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I am encountering an integral which involves logarithms, in particular,

\begin{equation} \int_{a}^b \cfrac{1}{(1+x) \, \left[\ln(1-x)-\ln(1+x)\right]} \, \mathrm{d}x, \end{equation} where $a$ and $b$ are finite real numbers.

Does this integral have an closed form solution ? It seems that integration by parts does not work...

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  • $\begingroup$ Perhaps you can expand as a Taylor series? $\endgroup$
    – Empy2
    Jul 12, 2013 at 9:05
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    $\begingroup$ If you substitute $v=\log\frac{1+x}{1-x}$, I think it becomes $\int \frac{dv}{v(1+e^v)}$. That doesn't look any easier, but it might be familiar to someone. $\endgroup$
    – Empy2
    Jul 12, 2013 at 10:09
  • $\begingroup$ It's doubtful this has an antiderivative. As @Michael points out, you can do some substitution to put it in simpler form, and then you can Taylor expand to obtain a series of exponential integrals. Beyond that, pffft. $\endgroup$
    – Ron Gordon
    Jul 12, 2013 at 15:35

2 Answers 2

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As it has already been mentioned in the comment section, the simple substitution $t=\ln\dfrac{1+x}{1-x}$

leads to an integral of the form $\displaystyle\int\frac{dt}{t~(e^t+1)},~$ which does not possess a closed form expression

in terms of elementary functions and constants: see Liouville's theorem and the Risch algorithm for more details. However, its definite counterpart, when evaluated over the entire positive real

semiaxis, yields $\displaystyle\int_0^\infty\frac{t^n}{e^t+1}~=~n!~\eta(n+1)~$ for $n>-1,$ and $~\displaystyle\int_0^\infty\frac{t^n}{e^t-1}=n!~\zeta(n+1)~$ for

$n>0:~$ see the Riemann $\zeta$ and Dirichlet $\eta$ functions for more information. Unfortunately, in our case $n=-1$, which means that the integral diverges, due to the fact that the $\Gamma$ function has a pole at the origin.

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Hint

With $\ds{x = {1 - t \over 1 + t}}$:

\begin{align} \int_{a}^{b}{1 \over \pars{1 + x}\bracks{\ln\pars{1 - x} - \ln\pars{1 + x}}} \,\dd x = -\int_{\pars{1- a}/\pars{1 + a}}^{\pars{1 - b}/\pars{1 + b}} {\dd t \over \pars{1 + t}\ln\pars{t}} \end{align}

A further procedure requires the particular values of $\ds{a}$ and $\ds{b}$. A useful expression is $\ds{\left.\int_{0}^{\infty}p^{q}\,\dd q\,\right\vert_{\ p\ \in\ \pars{0,1}} = -\,{1 \over \ln\pars{p}}}$.

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