An integral relating to Bernoulli polynomials

Question

Show that $$\int_{0}^{1}B_{2n+1}(x)(\cot({\pi}x)-2\sin(2{\pi}x))dx{\sim}0$$ where $B_{2n+1}(x)$ is the Bernoulli polynomials.

Answer 1

The statement is false. To show this we'll use a stronger estimate than the one used in this answer to your last question.

In equation $(2.11)$ of this paper (ScienceDirect link) the author gives the Fourier expansion

$$ B_{2n+1}(x) = 2(-1)^{n+1} (2n+1)! \sum_{k=1}^{\infty} \frac{\sin(2\pi k x)}{(2\pi k)^{2n+1}}, $$

which also serves as an asymptotic expansion when $n \to \infty$.

Integrating over only the first two terms yields

$$ \int_0^1 B_{2n+1}(x) (\cot(\pi x) - 2\sin(2\pi x))\,dx \sim (-1)^{n+1} \frac{2(2n+1)!}{(4\pi)^{2n+1}}, $$

which does not tend to $0$ as $n \to \infty$.

Here's a plot over the range $1 \leq n \leq 30$ with a logarithmic scale on the vertical axis, comparing the absolute values of the integral evaluated numerically (blue dots) and this asymptotic (red line).