Question

$\displaystyle \rm\int_{0}^1 { ln }^{2k} \left \lgroup \frac{ ln \left \lgroup \dfrac{1 - \sqrt{1 - {x}^{2} } }{x} \right \rgroup }{ ln \left \lgroup \dfrac{1 + \sqrt{1 - {x}^{2} } }{x} \right \rgroup } \right \rgroup \: dx$​​

Answer 1

Substitute $x\mapsto\sqrt{1-x^2}$, which transforms the integral to

$\displaystyle \int_0^1 \ln^{2k} \left(\frac{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}\right) \, dx = \int_0^1 \ln^{2k}\left(\frac{\ln\left(\frac{1-x}{\sqrt{1-x^2}}\right)}{\ln\left(\frac{1+x}{\sqrt{1-x^2}}\right)}\right) \frac{x}{\sqrt{1-x^2}} \, dx$

and factoring $\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}$ reduces this to

$\displaystyle = \int_0^1 \ln^{2k}\left(\frac{\ln\left(\sqrt{\frac{1-x}{1+x}}\right)}{\ln\left(\sqrt{\frac{1+x}{1-x}}\right)}\right) \frac x{\sqrt{1-x^2}} \, dx$

The inner logarithms differ only by a sign, so that

$\displaystyle = \int_0^1 \ln^{2k}(-1) \frac x{\sqrt{1-x^2}} \, dx$

Using the principal branch of the complex logarithm, we have

$\ln(-1) = \ln|-1| + i\arg(-1) = i\pi$

and hence

$\displaystyle \int_0^1 \ln^{2k} \left(\frac{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}{\ln\left(\frac{1-\sqrt{1-x^2}}x\right)}\right) \, dx = (i\pi)^{2k} \underbrace{\int_0^1 \frac x{\sqrt{1-x^2}} \, dx}_{=1} = \boxed{(-\pi^2)^k}$

where I assume k is an integer.