Give an asymptotic developement of $I_n=\int_0^1 (x^{n}-x^{n-2})\ln(1+x^n)dx.$

Question

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 ?

Answer 1

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 $\, n^2I_n \!=\! \int_0^\infty\! nf_n(e^{-t/n}) e^{-t/n}t\, dt. \,$

Answer 2

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.

Answer 3

There is a difference in title $(\color{red}{-})$ and body $(\color{red}{+})$; so, I worked both cases.

Let us consider first $$\int x^a \log \left(1+x^b\right)\,dx=\frac{x^{a+1} \log \left(1+x^b\right)}{a+1}-\frac b{1+a} \int \frac{x^{a+b}}{1+x^b}\,dx$$ $$\int \frac{x^{a+b}}{1+x^b}\,dx=\frac{x^{a+b+1}}{a+b+1} \,\,\, _2F_1\left(1,\frac{a+b+1}{b};\frac{a+2b+1}{b};-x^b\right)$$ Using the given bounds and simplifying leads to $$K_{a,b}=\int_0^1 x^a \log \left(1+x^b\right)\,dx=\frac{\log (2)-\Phi \left(-1,1,\frac{a+b+1}{b}\right)}{a+1}\qquad \text{if} \qquad \Re(b)>0$$ where appears the Hurwitz-Lerch transcendent function.

So, $$I_n=\int_0^1 (x^{n}-x^{n-2})\ln(1+x^n)\,dx=\frac{\log (2)-\Phi \left(-1,1,\frac{2 n+1}{n}\right)}{n+1}-\frac{\log (2)-\Phi \left(-1,1,\frac{2 n-1}{n}\right)}{n-1}$$ $$J_n=\int_0^1 (x^{n}+x^{n-2})\ln(1+x^n)\,dx=\frac{\log (2)-\Phi \left(-1,1,\frac{2 n+1}{n}\right)}{n+1}+\frac{\log (2)-\Phi \left(-1,1,\frac{2 n-1}{n}\right)}{n-1}$$

Now, consider the expansion of $\Phi (-1,1,2+\epsilon)$; it is $$(1-\log (2))+\left(\frac{\pi ^2}{12}-1\right) \epsilon +\left(1-\frac{3 \zeta (3)}{4}\right) \epsilon ^2+\left(\frac{7 \pi ^4}{720}-1\right) \epsilon ^3+\left(1-\frac{15 \zeta (5)}{16}\right) \epsilon ^4+O\left(\epsilon ^5\right)$$ Making $ \epsilon=\pm \frac 1n$ and continuing with Taylor, this gives $$I_n=\frac{4-\frac{\pi ^2}{6}-4 \log (2)}{n^2}+\frac{-540 \zeta (3)+2880-60 \pi ^2-7 \pi ^4-1440 \log (2)}{360 n^4}+O\left(\frac{1}{n^5}\right)$$

$$J_n=\frac{4 \log (2)-2}{n}+\frac{9 \zeta (3)-36+\pi ^2+24 \log (2)}{6 n^3}+O\left(\frac{1}{n^5}\right)$$

For $J_n$, this is the same result as in marty cohen's answer, since $$\frac{9 \zeta (3)-36+\pi ^2+24 \log (2)}{6}=2 \times 0.110304071913700 $$

This seems to work quite well even for small values of $n$ as shown in the table below $$\left( \begin{array}{ccccc} n & I_n^{exact} & I_n^{approx} & J_n^{exact} & J_n^{approx}\\ 2 & -0.1120487783 & -0.1115478099 & 0.4158382364 & 0.4138703791 \\ 3 & -0.0478495232 & -0.0478071470 & 0.2659498411 & 0.2657002461 \\ 4 & -0.0265505658 & -0.0265431189 & 0.1966526582 & 0.1965941828 \\ 5 & -0.0168863305 & -0.0168843897 & 0.1563016585 & 0.1562826096 \\ 6 & -0.0116869861 & -0.0116863381 & 0.1297937520 & 0.1297861210 \\ 7 & -0.0085688906 & -0.0085686341 & 0.1110165137 & 0.1110129899 \\ 8 & -0.0065519051 & -0.0065517901 & 0.0970062707 & 0.0970044656 \\ 9 & -0.0051721371 & -0.0051720804 & 0.0861468097 & 0.0861458088 \\ 10 & -0.0041867254 & -0.0041866953 & 0.0774800710 & 0.0774794804 \end{array} \right)$$

Answer 4

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 $-J_{-n}$.