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I want to show that $$ \int_0^1 \frac{x}{x^2+1}\ln\left(\cos^2(\ln x)\right) dx= -\frac{\ln^2 2}{2}$$

I tried unsuccesfully to prove this result, but everything I tried failed. Integration by parts and substitution fail in most of the cases, since the integrand blows up to $-\infty$ infinitely many times, so the resulting integrals doesn't converge.

By substituting $x=e^{-t}$ one gets to prove that $$ \int_0^\infty \frac{e^{-t}}{\cosh t}\ln\left(\cos^2 t\right) dt=-\ln^2 2$$ which converges much quicker then the previous one, but has the same discontinuity problems.

How to tackle this problem?

Zima
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  • How do you know the value of that integral? – Jochen Jan 28 '24 at 09:50
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    If you cannot find antiderivative, maybe try contour integration in the complex plane. – GEdgar Jan 28 '24 at 09:55
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    @Jochen I "milked" other known integrals, in the sense explained in this question, but when I tried to evaluate it without knowing the tricks used in its construction I failed, and I'm looking for a way to do it. – Zima Jan 28 '24 at 09:57
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    Since $\cos^2(t)=(1+\cos(2t))/2$ and $\int_0^{\infty}\frac{-\ln2}{1+e^{2t}},\mathrm{d}t=-\ln(2)^2/2$ it suffices to show $\int_0^\infty\frac{\ln(1+\cos(2t))}{1+e^{2t}},\mathrm{d}t=0$ – FShrike Jan 28 '24 at 11:25
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    Two methods of solution (including complex integration) - https://math.stackexchange.com/questions/4126627/how-to-calculate-the-integral-int-0-infty-frac-ln-cos2x1e2xdx/4436875#4436875 – Svyatoslav Jan 28 '24 at 15:52

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