<s>[INST] Asymptotics for $\sum_{k=0}^{n-1}(-1)^k \cot(\frac{2k+1}{4n}\pi)\log{(2\sin(\frac{2k+1}{4n}\pi))} $ I have been able to establish the identity
$$
I(n):=\int_0^1 \frac{y^{n-1}(1+y-y^{n+1}) - 1}{(1-y)(1+y^{2n})}\ dy = \frac{1}{n} \sum_{k=0}^{n-1}(-1)^k\cot(\frac{2k+1}{4n}\pi)\log{(2\sin(\frac{2k+1}{4n}\pi))}
$$
However, neither side seems conducive to discovering an asymptotic expansion as $n \to \infty.$
Purely on numerical evidence I conjecture that $I(n) \sim -\log(n).$
$$
\begin{array}{c|lcr}
n & I(n) &-\log{n} & -\log{n}/I(n)  \\
\hline
10^3 & -6.702 & -6.908 & 1.031  \\
10^4 & -9.004 & -9.2106 &1.023  \\
10^5 & -11.307 &-11.513 & 1.018   \\
\end{array}
$$
I seek the dominant and one subdominant asymptotic terms. (If the Lambert W function is involved, then maybe one term is sufficient.) Here are some ideas.
(1) On the LHS (left-hand side), letting $n \to \infty,$ the $y^n$ terms integrand go to zero and you are left with $-1/(1-y),$ and the integral over it will diverge.  However, plotting the integrand for large $n$, it is seen that the integrand follows the $-1/(1-y)$ curve until it sharply drops to zero near $y=1.$  It is probable that the position of the drop, $y_d,$ is analytically tractable, and one can approximate the integral as
$-\int_{0}^{y_d}dy/(1-y).$  However, I have no idea on how to approach a subdominant term.
(2) The RHS looks sort of like a Riemann sum, except for the pesky alternating sign, and the fact that odd integers appear in the arguments instead of every integer.  A long time ago I read that the Euler-Maclaurin formula has been extended to character sums, which this looks like.  I also know that Euler-Maclaurin can be used for asymptotic analysis.  A solution in this manner would be most edifying, since I know so little about it.
 [/INST] Note that
$$
I(n) = \int_0^1 {\frac{{(t^n  - 1)^2 }}{{(t - 1)(t^{2n}  + 1)}}dt}  + \int_0^1 {\frac{{t^{n - 1} }}{{t^{2n}  + 1}}dt} .
$$
Here
$$
\int_0^1 {\frac{{t^{n - 1} }}{{t^{2n}  + 1}}dt}  \le \int_0^1 {t^{n - 1} dt}  = \frac{1}{n}.
$$
Now
\begin{align*}
\int_0^1 {\frac{{(t^n  - 1)^2 }}{{(t - 1)(t^{2n}  + 1)}}dt} & =  
 - \int_0^1 {\frac{{1 - t^n }}{{1 - t}}dt} + \int_0^1 {\frac{{1 - t^{2n} }}{{1 + t^{2n} }}\frac{{t^n }}{{1 - t}}dt} \\ & =  - \sum\limits_{k = 1}^n {\frac{1}{k}}  +\int_0^1 {\frac{{1 - t^{2n} }}{{1 + t^{2n} }}\frac{{t^n }}{{1 - t}}dt}.
\end{align*}
We have
$$
 - \sum\limits_{k = 1}^n {\frac{1}{k}}  =  - \log n  -\gamma + \mathcal{O}\!\left(\frac{1}{n}\right).
$$
Furthermore,
\begin{align*}
&\int_0^1 {\frac{{1 - t^{2n} }}{{1 + t^{2n} }}\frac{{t^n }}{{1 - t}}dt}  = \int_0^{ + \infty } {\frac{{\tanh (ns)}}{{e^s  - 1}}e^{ - ns} ds} 
\\ &
 = \int_0^{ + \infty } {\frac{{t/n}}{{e^{t/n}  - 1}}e^{ - t} \frac{{\tanh t}}{t}dt} \\& 
 = \int_0^{ + \infty } {e^{ - t} \frac{{\tanh t}}{t}dt}  + \int_0^{ + \infty } {\left( {\frac{{t/n}}{{e^{t/n}  - 1}} - 1} \right)e^{ - t} \frac{{\tanh t}}{t}dt} .
