u} - \frac{1}{2} \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^s. \end{multline}

Hence our integral $I(s)$ is the sum of three integrals.

The first, \begin{multline} I_{1}(s) =\displaystyle \left.\left(\frac{1}{\log u} + \frac{1}{1-u} \right)u^{s+1} \right|_{0}^{1} - s \int_{0}^{1}\! \displaystyle \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^{s} \mathrm{d}u, \end{multline} using Gauss' formula, $$ I_{1}(s) = \frac{1}{2} + s \left( \psi(s+1)-\log \left(s+1\right) \right). $$ The second, applying Binet's formula, \begin{multline} I_{2} (s)= \displaystyle \int_{0}^{1}\! \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u} -\frac{1}{2} \right) \frac{u^{s}}{\log u} \mathrm{d}u \\ = -\log \Gamma(s+ 1)+ \left(s + 1/2 \right)\log \left(s+ 1 \right) - s - 1 + \frac{1}{2}\ln(2\pi). \end{multline} The third, using Gauss' formula once more \begin{equation} I_{3}(s) = - \frac{1}{2} \displaystyle \int_{0}^{1}\! \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^{s} \mathrm{d}u = \frac{1}{2}\left( \psi(s+1)-\log \left(s+1\right) \right). \end{equation} Consequently, $ I(s)=I_{1}(s)+I_{2}(s)+I_{3}(s)$ is given by

\begin{multline} I(s):= \displaystyle \int_0^{1} \left(\frac{1}{\log u}+ \frac{1}{1-u}-\frac{1}{2}\right) \frac{u^s}{1-u} \mathrm{d}u \\ = -\log \Gamma(s+1) + \left( s +\frac{1}{2}\right) \psi(s+1) - s - \frac{1}{2} + \frac{1}{2}\ln(2\pi). \end{multline}

We have, with $s=0$,

$$ \int_{0}^{1} \left(\frac{1}{\log x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} = -\frac{1}{2}+\frac{1}{2} \ln (2 \pi)-\frac{1}{2} \gamma. $$


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