Infinity norm of matrix defined using integral of Lagrange polynomials. Fix $n \geq 2$ an integer and let $x_{k+1} = k/n , k = 0, 1,\cdots, n-1.$
Let now $L_k$ be the Lagrange polynomial given by
$$L_k(x) = \prod_{j =1, j \neq k}^{n} \frac{x-x_j}{x_k - x_j}.$$
Define the matrix $A = [a_{i,j}]$ whose entries are given by
$$a_{i,j} := \int_{0}^{x_i}L_j(x)\, dx.$$
I'm interested in finding the infinity norm of $A$
$$\|A\|_\infty := \max_{ 1\leq i\leq n} \sum_{j=1}^n |a_{i,j}|.$$
My attempt: It seems (I used Mathematica for $n = 2, ..., 8$) that the maximum is reached in the last line of the matrix where we have $a_{n,j} \geq 0.$ Thus,  we have
$$\|A\|_\infty = \sum_{j=1}^n |a_{n,j}| = \sum_{j=1}^n\int_{x_1}^{x_n}L_j(x).$$
It's clear that $\sum_{j=1}^nL_j(x) =1.$ Then, we have
$$\|A\|_\infty = n-1.$$
Thank you for any hint.
 A: If you let
\begin{equation}
   \Lambda_n := \max_{0 \leq x \leq 1} \sum_{j=1}^{n} \left| l_j(x) \right|
\end{equation}
denote the Lebesgue constant, then it is easy to show that
\begin{equation}
  \left\| A \right\|_{\infty} \leq \Lambda_n.
\end{equation}
While there are few direct formulas for the Lebesgue constant, there are a few estimates on how it grows with $n$. For equidistant points, in particular, you have
\begin{equation}
  \Lambda_n = \frac{2^n}{e n \log(n)} \ \text{as} \ n \to \infty.
\end{equation}
I believe this bound is extremely pessimistic, but I was not able to significantly improve it so far. Hence my interest in the source for your hypothesis that $\left\| A \right\|_{\infty} = n - 1$.
Addendum: I don't think the hypothesis holds for equidistant nodes. If I set $n=15$, I get $\left\| A \right\|_{\infty} \approx 20.3$. From there, the norm very quickly grows.
Eisinberg, A.; Fedele, G.; Franzè, G., Lebesgue constant for Lagrange interpolation on equidistant nodes, Anal. Theory Appl. 20, No. 4, 323-331 (2004). ZBL1069.41002.
