Closed form of integral over fractional part $\int_0^1 \left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}\,dx$

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}

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

Proofs of the identity of the OP have already been given in answers.

However, it might be interesting so see the calculation which led me to the result which I didn't know in advance.

Part 1: Transformation of integral into a series

Substituting $x\to z-\sqrt{z^2-1}$ the integral becomes

$$i = \int_{1}^\infty \{z\}( \frac{z}{\sqrt{z^2-1}}-1)\,dx\tag{1}$$

Splitting the integral into the intervals $(k,k+1)$, $k=1,2,3,...$ we get

$$i= i_{s} := \lim_{n\to\infty} i_{s}(n) \tag{2a}$$

$$i_{s}(n):= \sum_{k=1}^n a(k)\tag{2b}$$

Letting $z = k + \xi$ we have $\{z\} = \xi$ and the summands become

$$a(k):=\int_0^1 \xi \left(\frac{k+\xi }{\sqrt{(k+\xi )^2-1}}-1\right)\,d\xi \\=\frac{1}{2} \left(k \sqrt{k^2-1}-(k+1) \sqrt{(k+1)^2-1}\right)+\frac{1}{2} \left(\log \left(k+\sqrt{(k+1)^2-1}+1\right)-\log \left(\sqrt{k^2-1}+k\right)\right)+(\sqrt{(k+1)^2-1}-\frac{1}{2})\tag{3} $$

Summing up from $k=1$ to $k=n$ the first two brackets telescope and one sum is left:

$$i_{s}(n) = p(n) + g(n)\tag{4a}$$

where

$$p(n)=\frac{1}{2} \left(-(n+1)\sqrt{(n+1)^2-1} -n+\log \left(n+\sqrt{n (n+2)}+1\right)\right)\tag{4b}$$

$$g(n) = \sum _{k=2}^{n+1} \sqrt{k^2-1}\tag{5}$$

where in $g$ we have omitted the summand with $k=1$ without altering the sum.

Part 2: asymptotics of $g(n)$

This is the tough part. In order to perform the limit (2a) we need the asymptotic behaviour of the terms in (4). We have to focus on $g(n)$ since the asymptotics of the other terms is simple to obtain.

Writing

$$\sqrt{k^2-1} = k \sqrt{1-\frac{1}{k^2}} = \sum _{m=0}^{\infty } (-1)^m \binom {\frac{1}{2}}{m}\frac{1}{k^{2 m-1}}\tag{6}$$

Performing the $k$-sum, according to $\sum _{k=2}^{n+1} 1/k^{2 m-1}=-1+H_{n+1}^{(2 m-1)}$ we get

$$g(n) = \sum _{m=0}^{\infty } (-1)^m \binom {\frac{1}{2}}{m}(-1+H_{n+1}^{(2 m-1)})\tag{7}$$

Notice that (7) is an exact formula. Now we can take the asymptotic limit (with respect to n) under the $m$-sum using the well-known asymptotics of $H_{n}^{(k)}$ leading to

$$H_{n+1}^{(2 m-1)}-1 \simeq h_0 +h_1 + h_2 \tag{8a}$$

$$h_0=-1, h_1=-\frac{1}{6} m n^{-2 m}+\frac{n^{-2 m}}{12}+\frac{1}{2} n^{1-2 m}-\frac{n^{2-2 m}}{2 m-2}, h_2 = \zeta (2 m-1)\tag{8b}$$

We now insert this into (7) and proceed carefully with the terms and the index $m$. We collect the contributions in $g_i(n)$.

The m-sum over $h_0$ gives $g_0 = 0$.

For the sum $(h_1+h_2)$ we consider first the two summands $m=0$ and $m=1$ separately

$$g_{1}(n) =\lim_{m\to 0}{ (-1)^m \binom {\frac{1}{2}}{m}(h_1+h_2 )}= \frac{n^2}{2}+\frac{n}{2}$$

$$g_{2}(n) =\lim_{m\to 1}{ (-1)^m \binom {\frac{1}{2}}{m}(h_1+h_2 )}= \frac{1}{24 n^2}-\frac{1}{4 n}-\frac{\log (n)}{2}-\frac{\gamma }{2}$$

Notice that for $m\to 1$ we had to keep the sum $(h_1+h_2)$ so that the pole in $\zeta$ at $m=1$ can compete with the other term with a pole.

The rest of the $m$-sum from $m=2$ is considered separately for $h_1$ and $h_2$.

For $h_1$ we could do the complete sum (Mathematica gives a lengthy expression including a hypergeometric function) but up to order $1/n^3$ we need only the term with $m=2$. This gives

$$g_{3}(n) =\lim_{m\to 2}{ (-1)^m \binom {\frac{1}{2}}{m}(h_1 )}= \frac{1}{16 n^2}-\frac{1}{16 n^3}$$

For $h_2$ the complete sum reads

$$g_{4}(n) =\sum_{m=2}^\infty (-1)^m \binom {\frac{1}{2}}{m}\zeta({2m-1})$$

Making use of the well-known integral representation of the $\zeta$-function

$$\zeta (2 m-1)=\int_0^{\infty } \frac{t^{2 m-2}}{\left(e^t-1\right) \Gamma (2 m-1)} \, dt\tag{9}$$

we can do the sum under the integral for which Mathematica gives the result

$$\sum _{m=2}^{\infty } \frac{(-1)^m \binom{\frac{1}{2}}{m} t^{2 m-2}}{\left(e^t-1\right) \Gamma (2 m-1)}= \frac{t-2 I_1(t)}{2 \left(e^t-1\right) t}\tag{10}$$

This gives $g_4(n)$ = $c_{g}$.

Hence we find for the asymptotic behaviour of $g(n)$

$$g_a(n) = g_1+g_2+g_3+g_4 \\ = c_{g}-\frac{1}{16 n^3}+\frac{n^2}{2}+\frac{5}{48 n^2}+\frac{n}{2}-\frac{1}{4 n}-\frac{\log (n)}{2}-\frac{\gamma }{2}\tag{11}$$

Part 3: harvest and final result

To obtain the complete asymptotic Expression according to (4) we need the asymptotics of $p(n)$ which is, however, easily calculated with the result

$$p_a(n) = -\frac{n^2}{2}+\frac{3}{16 n^2}-\frac{3n}{2}+\frac{\log (n)}{2}+\frac{3}{4}+\frac{\log (2)}{2}\tag{12}$$

Adding $p_a(n)$ and $g_a(n)$ the leading terms and the $\log$-terms cancel. Finally, taking the $\lim_{n\to\infty}$ gives

$$i_{s} = c_{g}-\frac{\gamma }{2}+\frac{3}{4}+\frac{\log (2)}{2}\tag{13}$$

which is the result of the OP.

Remark: the simplicity of the final expression surprised me: just a simple fraction, $\log(2)$, and $\gamma$, but at least $c_g$ is a non-trivial quantity which most probably is a new constant.