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I stumbled across the titular integral identity $(*)$ while working out an alternative method for evaluating $\displaystyle\int_0^\pi(\log\sin x)^2\,dx$. It's obtained by substituting $y=\tan\dfrac x2$, then expanding the logarithm on the LHS of $(*)$ and observing that, via $y\to\dfrac1y$,

  • $\displaystyle \int_0^\infty\frac{\log y}{1+y^2}\,dy=0$ and
  • $\displaystyle \int_0^\infty\frac{\log y\log\left(1+y^2\right)}{1+y^2}\,dy=\int_0^\infty\frac{(\log y)^2}{1+y^2}\,dy$,

so we can split up and simplify the LHS of $(*)$ to recover the integral $I$ below.

$$\begin{align*} \text{LHS} &= \int_0^\infty \frac{2\left(\log(2y)\right)^2 - 4\log(2y)\log\left(1+y^2\right) + \frac32\left(\log\left(1+y^2\right)\right)^2}{1+y^2}\,dy + \text{RHS} \\ \implies I &= \bbox[#00ff37,2pt]{\int_0^\infty \frac{4(\log2)^2-4(\log y)^2-8\log2\log\left(1+y^2\right)+3\left(\log\left(1+y^2\right)\right)^2}{1+y^2} \, dy = 0} \end{align*}$$

Besides evaluating $(1),(2),(3)$ termwise and combining their closed forms listed below, are there any other ways to more immediately determine that $I$ vanishes in order to arrive at $(*)$?

$$\begin{align*} \int_0^\infty \frac{(\log y)^2}{1+y^2} \, dy &= \pi\log2 \tag1\\ \int_0^\infty \frac{\log\left(1+y^2\right)}{1+y^2} \, dy &= \frac{\pi^3}8 \tag2\\ \int_0^\infty \frac{\left(\log\left(1+y^2\right)\right)^2}{1+y^2} \, dy &= \frac{\pi^3}6 + 2\pi (\log 2)^2 \tag3 \end{align*}$$

user170231
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1 Answers1

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Substitute $y=\tan t$ \begin{align} & \int_0^\infty\frac{\ln^2\frac{2y}{1+y^2}}{1+y^2} -\frac{\frac14\ln^2(1+y^2)}{1+y^2}\ dy\\ =&\int_0^{\pi/2}(\ln^2\overset{2t\to t}{\sin2t}-\ln^2\cos t) dt = \int_0^{\pi/2}(\ln^2{\sin t}-\ln^2\cos t) dt=0 \end{align}

Quanto
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