Limit of $\int_0^1\frac1x B_{2n+1}\left(\left\{\frac1x\right\}\right)dx$

Start from the definition of Bernoulli polynomials in terms of their generation function

$$\frac{t e^{xt}}{e^t - 1} = \sum_{n=1}^\infty \frac{t^n}{n!} B_n(x)$$

If one multiple both sides by $e^{-2\pi i kx}$ for any $k \in \mathbb{Z}$ and integrate, we get

$$ \sum_{n=1}^\infty \frac{t^n}{n!} \int_0^1 B_n(x) e^{-2\pi i k x} dx = \frac{t}{e^{t}-1} \int_0^1 e^{(t-2\pi i k) x} dx = \frac{t}{t-2\pi i k} $$ Comparing the coefficients of $t^n$ of two sides, we find for $n > 0$,

$$\int_0^1 B_n(x) e^{-2\pi ik x} dx = \begin{cases} 0, & k = 0,\\ - \frac{n!}{(2\pi ik)^n}, & k \ne 0 \end{cases} $$ As a result, for $n > 0$, the Bernoulli polynomial $B_n(x)$ has following Fourier series expansion over $(0,1)$.

$$B_n(\{x\}) = -\frac{n!}{(2\pi i)^n} \sum_{|k|>0} \frac{e^{2\pi i kx}}{k^n} \tag{*1}$$

For large $n$, the LHS of $(*1)$ is dominated by the two terms with $|k| = 1$. This means for any $x \in (0,1)$, if one fixes the parity of $n$ and send $n$ to $\infty$, we will have

$$(-1)^{\lfloor n/2 \rfloor - 1} \frac{(2\pi)^n}{2(n!)} B_n(\{x\}) \quad\to\quad \begin{cases} \cos(2\pi x), &n \text{ even}\\ \sin(2\pi x), &n \text{ odd} \end{cases} \tag{*2} $$

Let us switch to the evaluation of the integral $u_n$ for $n > 0$.
Let $u = 1/x$, we have

$$\begin{align} &\int_0^1 B_n\left(\left\{ \frac1x \right\}\right) \frac{dx}{x} = \lim_{N\to\infty} \int_{1/N}^1 B_n\left(\left\{ \frac1x \right\}\right) \frac{dx}{x}\\ =& \lim_{N\to\infty} \int_1^N B_n(\{u\}) \frac{du}{u} = \lim_{N\to\infty} \sum_{j=1}^{N-1} \int_j^{j+1} B_n(\{u\}) \frac{du}{u}\\ =& \lim_{N\to\infty} \int_0^1 B_n(u) \sum_{j=1}^{N-1} \frac{1}{u+j} du = \lim_{N\to\infty} \int_0^1 B_n(u) \sum_{j=1}^{N-1} \left( \frac{1}{u+j} - \frac{1}{j} \right) du\\ =& \int_0^1 B_n(u) \sum_{j=1}^\infty \left( \frac{1}{u+j} - \frac{1}{j} \right) du \end{align} $$ Compare the series in the last integral with following expansion of digamma function

$$\psi(1+z) = -\gamma + \sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+z}\right)$$

we find for $n > 0$, $$\int_0^1 B_n\left(\left\{ \frac1x \right\}\right) \frac{dx}{x} = -\int_0^1 B_n(u) \psi(1+u) du\tag{*3}$$

Combine $(*2)$ and $(*3)$, we find for large $k$

$$u_{2k+1} \sim (-1)^{k-1}\frac{2C(2k+1)!}{(2\pi)^{2k+1}} \quad\implies\quad |u_{2k+1}| \to \infty \;\;\text{ as }\;\; k \to \infty $$

where $C$ is a constant defined by an integral $$\begin{align} C & = -\int_0^1 \sin(2\pi x) \psi(1+x) dx = \int_1^\infty \frac{\sin(2\pi x)}{x} dx = \frac{\pi}{2} - \text{Si}(2\pi)\\ & \approx 0.1526447506622681689855415293400020129561... \end{align}$$

We may obtain a closed form for this integral.

