How do you prove the validity of this formula for $H(n)$? [closed]

Question

I'm looking for a proof of the identity

$$\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{2n}+\pi\int_{0}^{1} (1-u)\cot{\pi u}\left(1-\cos{2\pi n u}\right)\,du.$$

There is a generalization of this formula for $H_k(n)$, if somebody can provide me with insights on how it's derived, I'd really appreciate.

Remember there is really no nice formulas for $H(n)$, this is the very first one, so this deserves some effort from my fellow MOers.

So here's the real challenge: $H_{2k+1}(n)=\frac{1}{2n^{2k+1}}-\frac{(-1)^{k}(2\pi)^{2k+1}}{2}\int_{0}^{1}\sum_{j=0}^{k}\frac{B_{2k-2j}\left(2-2^{2k-2j}\right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n)\,du$,

where $f(u,n)=\cot{\pi u}\left[1-\cos{2\pi n(1-u)}\right]$

Answer 1

The OP's second display is a variant of an identity that appears in Section 3.7 of this recent preprint. It can be proved directly as follows (I will rename $k$ to $\ell$ in order to avoid confusion with the summation variable $k$ in the OP's first display). First, it is straightforward to see, e.g. by induction or by summing the relevant finite geometric series on the unit circle, that $$\sum_{k=1}^n\sin 2\pi ku =\frac{1}{2}\sin 2\pi nu+\frac{1}{2}(\cot{\pi u})(1-\cos{2\pi n u}).$$ Therefore, introducing the polynomials $$P_\ell(u):=(-1)^\ell\sum_{j=0}^{\ell}\frac{B_{2\ell-2j}\left(2-2^{2\ell-2j}\right)}{(2\ell-2j)!(2j+1)!}u^{2j+1},\qquad\ell\geq 0,$$ it suffices to prove that $$\int_0^1 P_\ell(u)(\sin 2\pi ku)\,du=\frac{-1}{(2\pi k)^{2\ell+1}},\qquad k\geq 1,\quad\ell\geq 0.\tag{1}$$ To see this, we proceed by induction on $\ell$. For $\ell=0$, $(1)$ follows easily from $P_0(u)=u$. Let us now assume that $\ell\geq 1$ and $(1)$ holds with $\ell-1$ in place of $\ell$. Using the identity (which also appears in Section 3.7 of the mentioned preprint) $$\frac{\sin xu}{\sin x}=\sum_{\ell=0}^\infty P_\ell(u)x^{2\ell},\qquad |x|<\pi,\tag{2}$$ we see readily that $P_\ell(0)=P_\ell(1)=0$ (for $\ell\geq 1$). Therefore, integrating by parts twice, \begin{align*}\int_0^1 P_\ell(u)(\sin 2\pi ku)\,du &=\frac{1}{2\pi k}\int_0^1 P'_\ell(u)(\cos 2\pi ku)\,du\\[8pt] &=\frac{-1}{(2\pi k)^2}\int_0^1 P''_\ell(u)(\sin 2\pi ku)\,du\\[8pt] &=\frac{1}{(2\pi k)^2}\int_0^1 P_{\ell-1}(u)(\sin 2\pi ku)\,du\\[8pt] &=\frac{1}{(2\pi k)^2}\cdot\frac{-1}{(2\pi k)^{2\ell-1}}=\frac{-1}{(2\pi k)^{2\ell+1}}. \end{align*} The proof is complete.

Added. For an alternative proof of $(1)$, observe that for $|x|<\pi$ we have by $(2)$ \begin{align*}\sum_{\ell=0}^\infty\left(\int_0^1 P_\ell(u)(\sin 2\pi ku)\,du\right)x^{2\ell}&=\int_0^1\left(\frac{\sin xu}{\sin x}\right)(\sin 2\pi ku)\,du\\[6pt] &=\frac{-2\pi k}{(2\pi k)^2-x^2}\\[6pt] &=\sum_{\ell=0}^\infty\frac{-x^{2\ell}}{(2\pi k)^{2\ell+1}}. \end{align*} Now compare the two sides.