Re-doing last post since it was incorrect and corrected by many people on here.

$$ \int_0^1 \frac{\tanh^{-1}(x)\ln x}{x(1-x^2)} \, dx $$

I have tried substitutions since $\tanh^{-1}(x)$ and $1-x^2$ are related by a derivative and both in the integrand. I am unsure how to evaluate this using residues or other methods. Substitutions of the form $y=\ln\tanh(u)$ did not help me at this point either. Thanks. I am not looking for something numerical, as I have done this already.

  • $\begingroup$ Hint: $\frac{d(\ln(\tanh(u)))}{d\tanh(u)}\frac{d\tanh(u)}{du}=\frac{1}{\tanh(u)}$. $\endgroup$ – user122283 Feb 24 '14 at 23:26
  • $\begingroup$ I tried that as I posted...What about the $sech^2$ you get from the derivative...? $\endgroup$ – Jeff Faraci Feb 24 '14 at 23:27
  • $\begingroup$ $$ \int_{0}^{1} \frac{\text{arctanh} (x) \ln x}{x(1-x^{2})} \ dx = \int_{0}^{1} \frac{\frac{1}{2}[\ln(1+x) - \ln(1-x)] \ln x}{x(1-x)(1+x)} \ dx$$ $$ = \frac{1}{2} \int_{0}^{1} \Big( \ln(1+x) \ln x - \ln(1-x) \ln x\Big)\Big(\frac{1}{x} + \frac{1}{2} \frac{1}{1-x} - \frac{1}{2} \frac{1}{1+x} \Big) \ dx $$ So you have 6 integrals to evaluate, and none of them should be overly difficult. You should be able to be evaluate all of them using power series, and a few of them have simple antiderivatives in terms of polylogarithms. $\endgroup$ – Random Variable Feb 25 '14 at 0:26
  • $\begingroup$ @RandomVariable That is exactly what I was looking for, thanks so much. $\endgroup$ – Jeff Faraci Feb 25 '14 at 1:17
  • $\begingroup$ @RandomVariable How did you do the integral $$ \int_0^1 \frac{\ln(x+1)\ln(x)}{x+1}dx=-\frac{\zeta(3)}{8}? $$ I managed to work out 2 of the others though. I am not sure how to use the power series approach in this method. Thanks a lot $\endgroup$ – Jeff Faraci Feb 25 '14 at 4:22

This is an answer to Jeff Faraci's question in the comments about showing that

$$\int_{0}^{1} \frac{\ln(x) \ln(1+x)}{1+x} \, dx = - \frac{\zeta(3)}{8} $$ using a power series approach.

This is one of those six integrals that I mentioned in the comments above, a couple of which may be more difficult to evaluate than it appeared at first glance.

The ordinary generating function of the harmonic numbers is $$\sum_{n=1}^{\infty} H_{n} x^{n} = -\frac{\ln(1-x)}{1-x}, \quad |x|<1.$$


$$ \begin{align} \int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1+x} \, dx &= -\int_{0}^{1} \ln x \sum_{n=1}^{\infty} H_{n} (-x)^{n} \, dx \\ &= -\sum_{n=1}^{\infty} (-1)^{n} H_{n} \int_{0}^{1} x^{n} \ln (x) \, dx \\ &= \sum_{n=1}^{\infty} (-1)^{n} \frac{H_{n}}{(n+1)^{2}} \\ &= \sum_{n=1}^{\infty} (-1)^n \frac{H_{n+1}-\frac{1}{n+1}}{(n+1)^{2}} \\ &= \sum_{n=1}^{\infty} (-1)^{n}\frac{H_{n+1}}{(n+1)^{2}} - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(n+1)^{3}} \\ &= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n^{2}} - 1-\Big(\frac{3 \zeta(3)}{4} -1 \Big) \\ &=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n^{2}} - \frac{3 \zeta(3)}{4} \end{align}$$

See robjohn's answer here for a way to show that $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n^{2}} = \frac{5}{8} \zeta(3). $$

Alternatively, you could use the generating function for $\frac{H_{n}}{n^{2}}$ that was derived in this answer.

I removed the link to the site where I evaluated $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{n^{2k}}$ using contour integration since MathJax no longer seems to work on that site.

But the general approach is described in the paper by Flajolet and Salvy that I linked to in the comments below.

  • $\begingroup$ Very nice...The link to the contour integration sure seems like you put a lot of time into it! Thanks a lot $\endgroup$ – Jeff Faraci Feb 25 '14 at 8:43
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    $\begingroup$ An overview of evaluating harmonic/Euler sums using contour integration is given in the following paper: projecteuclid.org/euclid.em/1047674270 $\endgroup$ – Random Variable Feb 25 '14 at 8:51
  • $\begingroup$ Thanks again for this one! I worked through all of this and it makes sense. Now I will check out that paper on "experimental mathematics" and the alternate evaluation of the sum. Thanks! $\endgroup$ – Jeff Faraci Feb 25 '14 at 21:20

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