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I was observing this question and was able to conjecture $$\int_{0}^{1}\ln^{2n}\left(\ln\left({1-\sqrt{1-x^2}\over x}\right)\over \ln\left({1+\sqrt{1-x^2}\over x}\right)\right)\mathrm dx=(-\pi^2)^n\tag1$$ Where $n\ge1$.

Making an attempt: $$x=\sin u \implies dx=\cos udu$$ $$\int_{0}^{\pi/2}\ln^{2n}\left(\ln\left({1-\cos u\over \sin u}\right)\over \ln\left({1+\cos u\over \sin u}\right)\right)\cos u\,du\tag2$$ simplify further to $$\int_{0}^{\pi/2}\ln^{2n}\left(\ln\tan\left({u\over 2}\right)\over \ln\cot\left({u\over 2}\right)\right)\cos u\,du\tag3$$ $$\int_{0}^{\pi/2}\ln^{2n}(-1)\cos u\,du\tag4$$ This is not making any sense here!

How can we prove $(1)?$

Parcly Taxel
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gymbvghjkgkjkhgfkl
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  • Why is that not making any sense? – DHMO Apr 19 '17 at 13:23
  • $ln(-1)$, this is not possible – gymbvghjkgkjkhgfkl Apr 19 '17 at 13:24
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    You conjectured $(1)$? How? – DHMO Apr 19 '17 at 13:25
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    Since $$ \frac{x}{1-\sqrt{1-x^2}} = \frac{x}{1-\sqrt{1-x^2}} \frac{1+\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = \frac{x(1+\sqrt{1-x^2})}{x^2} = \frac{1+\sqrt{1-x^2}}{x} $$, the argument of the log is always negative. Moreover, it's constant, since it's of the form $\log{u}/(\log{1/u}) = -1$. You can see this even without substituting. – Chappers Apr 19 '17 at 13:54

1 Answers1

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Since the principal value of $\ln(-1)$ is $i\pi$: $$\int_0^{\pi/2}\ln^{2n}(-1)\cos u\,du$$ $$=\int_0^{\pi/2}(i\pi)^{2n}\cos u\,du$$ $$=(i\pi)^{2n}\int_0^{\pi/2}\cos u\,du$$ $$=(i\pi)^{2n}=(-\pi^2)^n$$ However, as the complex logarithm is multivalued, this answer is not well-defined (as mentioned by Chappers in the comments).

Parcly Taxel
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  • $ln(-1)=i\pi$ that is very new to me, thank you so much. – gymbvghjkgkjkhgfkl Apr 19 '17 at 13:37
  • @Latte' This relies on Euler's identity $e^{i\pi}=-1$. – Parcly Taxel Apr 19 '17 at 13:37
  • The obvious is not so obvious after all! – gymbvghjkgkjkhgfkl Apr 19 '17 at 13:38
  • The integrand isn't real. The argument of log of the numerator is negative. – FDP Apr 19 '17 at 13:45
  • It's worth pointing out that the logarithm of $-1$ can equally well be chosen as $-i\pi$. Or $(2k+1)i\pi$ for any integer $k$, since $e^{2k\pi i}=1$. For numbers that are not positive reals, there is no canonical choice of logarithm without imposing some sort of continuity condition, and one cannot impose such a condition over the whole complex plane, even if we exclude zero. – Chappers Apr 19 '17 at 13:54
  • @Chappers I used principal value here. Would I have to make any further steps to justify the integral and this choice? – Parcly Taxel Apr 19 '17 at 13:57
  • Since the argument's the constant $-1$, there isn't any way to tell what the OP was intending (especially since they are not familiar with the logarithm of negative numbers; I suspect computer algebra software was involved in the evaluation somewhere...). But in any case, I do think it's also important to point out why the answer is not really well-defined in addition to explaining where the $-\pi^2$ comes from. – Chappers Apr 19 '17 at 14:12
  • Wolfram alpha always assumes that $\log$ is the principal logarithm. When dealing with real integrals we always assume the principal logarithm. – Zaid Alyafeai Apr 19 '17 at 15:42