Give an asymptotic developement of $I_n=\int_0^1 (x^{n}-x^{n-2})\ln(1+x^n)dx.$ Let $$I_n=\int_0^1 (x^n+x^{n-2})\ln(1+x^n)dx.$$
Give an asymptotic developement of $I_n$ at order $O(\frac{1}{n^3})$ when $n\to \infty $.
I wanted to use the fact that $$\sum_{k=1}^\infty \frac{(-1)^{k+1} x^{kn}}{k},$$
and thus $$I_n=\int_0^1(x^n+x^{n-2})\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{kn}dx,$$
but since the convergence of this series is a priori not uniform and the coefficient are not positive on $[0,1]$ I can't permute the sum and the integral. 
Any other idea ?
 A: If $\, f_n(x) := (x^n - x^{n-2}) \log(1+x^n), \,$ then
 $\, n f_n(e^{-x/n}) \to -2xe^{-x}\log(1+e^{-x}) \,$ as $\, n\to \infty. \,$
Use $\, x = e^{-t/n} \,$ in $\, I_n := \int_0^1\! f_n(x) \, dx \,$ with $\, dx = -e^{-t/n}t/n \,$ so
 $\, n^2I_n \!=\! \int_0^\infty\! nf_n(e^{-t/n}) e^{-t/n}t\, dt. \,$
A: Assuming that
you can reverse the order,
$\begin{array}\\
I_n
&=\int_0^1(x^n+x^{n-2})\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{kn}dx\\
&=\sum_{k=1}^\infty \int_0^1(x^n+x^{n-2})\frac{(-1)^{k+1}}{k}x^{kn}dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} \int_0^1(x^n+x^{n-2})x^{kn}dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} \int_0^1(x^{n(k+1)}+x^{n(k+1)-2})dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}(\frac1{n(k+1)+1}+\frac1{n(k+1)-1})\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}(\frac{2n(k+1)}{n^2(k+1)^2-1})\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{kn}(\frac{2(k+1)}{(k+1)^2-1/n^2})\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)n}(\frac{2}{1-1/(n^2(k+1)^2)})\\
&=\frac{2}{n}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)}\sum_{m=0}^{\infty}\frac{1}{n^{2m}(k+1)^{2m}}\\
&=\frac{2}{n}\sum_{m=0}^{\infty}\frac{1}{n^{2m}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)}\frac{1}{(k+1)^{2m}}\\
&=2\sum_{m=0}^{\infty}\frac{1}{n^{2m+1}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)^{2m+1}}\\
&=2(\frac{\ln(4/e)}{n}-\frac{0.110304071913...}{n^3}+O(\frac1{n^5}))
\qquad\text{(according to Wolfy)}\\
\end{array}
$
Note:
The  Inverse Symbolic Calculator 
did not find anything for
0.110304071913.
A: If $f(n, x) = O(|g(n, x)|)$ uniformly in $x$ on $a < x < b$ and $f$ and $g$ are measurable, then $\int_a^b f(n, x) dx = O(\int_a^b |g(n, x)| dx)$. We have
$$J_{n} = \int_0^1 x^n \ln(1 + x^n) dx =
\frac 1 n \int_0^1 \xi^{1/n} \ln(1 + \xi) d\xi, \\
|\xi^{1/n} - 1| \leq \frac 1 n |\! \ln \xi| \quad
 \text{when } 0 < n \land 0 < \xi < 1, \\
\left| J_n - \frac 1 n \int_0^1 \ln(1 + \xi) d\xi \right| \leq
 -\frac 1 {n^2} \int_0^1 \ln \xi \ln(1 + \xi) d\xi,$$
giving the leading term in the expansion. The complete asymptotic series for $J_n$ in powers of $n$ is
$$J_n \sim \sum_{k=1}^\infty
 \int_0^1 \ln^{k-1} \xi \ln(1 + \xi) d\xi \frac {n^{-k}} {(k-1)!}.$$
Similarly, for
$$K_n = \int_0^1 x^{n-2} \ln(1 + x^n) dx =
\frac 1 n \int_0^1 \xi^{-1/n} \ln(1 + \xi) d\xi,$$
we can take
$$|\xi^{-1/n} - 1| \leq \frac 2 {n \sqrt \xi} \quad
 \text{when } 2 < n \land 0 < \xi < 1.$$
The complete asymptotic series for $K_n$ is
$$K_n \sim \sum_{k=1}^\infty (-1)^{k-1}
 \int_0^1 \ln^{k-1} \xi \ln(1 + \xi) d\xi \frac {n^{-k}} {(k-1)!},$$
which is formally $-J_{-n}$.
