1. ## Difficult definite integrals.

I'm trying to get the exact value of following definite integrals without using a CAS or tables. Any help would be appreciated:

$\displaystyle \int_0^1 \frac{\ln x}{x^2 - 1} \, dx$

$\displaystyle \int_0^{+\infty} \frac{\ln x}{x^2 - 1} \, dx$

I can get answers using the Wolframalpha website and also from a table of integrals. But I'd really like a calculus approach. I can't see what contour to use using contour integration. I can't see how to 'differentiate under the integral'. I can't see any substitution that will help. I know they can be done otherwise there would be no answer in the tables. I'm stuck.

I've seen a few members who are absolutely amazing in the sorts of integrals they have solved. Hopefully someone can help me with these ones.

Thanks.

2. With a little of patience You can arrive to the general expression...

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

... and from (1) and the identity...

$\displaystyle \frac{1}{x^{2}-1} = -1 -x^{2} - x^{4} - ... - x^{2n} - ...$ (2)

... You have finally...

$\displaystyle \int_{0}^{1} \frac{\ln x}{x^{2} -1}\ dx = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = \frac{\pi^{2}}{8}$ (3)

Following a similar procedure You can solve also the second integral...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
With a little of patience You can arrive to the general expression...

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

... and from (1) and the identity...

$\displaystyle \frac{1}{x^{2}-1} = -1 -x^{2} - x^{4} - ... - x^{2n} - ...$ (2)

... You have finally...

$\displaystyle \int_{0}^{1} \frac{\ln x}{x^{2} -1}\ dx = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = \frac{\pi^{2}}{8}$ (3)

Following a similar procedure You can solve also the second integral...

Kind regards

$\chi$ $\sigma$
I would just like to add the following for the OP with regards to applying this procedure to the second integral:

It cannot be used directly because the series for $\frac{1}{x^2 - 1}$ only converges for |x| < 1.

So I suggest that as part of your attempt at a solution you consider $\int_1^{+\infty} \frac{\ln x}{x ^2 - 1} \, dx$ and make the substitution $x = \frac{1}{t}$.