A friend of mine, obtained a lower bound for the trace norm of matrices described in [this question](https://mathoverflow.net/questions/302424/is-this-lower-bound-for-a-norm-of-some-complex-matrices-true) (for the special case $a\_{ij} = \pm 1$). That lower bound is $ \frac{f(n)}{2\pi}$ where
$$
f(n) := \int\_0^\infty \log\left( \frac{(1+t)^n +(1-t)^n}{2} +n(n-1) t(1+t)^{n-2}\right)t^{- 3/2} \ \mathrm{d}t
$$
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?



Asymptotic behavior of an integral depending on an integer
This is an improvement of my previous post. I claim that
$$4\pi n-6\pi<f(n)<4\pi n-2\pi.$$
Starting from
$$f(n)-2\pi(n-1)=\int\_0^\infty \log\left(\frac{1+t}{2}+\frac{1+t}{2}\left(\frac{1-t}{1+t}\right)^n+n(n-1)\frac{t}{1+t}\right)\,t^{-3/2}\,dt,$$
we see that
$$f(n)-2\pi(n-1)<\int\_0^\infty \log\bigl(1+n^2 t\bigr)\,t^{-3/2}\,dt=2\pi n.$$
Hence the upper bound $f(n)<4\pi n-2\pi$ follows. For the lower bound, we assume $n\geq 2$ without loss of generality, and we start from
$$f(n)-2\pi(n-2)=\int\_0^\infty \log\left(\frac{(1+t)^2}{2}+\frac{(1+t)^2}{2}\left(\frac{1-t}{1+t}\right)^n+n(n-1)t\right)\,t^{-3/2}\,dt.$$
Combining this with the inequality
$$\frac{(1+t)^2}{2}+\frac{(1+t)^2}{2}\left(\frac{1-t}{1+t}\right)^n>1-(n-1)t,$$
which can be verified for $t<1/(n-1)$ and $t\geq 1/(n-1)$ separately, we see that
$$f(n)-2\pi(n-2)>\int\_0^\infty \log\bigl(1+(n-1)^2t\bigr)\,t^{-3/2}\,dt=2\pi(n-1).$$
Hence the lower bound $f(n)>4\pi n-6\pi$ follows also.


