<s>[INST] 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 ?
 [/INST] 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}
$
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0.110304071913.
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