# Bernoulli number type asymptotics

I find an interesting formula but I can not prove it. Show that
$$I_n=(-1)^{n+1}\int_0^1 B_{2n+1}(x)\cot(\pi x) \, dx\sim\frac{2(2n+1)!}{(2\pi)^{2n+1}}$$
where $B_n(x)$ is the Bernoulli Polynomials.

#### Solutions Collecting From Web of "Bernoulli number type asymptotics"

According to Corollary 1 of [1],

$$(-1)^{n+1} \frac{(2\pi)^{2n+1}}{2(2n+1)!} B_{2n+1}(x) \longrightarrow \sin(2\pi x)$$

uniformly on compact subsets of $\mathbb{C}$. It follows that

\begin{align} (-1)^{n+1} \frac{(2\pi)^{2n+1}}{2(2n+1)!} \int_0^1 B_{2n+1}(x) \cot(\pi x)\,dx &\to \int_0^1 \sin(2\pi x)\cot(\pi x)\,dx \\ &= 2\int_0^1 \sin(\pi x)\cos(\pi x)\cot(\pi x)\,dx \\ &= 1, \end{align}

and so

$$(-1)^{n+1} \int_0^1 B_{2n+1}(x) \cot(\pi x)\,dx \sim \frac{2(2n+1)!}{(2\pi)^{2n+1}}$$

as $n \to \infty$.

[1] Dilcher, K. Asymptotic behaviour of Bernoulli, Euler, and generalized Bernoulli polynomials. (ScienceDirect link)