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Other than being generally interested in this integral, it also appears in my $\zeta$ approach to my question here.

Is it possible to evaluate the following integral? I was thinking to perhaps use an infinite product approach for $\sec$, however, I couldn’t really get far with it. $$\int_{0}^{1} \ln \left( \sec \left( \frac{\pi x}{2} \right) \right) \ln (\sin (\pi x)) \, dx$$

If this integral is possible, is it also possible to evaluate the following two other integrals? $$\int_{0}^{1} \ln \left( \sec \left( \frac{\pi x}{2} \right) \right)^2 \ln (\sin (\pi x)) \, dx$$ $$\int_{0}^{1} x \ln \left( \sec \left( \frac{\pi x}{2} \right) \right) \ln (\sin (\pi x)) \, dx$$

KStarGamer
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1 Answers1

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Substitute $t= \frac{\pi x}{2} $ to get $$-\frac{2}{\pi} \int_0^{\frac{\pi}{2}}\ln(\cos t) \ln(\sin 2t) \ dt \\ = -\frac{2\ln 2}{\pi} \int_0^{\frac{\pi}{2}} \ln(\cos t) \ dt -\frac{2}{\pi}\int_0^{\frac{\pi}{2}} \ln(\sin t)\ln(\cos t) \ dt-\frac{2}{\pi} \int_0^{\frac{\pi}{2}} (\ln(\cos t))^2 \ dt$$ Now, using the results for

$$\int_0^{\frac{\pi}{2}} \ln(\cos t) \ dt$$

$$ \int_0^{\frac{\pi}{2}} \ln(\sin t) \ln(\cos t) \ dt $$

$$ \int_0^{\pi} (\ln(\sin t))^2 \ dt $$ (as $\int_0^{\frac{\pi}{2}} (\ln(\cos t))^2 \ dt = \frac 12 \int_0^{\pi} (\ln(\sin t))^2 \ dt $), the original integral evaluates to $$-\frac{\pi^2}{24} -(\ln 2)^2 $$

Vishu
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