Bernoulli number type asymptotics

Question

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.

Answer 1

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 $n \to \infty$.

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