An intriguing relationship between pairs of integrals

Question

For $n\in\mathbb N$, define $$I_n:=\displaystyle\int_0^{\pi/2}\ln^n(\sin x)\ln^n(\cos x)\ \mathrm dx,\\ J_n:=\displaystyle\int_0^{\pi/2}x\ln^n(\sin x)\ln^n(\cos x)\ \mathrm dx.$$ Then it appears numerically that $J_n=\dfrac\pi4I_n$ for all $n\in\mathbb N$.

Is there a way to prove that?

Note that at least for $n=1,2$, there are closed forms $$I_1=\frac{\pi}{2} \,\left (\ln ^2(2) -\frac{\pi^2}{24}\right),\ I_2=\frac{\pi}{2} \,\left (\ln ^4(2)+\frac{\pi^4}{160}-\ln(2)\zeta(3)\right).$$

Have you tried changing $\pi/2$ by a random number? – ajotatxe Nov 06 '19 at 21:32 — ajotatxe, Nov 06 '19 at 21:32

score 6 · Accepted Answer · edited Nov 07 '19 at 05:26

6

By changing $x=\frac{\pi}{2}-u$ in $J_n$, we find $$J_n=\frac\pi2I_n-J_n$$ the claim follows directly.

edited Nov 07 '19 at 05:26

Hello

2,133

answered Nov 06 '19 at 21:35

Paul Enta

14,113

1

Oh yes indeed, so easy... – Wolfgang Nov 06 '19 at 21:41

An intriguing relationship between pairs of integrals

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