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.


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 $+\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.