How to evaluate the following integral $\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$?

How to evaluate the following integral $$\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$$ It seems that it evaluates to$$\frac{1}{4}-\frac{\pi^2}{48}$$ Is this true? How would I prove it?

Solution 1:

Find this

$$I=\int_{0}^{\frac{\pi}{2}}\sin{x}\cos{x}\ln{(\cos{x})}\ln{(\sin{x})}dx$$

Solution

Since $$\sin(2x) = 2\sin(x)\cos(x)$$

then $$I=\dfrac{1}{8}\int_{0}^{\frac{\pi}{2}}\ln{(\sin^2{x})} \ln{(\cos^2{x})}\sin{(2x)}dx$$

Let $\cos{(2x)}=y$, and since $$\cos(2x) = 2\cos^2x - 1 = 1 - 2\sin^2x$$ we get $$I=\dfrac{1}{16}\int_{-1}^{1}\ln{\left(\dfrac{1-y}{2}\right)} \ln{\left(\dfrac{1+y}{2}\right)}dy$$

Let $\dfrac{1-y}{2}=z$, then we have \begin{align*}I&=\dfrac{1}{8}\int_{0}^{1}\ln{z}\ln{(1-z)}dz=\dfrac{-1}{8}\sum_{n=1}^{\infty}\dfrac{1}{n} \int_{0}^{1}z^n\ln{z}dz\\ &=\dfrac{1}{8}\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)^2}=\dfrac{1}{8}\sum_{n=1}^{\infty} \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-\dfrac{1}{8}\sum_{n=1}^{\infty} \dfrac{1}{(n+1)^2}\\ &=\dfrac{1}{4}-\dfrac{\pi^2}{48} \end{align*}

Solution 2:

Alternatively, using the general trigonometric form of beta function from equation $(14)$ Wolfram MathWorld - Beta Function we have $$\int_0^{\pi/2}\sin^nx\cos^mx\,dx=\frac{1}{2}\text{B}\left(\frac{n+1}{2},\frac{m+1}{2}\right)$$ Differentiating with respect to $m$ and $n$ once, then putting $m=1$ and $n=1$ we directly obtain the desired result $$\begin{align}\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx&=\frac{\text{B}\left(1,1\right)}{8}\bigg[\left(\psi_0(1)-\psi_0(2)\right)^2-\psi_1(2)\bigg]\\&=\frac{1}{4}-\frac{\pi^2}{48}\end{align}$$ Here I use equation $(26)$ from Wolfram MathWorld - Beta Function and also equation $(8)$ & equation $(15)$ from Wolfram MathWorld - Polygamma Function.