TITLE: Limit of $\int_0^1\frac1x B_{2n+1}\left(\left\{\frac1x\right\}\right)dx$
QUESTION [11 upvotes]: 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.

REPLY [5 votes]: 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 $+\infty$.