How to calculate the integral of $\operatorname{sgn}(\sin\pi/x)$ in the interval $(0,1)$?

Question

How can I calculate the integral of $\operatorname{sgn}(\sin\pi/x)$ in the interval $(0,1)$? I need to calculate this integral, thanks

Answer 1

Outline: In the interval $(1/2,1)$ our function is $-1$.

In the interval $(1/3,1/2)$, our function is $1$.

In the interval $(1/4,1/3)$, our function is $-1$.

In the interval $(1/5,1/4)$, our function is $1$.

And so on. The intervals have length $\frac{1}{1\cdot 2}$, $\frac{1}{2\cdot 3}$, $\frac{1}{3\cdot 4}$, and so on.

So the integral ought to be $$-\frac{1}{1\cdot2}+\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}-\frac{1}{5\cdot 6}+\cdots.$$

If we want a closed form, note that $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\cdots.\tag{1}$$ Now calculate $\int_0^{1} \ln(1+x)\,dx$. This is the same as what we obtain when we integrate the series (1) term by term.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\fermi\pars{\mu}\equiv\int_{0}^{1}\sgn\pars{\sin\pars{\mu \over x}}\,\dd x= \int_{1}^{\infty}{\sgn\pars{\sin\pars{\mu x}} \over x^{2}}\,\dd x}. \qquad\fermi\pars{\pi} = {\large ?}$

\begin{align} \fermi'\pars{\mu}&= \int_{1}^{\infty} {2\delta\pars{\sin\pars{\mu x}}\cos\pars{\mu x}x \over x^{2}}\,\dd x =2\int_{1}^{\infty}{\cos\pars{\mu x} \over x} \sum_{n = -\infty}^{\infty} {\delta\pars{x - n\pi/\mu} \over \verts{\mu\cos\pars{\mu x}}}\,\dd x \\[3mm]&={2 \over \verts{\mu}}\sum_{n = -\infty}^{\infty}{\sgn\pars{\cos\pars{n\pi}} \over n\pi/\mu}\,\Theta\pars{n - {\mu \over \pi}} ={2\sgn\pars{\mu} \over \pi}\sum_{n = -\infty}^{\infty}{\pars{-1}^{n} \over n} \Theta\pars{n - {\mu \over \pi}} \end{align}

\begin{align} \fermi\pars{\pi} - \overbrace{\fermi\pars{0^{+}}}^{\ds{\to 1}}& ={2 \over \pi}\sum_{n = -\infty}^{\infty}{\pars{-1}^{n} \over n} \int_{0^{+}}^{\pi}\Theta\pars{n\pi - \mu}\,\dd\mu=-2\ln\pars{2} \end{align}

$$\color{#44f}{\large% \int_{0}^{1}\sgn\pars{\sin\pars{\mu \over x}}\,\dd x =1 - 2\ln\pars{2}} $$

Answer 3

Usually, when you have a function defined by cases, one considers breaking the problem up along those cases.