# Simplifying $\prod\limits_{k\neq j=0}^{n-1}\frac1{\lambda_{n,k}-\lambda_{n,j}}$ for $\lambda_{n,k}=\exp\frac{i\pi(2k+1)}{n}$

I have been able to show that for $$n\in\Bbb N_{\geq2}$$ $$\phi(n)=\int_0^1\frac{dx}{x^n+1}=\sum_{k=0}^{n-1}\Gamma_{n,k}\log\frac{\lambda_{n,k}-1}{\lambda_{n,k}}$$ Where $$\lambda_{n,k}=\exp\frac{i\pi(2k+1)}{n}$$ And $$\Gamma_{n,k}=\prod_{k\neq j=0}^{n-1}\frac1{\lambda_{n,k}-\lambda_{n,j}}$$ And I was wondering: how do we simplify $$\Gamma_{n,k}$$ to ease the manual calculation of $$\phi(n)$$ values. The integral is always real, so I am sure there is a major way we can simplify $$\Gamma_{n,k}$$, but I have been so far unable to find it. I do suspect however that the product $$P_n=\prod_{k=0}^{n-1}\Gamma_{n,k}$$ May play a significant role in finding the simplification I seek.

For those interested, a proof.

Note that $$x^n+1$$ bay be factored as $$x^n+1=\prod_{k=0}^{n-1}(x-\lambda_{n,k})$$ Hence $$\phi(n)=\int_0^1\prod_{k=0}^{n-1}\frac1{x-\lambda_{n,k}}dx$$ Then define $$\Gamma_{n,k}$$ by saying that $$\prod_{k=0}^{n-1}\frac1{x-\lambda_{n,k}}=\sum_{k=0}^{n-1}\frac{\Gamma_{n,k}}{x-\lambda_{n,k}}$$ Multiplying both sides by $$\prod_{j=0}^{n-1}(x-\lambda_{n,j})$$: $$1=\sum_{k=0}^{n-1}\frac{\Gamma_{n,k}}{x-\lambda_{n,k}}\prod_{j=0}^{n-1}(x-\lambda_{n,j})$$ $$1=\sum_{k=0}^{n-1}\Gamma_{n,k}\prod_{k\neq j=0}^{n-1}(x-\lambda_{n,j})$$ So for any integer $$0\leq m\leq n-1$$ we may plug in $$x=\lambda_{n,m}$$ and simplify to get $$\Gamma_{n,m}=\prod_{m\neq j=0}^{n-1}\frac1{\lambda_{n,m}-\lambda_{n,j}}$$ And our result follows directly.

Perhaps another motivation for easing manual calculation of this product would be that $$\sum_{k=0}^{\infty}\frac{(-1)^k}{nk+1}=\phi(n)$$ Which brings about a plethora of interesting closed forms.

Edit: A little progress

We define $$c_{n,j}=\operatorname{Re}\lambda_{n,j}=\cos\frac{\pi(2j+1)}{n}$$ And $$s_{n,j}=\operatorname{Im}\lambda_{n,j}=\sin\frac{\pi(2j+1)}{n}$$ So $$\log\frac{\lambda_{n,k}-1}{\lambda_{n,k}}=\log\left(1-\lambda_{n,k}^{-1}\right)=\log\left(1-c_{n,k}+is_{n,k}\right)$$ And we also see that \begin{align} \prod_{k\neq j=0}^{n-1}\frac1{\lambda_{n,k}-\lambda_{n,j}}&=\prod_{k\neq j=0}^{n-1}\frac1{e^{i\pi(2k+1)/n}-e^{i\pi(2j+1)/n}}\\ &=\prod_{k\neq j=0}^{n-1}\frac{e^{-i\pi(2k+1)/n}}{1-e^{i\pi(2j-2k)/n}}\\ &=e^{i(2k+1)(2-n)/n}\prod_{k\neq j=0}^{n-1}\frac12\left(1+i\cot\frac{\pi(j-k)}n\right)\\ \Gamma_{n,k}&=\frac{\lambda_{n,k}^{2-n}}{2^{n-2}}\prod_{k\neq j=0}^{n-1}\left(1+i\cot\frac{\pi(j-k)}n\right) \end{align} But the remaining product I do not know how to deal with.

Defining the polynomial \begin{align} P(x)&=x^n+1\\ &=\prod_{j=0}^{n-1}\left( x- \lambda_{n,j}\right) \end{align} we can express its derivative at $$x=\lambda_{n,k}$$ as: \begin{align} P'(\lambda_{n,k})&=\prod_{k\neq j=0}^{n-1}\left( \lambda_{n,k}-\lambda_{n,j} \right)\\ &=\frac{1}{\Gamma_{n,k}} \end{align} But we have also $$P'(x)=nx^{n-1}=n\tfrac{x^n}{x}$$. Thus, as $$\left(\lambda_{n,k} \right)^n=-1$$, $$\begin{equation} P'(\lambda_{n,k})=n\frac{-1}{\lambda_{n,k}} \end{equation}$$ Finally, $$\begin{equation} \Gamma_{n,k}=-\frac{\lambda_{n,k}}{n} \end{equation}$$ This trick comes rather naturally if the integral is evaluated by the residue method, for the function $$f(z)=(1+z^n)^{-1}\ln\left(\tfrac z{1-z}\right)$$ along the keyhole contour.
In fact, if we may write a function $$f$$ as a product over its roots, i.e. $$f(x)\equiv\prod_{f(\omega)=0}(x-\omega)$$ where each root $$\omega$$ contributes exactly one term, then we may also write $$\frac1{f(x)}=\sum_{f(\omega)=0}\frac{b(\omega)}{x-\omega}$$ for some coefficients $$b(\omega)$$ which we can show to be $$b(\omega)=\prod_{{f(\alpha)=0}\atop{\alpha\ne \omega}}\frac1{\omega-\alpha}.$$ At the same time however, we have that the equality $$f'(x)=\sum_{f(\omega)=0}\frac{f(x)}{x-\omega}=\sum_{f(\omega)=0}\prod_{{f(\alpha)=0}\atop{\alpha\ne \omega}}(x-\alpha)$$ holds under the assumption that $$f(x)=0\Rightarrow f'(x)\ne0$$.
So for any root $$\phi$$ we plug in $$x=\phi$$ to see that $$f'(\phi)=\prod_{{f(\alpha)=0}\atop{\alpha\ne \phi}}(\phi-\alpha)$$ which implies that $$b(\omega)=\frac{1}{f'(\omega)}$$ and $$\frac1{f(x)}=\sum_{f(\omega)=0}\frac1{(x-\omega)f'(\omega)}.$$ Then @PaulEnta's results are easily derived form there.