Question

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

Answer 1

Substitute
`x\mapsto√(1-x^2)`, which transforms the integral to

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

and factoring
`√(1-x^2)=√((1-x)(1+x))` reduces this to

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

The inner logarithms differ only by a sign, so that

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

where I assume k is an integer.