Tough Integrals to Crack

Calculus Level 5

2 ∫ 0 π 6 ln ⁡ 2 ( 2 sin ⁡ x ) d x ∫ π 6 π 2 ln ⁡ 2 ( 2 sin ⁡ x ) d x = ? \large \dfrac{\displaystyle2\int_0^{\frac{\pi}{6}} \ln^2\left(2\sin{x}\right)\,dx}{\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \ln^2\left(2\sin{x}\right)\,dx}\,=\,?


The answer is 7.

Digvijay Singh by Digvijay Singh , 21, India

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1 solution

Mark Hennings
Nov 1, 2018

We note that ∫ 0 1 6 π ln ⁡ 2 ( 2 sin ⁡ x ) d x = − 1 2 L s 3 ( 1 3 π ) ∫ 0 1 2 π ln ⁡ 2 ( 2 sin ⁡ x ) d x = − 1 2 L s 3 ( π ) \int_0^{\frac16\pi}\ln^2(2\sin x)\,dx \; = \; -\tfrac12\mathrm{Ls}_3(\tfrac13\pi) \hspace{2cm} \int_0^{\frac12\pi}\ln^2(2\sin x)\,dx \; = \; -\tfrac12\mathrm{Ls}_3(\pi) where L s n \mathrm{Ls}_n is the log-sine integral L s n ( τ ) = − ∫ 0 τ ln ⁡ n − 1 ( 2 sin ⁡ 1 2 x ) d x \mathrm{Ls}_n(\tau) \; = \; -\int_0^\tau \ln^{n-1}\big(2\sin\tfrac12x\big)\,dx This paper shows, amongst much else, that L s 3 ( π ) = − 1 12 π 3 L s 3 ( 1 3 π ) = − 7 108 π 3 \mathrm{Ls}_3(\pi) \; = \; -\tfrac{1}{12}\pi^3 \hspace{2cm} \mathrm{Ls}_3(\tfrac13\pi) = -\tfrac{7}{108}\pi^3 which makes the desired ratio of integrals equal so 2 L s 3 ( 1 3 π ) L s 3 ( π ) − L s 3 ( 1 3 π ) = 2 × 7 108 1 12 − 7 108 = 7 \frac{2\mathrm{Ls}_3(\frac13\pi)}{\mathrm{Ls}_3(\pi) - \mathrm{Ls}_3(\frac13\pi)} \; = \; \frac{2 \times \frac{7}{108}}{\frac{1}{12} - \frac{7}{108}} \; = \; \boxed{7}

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