TITLE: Integrate $\int^{\frac{\pi}{2}}_{0} \{ \cot x\}$ dx
QUESTION [14 upvotes]: So I have been trying to solve this question, 
$$
\int_{0}^{\pi/2}\left\{\,\cot\left(x\right)\,\right\}\,\mathrm{d}x
$$
where $\left\{\right\}$ means the fractional part.
$\left(\,x - \left\lfloor\,x\,\right\rfloor\,\right)$.
Progress so far, 

MATLAB gives out the following answer when
int(cot(x)-fix(cot(x)),0,pi/2) is run, ans = int(cot(x) - fix(cot(x)), x, 0, pi/2)
The tutorial here is somewhat tries to solve a different version of the problem.
Writing $ \{ \text{cotx} \} $ as $ \text{cotx} - \newcommand{\floor}[1]{\lfloor #1 \rfloor} \lfloor \text{cot} x \rfloor $ gives us $\int^{\frac{\pi}{2}}_{0} \text{cotx} - \newcommand{\floor}[1]{\lfloor #1 \rfloor} \int^{\frac{\pi}{2}}_{0}{\lfloor \text{cot} x \rfloor}$ where the first part doesn't converge ( $ \infty $ ). 

Can someone help me?

REPLY [6 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\pi/2}\braces{\cot\pars{x}}\dd x & =
\int_{0}^{\pi/2}\braces{\tan\pars{x}}\dd x
\,\,\,\stackrel{\tan\pars{x}\ \mapsto\ x}{=}\,\,\,
\int_{0}^{\infty}{\braces{x} \over x^{2} + 1}\,\dd x
\\[5mm] &=
\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} -
\int_{0}^{\Lambda}{\left\lfloor\,{x}\,\right\rfloor \over x^{2} + 1}\,\dd x}
\\[5mm] & =
\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} -
\int_{0}^{\left\lfloor\,\Lambda\,\right\rfloor}
{\left\lfloor\,{x}\,\right\rfloor \over x^{2} + 1}\,\dd x -
\int_{\left\lfloor\,\Lambda\,\right\rfloor}^{\Lambda}
{\left\lfloor\,{x}\,\right\rfloor \over x^{2} + 1}\,\dd x}
\end{align}

Note that
  $\ds{\lim_{\Lambda \to \infty}\int_{\left\lfloor\,\Lambda\,\right\rfloor}^{\Lambda}
{\left\lfloor\,{x}\,\right\rfloor \over x^{2} + 1}\,\dd x = 0}$.

Then,
\begin{align}
\int_{0}^{\pi/2}\braces{\cot\pars{x}}\dd x & =
\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} -
\sum_{n = 1}^{\left\lfloor\,\Lambda\,\right\rfloor - 1}
\int_{n}^{n + 1}{n \over x^{2} + 1}\,\dd x}
\\[5mm] & =
\lim_{\Lambda \to \infty}\braces{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} -
\sum_{n = 1}^{\left\lfloor\,\Lambda\,\right\rfloor - 1}
n\bracks{\arctan\pars{n + 1} - \arctan\pars{n}}}
\\[5mm] & =
\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} -
\sum_{n = 1}^{\left\lfloor\,\Lambda\,\right\rfloor}
\pars{n - 1}\arctan\pars{n} +
\sum_{n = 1}^{\left\lfloor\,\Lambda\,\right\rfloor - 1}n\arctan\pars{n}}
\\[5mm] & =
\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} -
\left\lfloor\,\Lambda\,\right\rfloor
\arctan\pars{\left\lfloor\,\Lambda\,\right\rfloor} +
\sum_{n = 1}^{\left\lfloor\,\Lambda\,\right\rfloor}\arctan\pars{n}}
\\[5mm] & =
\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} -
\left\lfloor\,\Lambda\,\right\rfloor
\arctan\pars{\left\lfloor\,\Lambda\,\right\rfloor} +
{\pi \over 2}\,\left\lfloor\,\Lambda\,\right\rfloor -
\sum_{n = 1}^{\left\lfloor\,\Lambda\,\right\rfloor}\arctan\pars{1 \over n}}
\\[5mm] & =
\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} +
\left\lfloor\,\Lambda\,\right\rfloor
\arctan\pars{1 \over \left\lfloor\,\Lambda\,\right\rfloor} -
\sum_{n = 1}^{\left\lfloor\,\Lambda\,\right\rfloor}\arctan\pars{1 \over n}}
\\[1cm] & =\
\underbrace{\overbrace{\quad\lim_{\Lambda \to \infty}\bracks{%
{1 \over 2}\ln\pars{\Lambda^{2} + 1} +
\left\lfloor\,\Lambda\,\right\rfloor
\arctan\pars{1 \over \left\lfloor\,\Lambda\,\right\rfloor} -
H_{\left\lfloor\,\Lambda\,\right\rfloor}}\quad}^{\ds{=\ 1 - \gamma}}}
_{\ds{~H_{z}:\ Harmonic\ Number.\ \gamma:\ Euler-Mascheroni\ Constant~}}
\\[2mm] & +
\sum_{n = 1}^{\infty}\bracks{{1 \over n} - \arctan\pars{1 \over n}}
\end{align}

Therefore, the original question is reduced to

\begin{align}
\bbx{\int_{0}^{\pi/2}\braces{\cot\pars{x}}\dd x =
1 - \gamma +
\sum_{n = 0}^{\infty}\bracks{{1 \over n + 1} - \arctan\pars{1 \over n + 1}}}
\label{1}\tag{1}
\end{align}
The 'remaining' sum is a identity ( see $\ds{\mathbf{\color{#000}{6.1.27}}}$ in A & S Table ). Namely,
$$
\sum_{n = 0}^{\infty}\bracks{{1 \over n + 1} - \arctan\pars{1 \over n + 1}} =
\mrm{arg}\pars{\Gamma\pars{1 + \ic}} + \gamma
$$
$$
\mbox{such that}\quad
\bbox[15px,#ffe,border:1px dotted navy]{\ds{\int_{0}^{\pi/2}\braces{\cot\pars{x}}\dd x =
1 + \mrm{arg}\pars{\Gamma\pars{1 + \ic}}}} \approx 0.6984
$$