Integrate $\int_{-1}^{1} \frac{\log(1-x^2)}{(1-\beta x)^2}dx$

Question

I would like to integrate the following:

$$\int_{-1}^{1} dx \frac{\log(1-x^2)}{(1-\beta x)^2}$$

where $\beta$ is real and greater than zero. While it looks like this diverges I have a reference which has a closed finite form.

How do I integrate this?

Answer 1

I suppose that the problem with Mathematica came from the fact that you computed first the antiderivative $$I(x)=\int \frac{\log \left(1-x^2\right)}{(1-\beta x)^2}\,dx$$ which gives $$ I(x)=\frac1\beta\left(\frac{\log \left(1-x^2\right)}{1-\beta x}-\frac{2 \log (1-\beta x)}{\beta ^2-1}+\frac{\log (1-x)}{\beta -1}-\frac{\log (x+1)}{\beta +1}\right)$$ and that you tried to compute $I(1)$ and $I(-1)$ which are undefined.

Asking directly for the definite integral would have given the result.

Starting from the antiderivative, you could have computed $I(1-\epsilon)-I(-1+\epsilon)$ and developing the result as a Taylor series built around $\epsilon=0$ would give $$\frac{4}{\beta(1 -\beta ^2) }\left(\beta \log (2)- \tanh ^{-1}(\beta )\right)-\frac{2 \left(\beta ^2+1\right)}{\left(\beta ^2-1\right)^2}\epsilon \log \left(\frac{2 \epsilon }{e}\right)+O\left(\epsilon ^2\right)$$ and then the limit $$\int_{-1}^{+1} \frac{\log \left(1-x^2\right)}{(1-\beta x)^2}\,dx=\frac{4}{\beta(1 -\beta ^2) }\left(\beta \log (2)- \tanh ^{-1}(\beta )\right)$$

Answer 2

Let $x\to -x$ over $(-1,0)$

\begin{align} \int_{-1}^1\frac{\ln(1-x^2)}{(1-\beta x)^2}dx = & \ 2\int_0^1\ln(1-x^2) \frac{1+\beta^2x^2}{(1-\beta^2x^2)^2} dx\\ = & \ 2\int_0^1 \ln(1-x^2)\ d\left( \frac{x}{1-\beta^2x^2} - \frac{1}{1-\beta^2}\right)\\ \overset{ibp}= & \ \frac4{1-\beta^2}\int_0^1\frac1{1+x}-\frac1{1-\beta^2x^2}\ dx\\ =&\ \frac4{1-\beta^2}\left(\ln2-\frac1{2\beta}\ln \frac{1+\beta}{1-\beta}\right) \end{align}