Closed form of integral over fractional part $\int_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}\,dx$ Recenly, several interesting questions have been posted asking for closed forms of integrals over the fractional part of certain functions.
For me the story started with Evaluation of $\int_{0}^{1}\int_{0}^{1}\{\frac{1}{\,x}\}\{\frac{1}{x\,y}\}dx\,dy\,$ which after a long and instructive journey I could solve completely. Another example was symmetric double-integral on fractional part. These are examples of double integrals. There are as well many single integrals, and, as we can see below, the field of single integrals is by far not exhausted.    
This time my result is given in the beginning and a proof is asked for.
Let $\{z\}$ be the fractional part of $z$.
Prove that:
$$i := \int_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\} = i_{s} $$
where
$$i_{s}=c_{g}-\frac{\gamma }{2}+\frac{3}{4}+\frac{\log (2)}{2} \simeq 0.28000699470709318696$$
Here $\gamma$ is the Euler-Mascheroni constant and
$$c_{g} = \int_0^{\infty } \frac{t-2 I_1(t)}{2 \left(e^t-1\right) t} \, dt \simeq -0.52795876312211303745$$
where 
$I_{n}(t)$ is the modified Bessel function of the first kind.
$c_{g}$ is a (probably) new constant which appears in the asymptotic expansion of the sum
$$g(n) = \sum _{k=1}^n \sqrt{k^2-1} $$
 A: It is helpful to derive the asymptotic expansion of $g$ first. We can use the binomial series to find
\begin{align}
g(n) &= \sum \limits_{k=2}^n k \sqrt{1-k^{-2}} = \sum \limits_{k=2}^n k \sum \limits_{j=0}^\infty {1/2\choose j} (-k^{-2})^j \\
&= \frac{n(n+1)}{2} - 1 - \frac{H_n}{2} + \frac{1}{2} + \sum \limits_{j=2}^\infty {1/2\choose j} (-1)^j \sum \limits_{k=2}^n k^{1-2j}
\end{align}
with the harmonic numbers $H_n$. The monotone convergence theorem now yields the asymptotic equivalence
$$ g(n) \sim \frac{n(n+1)}{2} - \frac{H_n}{2} + c_g + \mathcal{o}(1)$$
as $n \to \infty$ . The constant term can be written as
$$ c_g = - \frac{1}{2} + \sum \limits_{j=2}^\infty {1/2\choose j} (-1)^j [\zeta(2j-1) - 1] = \sum \limits_{k=2}^\infty \left(\sqrt{k^2-1} - k + \frac{1}{2k}\right) \, ,$$
which agrees with the integral representation after using the series expansion of $I_1$.
In order to find $i$ we use the substitution $x = t - \sqrt{t^2-1}$ :
\begin{align}
i &= \int \limits_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\} \, \mathrm{d} x = \int \limits_1^\infty \{t\} \left(\frac{t}{\sqrt{t^2-1}}-1\right) \, \mathrm{d} t \\
&= \sum \limits_{n=1}^\infty \int \limits_n^{n+1} (t-n) \left(\frac{t}{\sqrt{t^2-1}}-1\right) \, \mathrm{d} t \\
&= \frac{1}{2} \sum \limits_{n=1}^\infty \left[\ln\left(\sqrt{(n+1)^2-1}+n+1\right) - \ln\left(\sqrt{n^2-1}+n\right)\right. \\
&\phantom{= \frac{1}{2} \sum \limits_{n=1}^\infty\left[\right.} \left.- (n+1)\sqrt{(n+1)^2-1} + n \sqrt{n^2-1} + 2\sqrt{(n+1)^2 - 1} - 1  \right] \, .
\end{align}
The remaining series is (mostly) telescoping and we obtain
\begin{align}
i &= \frac{1}{2} \lim_{N \to \infty} \left[\ln\left(\sqrt{N^2-1} + N\right) - N \sqrt{N^2-1} + 2 g(N) - N + 1\right] \\
&= \frac{1}{2} \lim_{N \to \infty} \left[\ln\left(1+\sqrt{1-N^{-2}}\right) + \ln(N) - H_N  + N \left(N+1 -  \sqrt{N^2-1} - 1\right) + 2 c_g + 1\right] \\
&= \frac{1}{2} \left[\ln(2) - \gamma  + \frac{1}{2} + 2 c_g + 1\right] \\
&= \frac{3}{4} + \frac{\ln(2)-\gamma}{2} + c_g \, .
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

An alternative:

\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{1}\braces{{1 \over 2}\pars{x + {1 \over x}}}\dd x}}
\,\,\,\stackrel{x\ =\ 1 - t/\root{t^{2} - 1}}{=}\,\,\,
\int_{\infty}^{1}\braces{t}\pars{1 - {t \over \root{t^{2} - 1}}}\dd t
\\[5mm] & =
\underbrace{\int_{1}^{\infty}\pars{{t^{2} \over \root{t^{2} - 1}} - t - {1 \over 2t}}\dd t}_{\ds{{1 \over 4} + {1 \over 2}\,\ln\pars{2}}}
\\[2mm] + &\
\lim_{{\large N \to \infty} \atop {\large N\ \in\ \mathbb{N}}}\bracks{{1 \over 2}\,\ln\pars{N} -
\int_{1}^{N}\left\lfloor{t}\right\rfloor \pars{{t \over \root{t^{2} - 1}} - 1}\dd t}
\label{1}\tag{1}
\end{align}

\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{1}^{N}\left\lfloor{t}\right\rfloor \pars{{t \over \root{t^{2} - 1}} - 1}\dd t}} =
\sum_{k = 1}^{N - 1}\int_{k}^{k + 1}k\pars{{t \over \root{t^{2} - 1}} - 1}\dd t
\\[5mm] = &\
\sum_{k = 1}^{N - 1}k\pars{\root{k^{2} + 2k} - \root{k^{2} - 1} - 1}
\\[5mm] = &\
\sum_{k = 1}^{N - 1}k\pars{{ 2k + 1\over \root{k^{2} + 2k} + \root{k^{2} - 1}} - 1 - {1 \over 2k^{2}}} + {1 \over 2}
\overbrace{\bracks{\sum_{k = 1}^{N - 1}{1 \over k} - \ln\pars{N - 1}}}
^{\ds{\stackrel{\mrm{as}\ N\ \to\ \infty}{\LARGE\to}\gamma}}
\\[2mm] + &\
{1 \over 2}\,\ln\pars{N - 1}\label{2}\tag{2}
\end{align}


\eqref{1} and \eqref{2} lead to $\ds{\pars{~\mbox{as}\ N \to \infty~}}$:

\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{1}\braces{{1 \over 2}\pars{x + {1 \over x}}}\dd x}}
\\[5mm] = &\
{1 \over 4} + {1 \over 2}\,\ln\pars{2} - {1 \over 2}\,\gamma\ -\
\underbrace{\sum_{k = 1}^{\infty}\pars{{2k^{2} +  k \over \root{k^{2} + 2k} + \root{k^{2} - 1}} - k - {1 \over 2k}}}_{\ds{\approx 0.0279588}}
\\[5mm] \approx &\ \bbx{0.2800070}
\end{align}
