Recall an integral

Actually, I haven’t taken calculus seriously for many years, though it is full of techniques and skills, I would not enjoy spending time on it. And this time I accidentally met some problem that amazed me.

Just for fun.
1.

\displaystyle\int_0^{\pi}\log(\sin(x))dx = -\pi\log 2

2.

\displaystyle\int_0^{\frac{\pi}{2}}\log^2(\sin(x)dx =\frac{ \pi}{2}^2\log 2 + \frac{\pi^3}{24}

3.

\displaystyle\int_0^{\frac{\pi}{2}}\log(\sin(x))\log(\cos(x))dx = \frac{\pi}{2}^2\log 2 - \frac{\pi^3}{48}

4.

\displaystyle\int_0^{\frac{\pi}{2}}\log(\tan(x)) dx= 0

5.

\displaystyle\int_0^{\frac{\pi}{2}}\log^2(\tan(x))dx = \frac{\pi^3}{8}

Proof.

Just consider the Fourier coefficients of the function, while x\in(0,\pi/2)

\displaystyle\log(\sin(x)) = -\log 2 -\sum_{n\ge 1} \frac{\cos(2nx)}{n}

\displaystyle\log(\cos(x))=-\log(2) -\sum_{n\ge 1}\frac{(-1)^n\cos(2nx)}{n}

QED

About YiMin

This is just a nerd PhD student of Math@UT Austin.