Proposition. Let $k=1,2,3,\ldots$. Then

$$ \int_0^1 \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx=(-1)^{k+1}\frac{(2k+1)!}{(2\pi)^{2k+1}}\pi \: \zeta(2k+1)-\sum_{j=0}^{2k}\!\frac{ {{2k+1}\choose j} B_j}{2k+1-j} \quad (*) $$

Proof. Recall the celebrated Fourier expansion (see this very nice paper, p. 9) $$ B_{2k+1}(\{x\}) = (-1)^{k+1}\frac{2(2k+1)!}{(2\pi)^{2k+1}}\sum_{p=1}^{\infty} \frac{\sin (2\pi px)}{p^{2k+1}}, \quad 0\leq x \leq 1, \tag1 $$

thus, on the one hand, you may write

$$\begin{align} \int_0^{+\infty} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx & = (-1)^{k+1}\frac{2(2k+1)!}{(2\pi)^{2k+1}}\sum_{p=1}^{\infty} \frac{1}{p^{2k+1}}\int_0^{+\infty} \frac{\sin (2\pi p\left\{ 1/x \right\})}{x}dx\\ & = (-1)^{k+1}\frac{2(2k+1)!}{(2\pi)^{2k+1}}\sum_{p=1}^{\infty} \frac{1}{p^{2k+1}}\int_0^{+\infty} \frac{\sin (2\pi p/x )}{x}dx\\ & = (-1)^{k+1}\frac{2(2k+1)!}{(2\pi)^{2k+1}}\sum_{p=1}^{\infty} \frac{1}{p^{2k+1}}\int_0^{+\infty} \frac{\sin (2\pi p\:x)}{x}dx\\ & = (-1)^{k+1}\frac{2(2k+1)!}{(2\pi)^{2k+1}}\sum_{p=1}^{\infty} \frac{1}{p^{2k+1}}\cdot\frac{\pi}{2}\\ & = (-1)^{k+1}\frac{2(2k+1)!}{(2\pi)^{2k+1}}\frac{\pi}{2} \: \zeta(2k+1). \tag2 \end{align}$$ On the other hand, you have $$\begin{align} \int_0^{+\infty} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx & = \int_0^{1} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx +\int_1^{+\infty} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx \\ & = \int_0^{1} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx +\int_1^{+\infty} \frac{B_{2k+1}\left(1/x\right)}{x}dx\\ & = \int_0^{1} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx +\int_0^{1} \frac{B_{2k+1}\left(x\right)}{x}dx \tag3 \\ & = \int_0^{1} \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx +\sum_{j=0}^{2k}\!\frac{ {{2k+1}\choose j} B_j}{2k+1-j}, \tag4 \end{align} $$ using $$ B_n(x)=\sum_{j=0}^{n}{{n}\choose j} B_j\:x^{n-k}. $$ Then $(4)$ together with $(2)$ gives $(*)$.

Using the uniform convergence on $[0,1]:$ $$ (-1)^{k+1} \frac{(2\pi)^{2k+1}}{2(2k+1)!} B_{2k+1}(x) \longrightarrow \sin(2\pi x) $$ (proved here, Corollary 1, pdf p. 3) it follows that $$ (-1)^{k+1} \int_0^1 \frac{B_{2k+1}\left(x\right)}{x}dx \sim \frac{2(2k+1)!}{(2\pi)^{2k+1}}\int_0^1 \frac{\sin(2\pi x)}{x}dx \tag5 $$ as $k \to +\infty$. Consequently, as $k \to +\infty$, combining $(5)$, $(3)$ and $(2)$, with $ \displaystyle \zeta(2k+1)=1+o\left(\frac1k\right)$, leads to $$ |u_k|=\left|\int_0^1 \frac{B_{2k+1}\left(\left\{ 1/x \right\}\right)}{x}dx\right| \sim \frac{2(2k+1)!}{(2\pi)^{2k+1}}\left( \frac{\pi}{2}-\int_0^1 \frac{\sin(2\pi x)}{x}dx\right) $$ which tends to $+\infty$.