Integrate $\int_{0}^{\infty} \frac{r^2\log(r^2 - 1)}{(r^2 - 1)^2} \, dr$

Question

How do I calculate:

$$\int_{0}^{\infty} \frac{r^2 \log(r^2 - 1)}{(r^2 - 1)^2} \, dr$$

Basically, I have been attempting this integration for a while using multiple methods so any hints or suggestions would be appreciated. Thanks!

Answer 1

Your function is not locally integrable at $(0,+\infty) $. it should be $\int_1... $.

if it is the case, use by parts integration to transform $\ln $ to a fraction.

Answer 2

$$\int_{0}^{\infty} \frac{r^2 \log(r^2 - 1)}{(r^2 - 1)^2} dr$$ take $\log(r^2 - 1)$ =$t$ then $(r^2 - 1)$=$e^t$ and $r^2=e^t+1$ also $\frac{2r}{(r^2-1)}dr$=$dt$

Now, $$\int \frac{\sqrt(e^t+1)t}{2e^t} dt$$ can you carry on from here?

Answer 3

There are a couple of headaches involved with this integral. The worst is that for $r<1$ the integrand has an imaginary part which is $\pm i\pi$ depending on choiice of ranch for the log, and the integral from $0$ to $1$ of the imaginary part looks like i\pi$\int_1^\infty (1-s^2)^-2ds$(after sustituting $r=1/s$, and this does not converge.

the other headache is that neither of the real part integrals converge, and even the sum of the integrand obtained from the substitution with the original integrand (which is tempting because the limits of integration will both be $1$ to $\infty$ does not converge.

Are you sure you have the problem right?