Integral of sign function

Question

Define $f: \mathbb{R}\rightarrow\mathbb{R}$ as - $f(x)=\sum_{n=1}^\infty \frac{sgn(sin(2^nx))}{2^n}$.

Prove that $f$ is integrable on $[0,\frac{2\pi}{3}]$ and calculate $\int_{[0,\frac{2\pi}{3}]}f$.

Any ideas?

Answer 1

So, a simple way to prove integrability and calculating the integral is to use this theorem :

If $(f_n)$ is a uniformly convergent sequence on $[a, b]$ with limit $f$, then Riemann integrability of all $(f_n)$ implies Riemann integrability of $f$, and $$\int_a^b f(t) dt = \lim_{n\to \infty} \int_a^b f_n(t) dt$$

Define $$f_n(x) = \sum_{k=1}^n \frac{\text{sign}( \sin (2^k x))}{2^k}$$

The $(f_n)$ are Riemann integrable (they are piecewise continuous), and $$\sup_{x \in [0, \frac{2\pi}{3}]} |f_n(x) - f(x)| \leq \frac{1}{2^{n-1}}$$

So the $f_n$ converge uniformly to $f$.

Hence $f$ is integrable and $$\int_0^{\frac{2\pi}{3}} f(x) dx = \lim_{n\to \infty} \int_0^{\frac{2\pi}{3}}f_n(x) dx$$

Now

$$\int_0^{\frac{2\pi}{3}}f_n(x) dx = \int_0^{\frac{2\pi}{3}} \sum_{m=1}^{n} \frac{\text{Sign}(\sin(2^k x))}{2^k} dx$$

$$ = \sum_{k=1}^{n} \frac{1}{2^k} \int_0^{\frac{2\pi}{3}} \text{Sign}(\sin(2^k x)) dx$$

As $\text{Sign}(\sin(2^k x))$ is $\frac{ \pi}{2^{k-1}}$ periodic, with integral equal to $0$ on a whole period, we have that

$$\int_0^{\frac{2\pi}{3}} \text{Sign}(\sin(2^{2k} x)) dx = \int_0^{\frac{2\pi}{3 \times 2^{2k}}} \text{Sign}(\sin(2^{2k} x)) dx = \int_0^{\frac{2\pi}{3 }} \text{Sign}(\sin( x)) dx = -\frac{\pi}{6}$$

and

$$\int_0^{\frac{2\pi}{3}} \text{Sign}(\sin(2^{2k+1} x)) dx = \int_0^{\frac{\pi}{3 \times 2^{2k+1}}} \text{Sign}(\sin(2^{2k+1} x)) dx = \int_0^{\frac{\pi}{3 }} \text{Sign}(\sin( x)) dx = \frac{\pi}{6}$$

So finally,

$$\int_0^{\frac{2\pi}{3}}f_n(x) dx = \sum_{k=1}^{n} \frac{1}{2^k} \frac{\pi (-1)^{k+1}}{6} \int_0^{\frac{2\pi}{3}} \text{Sign}(\sin(2^k x)) dx$$

$$ = -\frac{\pi}{6} \sum_{k=1}{n} \left( \frac{1}{2} \right)^k \to -\frac{\pi}{6}$$