\end{align*}
Here
\begin{align*}
\int_0^{ + \infty } {\left| {\frac{{t/n}}{{e^{t/n}  - 1}} - 1} \right|e^{ - t} \frac{{\tanh t}}{t}dt} & \le \int_0^{ + \infty } {\frac{t}{{2n}}e^{ - t} \frac{{\tanh t}}{t}dt} \\ & = \frac{1}{{2n}}\int_0^{ + \infty } {e^{ - t} \tanh tdt}  = \mathcal{O}\!\left( {\frac{1}{n}} \right),
\end{align*}
since $\left| {\frac{x}{{e^x  - 1}} - 1} \right| \le \frac{x}{2}
$ for all $x>0$. Thus, in summary,
$$I(n)=-\log n+C+\mathcal{O}\!\left( {\frac{1}{n}} \right)$$ as $n\to +\infty$, where
$$
C =  - \gamma  + \int_0^{ + \infty } {e^{ - t} \frac{{\tanh t}}{t}dt}  = -\gamma +2\log \left( {\frac{{2\Gamma \left( {\frac{5}{4}} \right)}}{{\Gamma \left( {\frac{3}{4}} \right)}}} \right)
=0.20597 \ldots \, .
$$
It is possible to obtain a complete asymptotic expansion as follows. We observe that
$$
\int_0^1 \frac{{t^{n - 1} }}{{t^{2n}  + 1}} =  \frac{\pi }{{4n}}.
$$
It is known that
$$
 - \sum\limits_{k = 1}^n {\frac{1}{k}} \sim - \log n - \gamma  - \frac{1}{{2n}} + \sum\limits_{k = 1}^\infty  {\frac{{B_{2k} }}{{2kn^{2k} }}} ,
$$
where $B_{2k}$ denotes the Bernoulli numbers. Also, using the generating function of the Bernoulli numbers, it is found that
\begin{align*}
& \int_0^{ + \infty } {\left( {\frac{{t/n}}{{e^{t/n}  - 1}} - 1} \right)e^{ - t} \frac{{\tanh t}}{t}dt} \\ & \sim  - \frac{1}{{2n}}\int_0^{ + \infty } {e^{ - t} \tanh tdt}  + \sum\limits_{k = 1}^\infty  {\left[ {\frac{{B_{2k} }}{{(2k)!}}\int_0^{ + \infty } {t^{2k-1} e^{ - t} \tanh tdt} } \right]\frac{1}{{n^{2k} }}} .
\\ &
 =  - \frac{{\pi  - 2}}{{4n}} + \sum\limits_{k = 1}^\infty  {\frac{B_{2k}}{2k}\frac{ \zeta \left(2k,\frac{1}{4}\right)  -2 \zeta \left( {2k ,\frac{3}{4}} \right) + \zeta \left( {2k,\frac{5}{4}} \right) }{{16^k }}\frac{1}{{n^{2k} }}} .
\end{align*}
Here $\zeta(s,a)$ is the Hurwitz zeta function. Therefore,
\begin{align*}
I(n) & \sim - \log n + C + \sum\limits_{k = 1}^\infty  {\frac{B_{2k}}{2k}\left( 1+\frac{ \zeta \left(2k,\frac{1}{4}\right)  -2 \zeta \left( {2k ,\frac{3}{4}} \right) + \zeta \left( {2k,\frac{5}{4}} \right) }{{16^k }}\right)\frac{1}{{n^{2k} }}} 
\\ & = - \log n + C + \sum\limits_{k = 1}^\infty
\frac{{B_{2k} }}{{16^k k}}\left( {\zeta \left( {2k,\tfrac{1}{4}} \right) - \zeta \left( {2k,\tfrac{3}{4}} \right)} \right)
\frac{1}{{n^{2k} }},
\end{align*}
as $n\to +\infty$.
</s>