# How to compute $\int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x$?

By my post, I had found that \begin{aligned} \frac{\pi^{3}}{8}&= \int_{0}^{\infty} \frac{\ln ^{2} x}{x^{2}+1} d x \\&\stackrel{x\mapsto\tan x}{=} \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\tan x) d x \\ &= \int_{0}^{\frac{\pi}{2}}[\ln (\sin x)-\ln (\cos x)]^{2} d x \\ &= \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\sin x) d x-2 \int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x +\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\cos x) d x \\ &=2\left[\frac{\pi^{2}}{24}\left(\pi^{2}+3 \ln ^{2} 4\right)\right]-2 I, \end{aligned} the last line comes from the post.

Rearranging yields the result \begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x \\ &=\frac{1}{2}\left(\frac{\pi^{3}}{12}+\pi \ln^{2} 2-\frac{\pi^{3}}{8}\right) \\ &=-\frac{\pi^{3}}{48}+\frac{\pi}{2} \ln ^{2} 2 \end{aligned} Is there any other solution? Comments and solutions are highly appreciated.

Fourier series provide an alternative and simple way. Over $$(0,\pi)$$ we have $$\log\sin x = -\log 2-\sum_{m\geq 1}\frac{\cos(2m x)}{m} \tag{1}$$ pointwise and uniformly over any compact subset of $$(0,\pi)$$. Similarly, over $$(0,\pi/2)$$ $$\log\cos x = -\log 2 + \sum_{n\geq 1}\frac{\cos(2n x)}{n}(-1)^{n+1} \tag{2}$$ holds pointwise and uniformly over compact subsets. Since for any $$m,n\in\mathbb{N}^+$$

$$\int_{0}^{\pi/2}\cos(2n x)\cos(2mx)\,dx =\frac{\pi}{4}\delta(m,n)\tag{3}$$ we have

$$\int_{0}^{\pi/2}\log(\sin\theta)\log(\cos\theta)\,d\theta = \frac{\pi}{2}\log^2 2-\frac{\pi}{4}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\color{red}{\frac{\pi}{2}\log^2 2-\frac{\pi^3}{48}}.\tag{4}$$

Yet another way is to consider that the integral is

$$\lim_{\alpha,\beta\to 0^+}\frac{\partial^2}{\partial a\partial b}\int_{0}^{\pi/2}(\sin\theta)^{\alpha}(\cos\theta)^{\beta}\,d\theta$$ where the last integral is a value of the Beta function.

If you are familiar with beta function and feymman's trick then

$$B(x,y) = 2\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \cos^{2y-1}(t) \ dt$$

$$\frac{\partial }{\partial x} B(x,y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t)\ln(\sin t) \cos^{2y-1}(t) \ dt$$

$$\frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} B(x,y) \right) = 8\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \ln(\sin t ) \ln(\cos t) \cos^{2y-1}(t) \ dt$$

and we know that $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

so differentiate and put $$x =\frac{1}{2} , y = \frac{1}{2}$$ $$\frac{\partial^2 B}{\partial x\partial y}=(\psi(x)-\psi(x+y))(\psi(y)-\psi(x+y))B(x,y)+B(x,y)(-\psi'(x+y)))$$

$$\psi \left(\frac{1}{2} \right) = -2\ln 2 - \gamma$$

$$\psi(1) = -\gamma$$

$$\psi^{(1)}(1) = \frac{\pi^2}{6}$$

$$B\left( \frac{1}{2} , \frac{1}{2} \right) = \frac{\Gamma \left( \frac{1}{2} \right)^2}{\Gamma(1)} = \pi$$

thus you'll have the integral $$= \frac{\pi}{8} \left(4\ln(2)^2 - \frac{\pi^2}{6} \right)$$

Hint:

First, establish

$$\int_0^{\pi/2} \ln(2\sin x)\ln(2\cos x)dx =-\frac12\int_0^{\pi/2} \ln^2(2\sin x)dx$$ and the use the known $$\int_0^{\pi/2} \ln^2(2\sin x)dx=\frac{\pi}{4}\zeta(2)$$.

In fact $$\begin{eqnarray} &&\int_0^{\pi/2}\ln(\sin x)\ln(\cos x)dx\\ &=&\frac12\int_0^{\pi/2}\bigg[\ln(\sin x)+\ln(\cos x)\bigg]^2dx-\frac12\int_0^{\pi/2}\bigg[\ln^2(\sin x)+\ln^2(\cos x)\bigg]dx\\ &=&\frac12\int_0^{\pi/2}\ln^2(\sin x\cos x)dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\ &=&\frac12\int_0^{\pi/2}\ln^2(\frac12\sin 2x)dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\ &=&\frac14\int_0^{\pi}(\ln2-\ln\sin x)^2dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\ &=&\frac12\int_0^{\pi/2}(\ln2-\ln\sin x)^2dx-\int_0^{\pi/2}\ln^2(\sin x)dx\\ &=&\frac12\bigg(\frac\pi2\ln^22-2\ln 2\int_0^{\pi/2}\ln\sin xdx+\int_0^{\pi/2}\ln^2\sin xdx\bigg)-\frac12\int_0^{\pi/2}\ln^2(\sin x)dx\\ &=&\frac{\pi}{2}\ln^2 2-\frac{\pi^3}{48}. \end{eqnarray}$$ Here the famous results $$\int_0^{\pi/2}\ln(\sin x)dx=-\frac\pi2\ln2,\int_0^{\pi/2}\ln^2(\sin x)dx=\frac{1}{24} \left(\pi^3 + 3\pi \ln^2(4)\right)$$ are used.