a couple more definite integrals

**Random Variable** · March 29th 2011, 05:54 PM

$\displaystyle \int_{0}^{1} \ln(1+x) \ln(1-x) \ dx$

If the above integral is too easy, also try to evaluate $\displaystyle \int_{0}^{1} \ln(1+x) \ln (x) \ln(1-x) \ dx$ (though I personally wouldn't recommend it

).

**chisigma** · March 31st 2011, 07:08 AM

With the substitution $t=1-x$ the integral becomes...

$\displaystyle I= \int_{0}^{1} \ln t\ \ln (2-t)\ dt$ (1)

Now is...

$\displaystyle \ln (2-t)= \ln 2 - \sum_{n=1}^{\infty} \frac{t^{n}}{n\ 2^{n}}$ (2)

... and remembering the general formula...

$\displaystyle \int_{0}^{1} t^{n}\ \ln t\ dt = -\frac{1}{(n+1)^{2}}$ (3)

... You obtain...

$\displaystyle I= -\ln 2 + \sum_{n=1}^{\infty} \frac{1}{n\ (n+1)^{2}\ 2^{n}}$ (4)

To extablish if the sum of the series in (4) can be expressed as function of known constants it is necessary some more work...

Kind regards

$\chi$ $\sigma$

**Random Variable** · March 31st 2011, 08:44 AM

If you make a slightly different substitution, you'll end up having to evaluate a much simpler series.

$\displaystyle \int^{1}_{0} \ln (1+x) \ln(1-x) \ dx = \frac{1}{2} \int^{1}_{-1} \ln (1+x) \ln(1-x) \ dx$

Now let $\displaystyle 2t = 1+x$

$\displaystyle = \int^{1}_{0} \ln (2t) \ln(2-2t) \ dt = \int^{1}_{0} \big(\ln 2 + \ln t\big) \big(\ln 2 + \ln(1-t) \big) \ dt$

$\displaystyle = \ln^{2}(2) \int ^{1}_{0} \ dt + \ln 2 \int^{1}_{0} \ln(1-t) \ dt + \ln 2 \int^{1}_{0} \ln t \ dt + \int^{1}_{0} \ln t \ln(1-t) \ dt$

The first three integrals are straightforward, and the last one is equivalent to evaluating $\displaystyle \sum^{\infty}_{n =1} \frac{1}{n(n+1)^2}$ which is much easier to evaluate than $\displaystyle \sum^{\infty}_{n =1} \frac{1}{n(n+1)^{2}2^{n}}$ .

The final answer is $\displaystyle \ln^{2}(2) - 2 \ln 2 + 2 - \frac{\pi^{2}}{6}$

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