Limit of $\int_0^1\frac1x B_{2n+1}\left(\left\{\frac1x\right\}\right)dx$ Set $$u_n= \int_0^1 \frac{B_{2n+1}\left(\left\{\frac1x\right\}\right)}{x}dx\tag1$$
where $\left\{t\right\}=t-\lfloor t \rfloor$ denotes the fractional part of $t$ and where $B_n(\cdot)$ are the Bernoulli polynomials:
$$ \begin{aligned}B_{1}(x)&=x-\frac12\\B_{3}(x)&=x^3-\frac32x^2+\frac12x\\\ldots\,&= \ldots\end{aligned}$$
I'm interested in finding $\lim\limits_{n \to +\infty}|u_n|.$ 
I've tried to see what Wolfram|Alpha gives with no success. My guess is that $$\lim\limits_{n \to +\infty}|u_n|=+\infty. \tag2$$
Could you prove/disprove $(2)$? Thank you.
 A: 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
$$\begin{align}
C & = -\int_0^1 \sin(2\pi x) \psi(1+x) dx  = \int_1^\infty \frac{\sin(2\pi x)}{x} dx = \frac{\pi}{2} - \text{Si}(2\pi)\\
& \approx 0.1526447506622681689855415293400020129561...
\end{align}$$
