Asymptotic expansion of $ u_n = \int_0^1 \ln(1+t^n) \mathrm dt $ I would like to know how I can compute the asymptotic expansion of:
$$ u_n = \int_0^1 \ln(1+t^n) \mathrm dt $$
Using the dominated convergence theorem, we get:
$$ u_n \sim \frac{\pi^2}{12n}$$
How can I get the next terms of the asymptotic expansion?
 A: First note that we have $$\log(1+t^n) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1} t^{nk}}{k}$$
Hence,
$$I(n) = \int_0^1 \log(1+t^n) dt = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}k \int_0^1 t^{nk} dt$$
We have
$$\int_0^1 t^{nk} dt = \dfrac1{nk+1}$$
This gives us
$$I(n) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k(nk+1)} = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{nk^2(1+1/(nk))}$$
Use geometric/Taylor series to get that$$\dfrac1{1 + \dfrac1{nk}} = \sum_{l=0}^{\infty} \left(- \dfrac1{nk}\right)^l$$
Hence,
\begin{align}
I(n) & = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{nk^2} \left( \sum_{l=0}^{\infty} \left(- \dfrac1{nk}\right)^l\right) = \sum_{l=0}^{\infty} \dfrac1n \left(- \dfrac1n\right)^l \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^{2+l}}
\end{align}
Now note that
$$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^{2+l}} = \left(\dfrac{2^{l+1}-1}{2^{l+1}} \right) \zeta(l+2)$$
Hence, we get the asymptotic expansion of $I(n)$ as
$$\color{red}{\boxed{
I(n) = \displaystyle \sum_{l=0}^{\infty} \dfrac{(-1)^l}{n^{l+1}}\left(\dfrac{2^{l+1}-1}{2^{l+1}} \right) \zeta(l+2)}}$$
Considering the $l=0$ term, we get the leading order estimate as you have:
$$\color{blue}{\boxed{
I(n) \sim \dfrac{(-1)^0}{n^{0+1}}\left(\dfrac{2^{0+1}-1}{2^{0+1}} \right) \zeta(0+2) = \dfrac{1}{n}\left(\dfrac{2-1}{2} \right) \zeta(2) = \dfrac{\zeta(2)}{2n} = \dfrac{\pi^2}{12n}}}$$
