Show that $$\displaystyle{\int_0^1 \left \lbrace (-1)^{\left \lceil \frac{1}{x} \right \rceil }\frac{1}{x} \right \rbrace \, {\rm d}x = \frac{\pi}{2}}$$
My try
$$\displaystyle{\int\limits_0^1 {\left\{ {{{\left( { – 1} \right)}^{\left[ {1/x} \right]}}\frac{1}{x}} \right\}dx} = \mathop = \limits^{x \to 1/x} = \int\limits_1^\infty {\left\{ {{{\left( { – 1} \right)}^{\left[ x \right]}}x} \right\}\frac{1}{{{x^2}}}dx} = }$$ $$\displaystyle{\sum\limits_{n = 1}^\infty {\left( {\int\limits_{2n – 1}^{2n} {\left\{ {{{\left( { – 1} \right)}^{\left[ x \right]}}x} \right\}\frac{1}{{{x^2}}}dx} + \int\limits_{2n}^{2n + 1} {\left\{ {{{\left( { – 1} \right)}^{\left[ x \right]}}x} \right\}\frac{1}{{{x^2}}}dx} } \right)} = }$$
$$\displaystyle{ = \sum\limits_{n = 1}^\infty {\left( {\int\limits_{2n – 1}^{2n} {\left\{ { – x} \right\}\frac{1}{{{x^2}}}dx} + \int\limits_{2n}^{2n + 1} {\left\{ x \right\}\frac{1}{{{x^2}}}dx} } \right)} = }$$ $$\displaystyle{\sum\limits_{n = 1}^\infty {\left( {\int\limits_{2n – 1}^{2n} {\left( {1 – \left\{ x \right\}} \right)\frac{1}{{{x^2}}}dx} + \int\limits_{2n}^{2n + 1} {\left\{ x \right\}\frac{1}{{{x^2}}}dx} } \right)}}$$
I'm not sure about the closed form; my friends say it's either $\log\left(\frac{\pi}{2}\right)$, $1 – \log\left(\frac{\pi}{2}\right)$, or $\frac{\pi}{2}$.
Best Answer
Whatever we do, the ceiling function is going to cause difficulties with the integration at some point (as it's discontinuous). So let's start by doing something about that. Let $N$ be a positive integer and $y=\frac{1}{x}$, and define $$I_N=\int_\frac{1}{2N+1}^1 \left \lbrace (-1)^{\left \lceil \frac{1}{x} \right \rceil }\frac{1}{x} \right \rbrace dx = \int_1^{2N+1} \frac{1}{y^2}\left \lbrace (-1)^{\left \lceil y \right \rceil }y \right \rbrace dy $$
Now we can get rid of the ceiling function and the power of $-1$ by considering cases when $2n-1< y \le 2n$ and when $2n< y\le 2n+1$:
$$\begin{align*}I_N&=\sum_{n=1}^N \left[ \int_{2n-1}^{2n} \frac{1}{y^2}\left \lbrace (-1)^{\left \lceil y \right \rceil }y \right \rbrace dy + \int_{2n}^{2n+1} \frac{1}{y^2}\left \lbrace (-1)^{\left \lceil y \right \rceil }y \right \rbrace dy \right] \\ &=\sum_{n=1}^N \left[ \int_{2n-1}^{2n} \frac{1}{y^2}\left \lbrace y \right \rbrace dy + \int_{2n}^{2n+1} \frac{1}{y^2}\left \lbrace -y \right \rbrace dy \right] \\ &=\sum_{n=1}^N \left[ \int_{2n-1}^{2n} \frac{1}{y^2}\left \lbrace y \right \rbrace dy + \int_{2n}^{2n+1} \frac{1}{y^2}\left \lbrace 1-y \right \rbrace dy \right] \end{align*}$$
Now, with the substitution $u=y-(2n-1)$, $$\int_{2n-1}^{2n} \frac{1}{y^2}\left \lbrace y \right \rbrace dy = \int_0^1 \frac{u}{(u+2n-1)^2} du=\log\frac{2n}{2n-1}-\frac{1}{2n}$$
and with $v=y-2n$, $$\int_{2n}^{2n+1} \frac{1}{y^2}\left \lbrace 1-y \right \rbrace dy =\int_0^1 \frac{1-v}{(v+2n)^2} dv=\log\frac{2n}{2n+1}+\frac{1}{2n}$$
so that $$\begin{align*} I_N&=\sum_{n=1}^N \left[\log\frac{2n}{2n-1}-\frac{1}{2n} + \log\frac{2n}{2n+1}+\frac{1}{2n}\right] \\ &=\sum_{n=1}^N \log\frac{4n^2}{4n^2-1} \end{align*}$$
Letting $N\to \infty$, this is the logarithm of the well-known Wallis product formula for $\frac{\pi}{2}$; so the value of the original integral is $\boxed{\log\frac{\pi}{2}}$.