A friend of mine, obtained a lower bound for the trace norm of matrices described in this question (for the special case $a_{ij} = \pm 1$). That lower bound is $ \frac{f(n)}{2\pi}$ where $$ f(n) := \int_0^\infty \log\left( \frac{(1+t)^n +(1t)^n}{2} +n(n1) t(1+t)^{n2}\right)t^{ 3/2} \ \mathrm{d}t $$ Numerical computaions suggest that $$ f(n) = 4 \pi n + o(n) $$ How to justify it? Moreover, is it possible to obtain a good rate of convergence?

$\begingroup$ Does anything useful happen if you apply L'Hospital's rule and differentiate under the integral sign? $\endgroup$ – Nate Eldredge Jun 22 '18 at 15:25

$\begingroup$ @NateEldredge The function under integral divided by $n$ converges pointwise to $log(1+t)$ but it seems that there is no dominating function to use convergence theorems for integrals. $\endgroup$ – Mahdi Jun 22 '18 at 16:13

4$\begingroup$ Can't you just consider the difference $f(n)  4n\pi = f(n)  \int_0^\infty \log((1+t)^n)t^{3/2}\,dt$, combined into a single integral that looks like $\int_0^\infty \log(g(n,t))t^{3/2}\,dt$, where $g(n,t)$ is bounded below away from $0$ and bounded above by say $n^2$? That would show that $f(n) = 4n\pi + O(\log n)$. $\endgroup$ – Greg Martin Jun 23 '18 at 6:54

3$\begingroup$ @GregMartin In fact using this idea and and a change of variable $nt=u$ gives $f(n)=4\pi n +O(1)$. $\endgroup$ – Mostafa Jun 23 '18 at 7:56

1$\begingroup$ numerically it seems that $f(n)=4\pi n6\pi+O(1/\sqrt{n})$ $\endgroup$ – Henri Cohen Jun 23 '18 at 18:18
This is an improvement of my previous post. I claim that $$4\pi n6\pi<f(n)<4\pi n2\pi.$$ Starting from $$f(n)2\pi(n1)=\int_0^\infty \log\left(\frac{1+t}{2}+\frac{1+t}{2}\left(\frac{1t}{1+t}\right)^n+n(n1)\frac{t}{1+t}\right)\,t^{3/2}\,dt,$$ we see that $$f(n)2\pi(n1)<\int_0^\infty \log\bigl(1+n^2 t\bigr)\,t^{3/2}\,dt=2\pi n.$$ Hence the upper bound $f(n)<4\pi n2\pi$ follows. For the lower bound, we assume $n\geq 2$ without loss of generality, and we start from $$f(n)2\pi(n2)=\int_0^\infty \log\left(\frac{(1+t)^2}{2}+\frac{(1+t)^2}{2}\left(\frac{1t}{1+t}\right)^n+n(n1)t\right)\,t^{3/2}\,dt.$$ Combining this with the inequality $$\frac{(1+t)^2}{2}+\frac{(1+t)^2}{2}\left(\frac{1t}{1+t}\right)^n>1(n1)t,$$ which can be verified for $t<1/(n1)$ and $t\geq 1/(n1)$ separately, we see that $$f(n)2\pi(n2)>\int_0^\infty \log\bigl(1+(n1)^2t\bigr)\,t^{3/2}\,dt=2\pi(n1).$$ Hence the lower bound $f(n)>4\pi n6\pi$ follows also.