Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What technique we should use for this integral: $$\int_0^1\frac{\ln x\ln (1-x^2)}{1-x^2}\text{d}x$$ Can anyone give a brief way to evaluate this?

share|cite|improve this question
A brief way? Do you have a lengthy way? – Gerry Myerson Mar 2 '13 at 11:40
$\frac{1}{4} \left(-\pi ^2 \text{Log}[2]+7 \text{Zeta}[3]\right)$ according to Wolfram!! – Santosh Linkha Mar 2 '13 at 11:43
Where is this integral from? Is there any context? – Chris Mar 2 '13 at 12:03
A related problem. – Mhenni Benghorbal Mar 2 '13 at 13:50
Another related problem. – Mhenni Benghorbal Mar 2 '13 at 13:59
up vote 8 down vote accepted

One possible, although not very short, way is to reduce the integral to derivatives of the beta integral: $$\begin{eqnarray} \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x &\stackrel{u=x^2}{=}& \frac{1}{4} \int_0^1 \frac{\log(u)}{\sqrt{u}} \frac{\log(1-u)}{1-u} \mathrm{d}u \\ &=& \frac{1}{4} \lim_{\alpha\to \frac{1}{2}} \lim_{\beta \downarrow 0^+} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}}{\mathrm{d} \beta} \int_0^1 u^{\alpha-1} (1-u)^{\beta-1}\mathrm{d}u \\ &=& \frac{1}{4} \lim_{\alpha\to \frac{1}{2}} \lim_{\beta \downarrow 0^+} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}}{\mathrm{d} \beta} \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)} \\ \\ &=& \frac{1}{4} \lim_{\alpha\to \frac{1}{2}} \lim_{\beta \downarrow 0^+} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}}{\mathrm{d} \beta} \frac{\Gamma(\alpha) \Gamma(\beta+1)}{\beta \, \Gamma(\alpha+\beta)} \end{eqnarray} $$ Now we would do a series expansion around $\beta=0$: $$ \frac{\Gamma(\alpha) \Gamma(\beta+1)}{\beta \, \Gamma(\alpha+\beta)} = \frac{1}{\beta} + \left(-\gamma - \psi(a)\right) + \frac{\beta}{2} \left( \frac{\pi^2}{6} + \gamma^2+2 \gamma \psi(a) + \psi(a)^2 -\psi^{(1)}(a) \right) + \mathcal{o}(\beta) $$ Differentation with respect to $\alpha$ kills the singular term, one we find: $$ \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x = \frac{1}{4} \lim_{\alpha \to \frac{1}{2}} \left( \gamma \psi^{(1)}(a) + \psi(a) \psi^{(1)}(a) - \frac{1}{2} \psi^{(2)}(a) \right) $$ Now, using $\psi(1/2) = -\gamma - 2 \log(2)$, $\psi^{(1)}(1/2) = \frac{\pi^2}{2}$ and $\psi^{(2)}(1/2) = -14 \zeta(3)$ we finally arrive at the answer: $$ \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x = \frac{1}{4} \left( 7 \zeta(3) - \pi^2 \log(2) \right) $$

Notice also that $$ \frac{\log(1-x^2)}{1-x^2} = -\sum_{n=1}^\infty H_n x^{2n} $$ This leads, together with $\int_0^1 x^{2n} \log(x) \mathrm{d}x = -\frac{1}{(2n+1)^2}$: $$ \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x = \sum_{n=1}^\infty \frac{H_n}{(2n+1)^2} $$ Sum of these types have been discussed before.

Edit: In fact exactly this sum was asked about by the OP, as noted by @MhenniBenghorbal in the comments, and @joriki provided an excellent answer.

share|cite|improve this answer
Thank you Sasha! Is there special meaning for the downarrow? – Ryan Mar 3 '13 at 1:50
@Ryan The downarrow stands here for the limit from above. $lim_{\beta\downarrow 0} f(\beta)$ is also sometimes written as $\lim_{\beta \to 0^+} f(\beta)$. – Sasha Mar 3 '13 at 1:52

Not sure about an integration method by hand, but here's a Monte Carlo calculation of your integral in Matlab:

for j=1:m
    z(j) = i;
integral = mean(z)
variance = var(z)

integral = 0.3932

variance = 1.5804e-04

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.