Asymptotic expansion at order 2 of $\int_0^1 \frac{x^n}{1+x} \, dx$

I'd like to get an asymptotic expansion of $\int_0^1 \frac{x^n}{1+x} \, dx$ at order two in $\frac{1}{n}$.

I'm able to prove that $$\lim\limits_{n \to \infty} n \int_0^1 \frac{x^n}{1+x} \, dx = \frac{1}{2}$$ which provides an asymptotic expansion at order $1$. How can I go one step further? Even better, is there a way to get an asymptotic expansion at any order $m$?

• For maximum $M_n=\max(U_1,\dots,U_n)$ of $n$ standard uniforms, $\mu_n=E(M_n)=\frac n{n+1}$ and $\sigma_n^2=var(M_n)=\frac n{(n+1)^2(n+2)}$ rewrite $$I_{n}=\frac 1 {n+1}E\left(\frac 1 {1+M_{n+1}}\right)\approx \frac 1 {n+1}\left(\frac 1{1+\mu_{n+1}}+\frac {\sigma_{n+1}^2}{4(1+\mu_{n+1})^3}\right)$$ and expand. – A.S. Dec 26 '15 at 19:41
• Another method: Substitute $x = e^{-t}$ and apply Watson's lemma. – Antonio Vargas Dec 26 '15 at 23:56

Method 1. An elementary approach. You may just integrate by parts twice, \begin{align} I_n&=\int_0^1\frac{x^n}{1+x}\:dx \\&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{1+x}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\ &=\frac1{2(n+1)}+\frac{1}{n+1}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\ &=\frac1{2(n+1)}+\frac{1}{n+1}\left(\left. \frac{x^{n+2}}{(n+2)}\frac{1}{(1+x)^2}\right|_0^1+\frac{2}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx \right)\\ &=\frac1{2(n+1)}+\frac1{4(n+1)(n+2)}+\frac{2}{(n+1)^2}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx \end{align} but $$0\leq \int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx\leq\int_0^1x^{n+1}dx=\frac{1}{(n+2)}$$ thus $$\frac{2}{(n+1)^2}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx=\mathcal{O}\left(\frac{1}{n^3} \right)$$ Finally, as $n \to \infty$,

$$I_n=\int_0^1\frac{x^n}{1+x}dx=\frac1{2n}-\frac{1}{4n^2}+\mathcal{O}\left(\frac{1}{n^3} \right).$$

 Method 2. One may use the standard integral representation of the digamma function and its asymptotics, as $n \to \infty$,

\begin{align} I_n=\int_0^1\frac{x^n}{1+x}\:dx&=\int_0^1\frac{x^n-x^{n+1}}{1-x^2}\:dx\\ &=\frac12\int_0^1\frac{(1-t^{n/2})-(1-t^{(n-1)/2})}{1-t}\:dt\\ &=\frac12\psi\left(\frac{n}{2}+1\right)-\frac12\psi\left(\frac{n}{2}+\frac12\right)\\ &=\frac1{2n}-\frac{1}{4n^2}+\mathcal{O}\left(\frac{1}{n^3} \right). \end{align}

• Yes, I know. I was only suggesting that you consider adding an explanation to enhance your answer. Happy Holidays! - Mark – Mark Viola Dec 26 '15 at 19:49
• Thanks Mark! In fact, one can see that the remainder $r_n$ is such that $0<r_n<\frac2{(n+1)^2(n+2)}$, which is sufficient to qualify it as $\displaystyle \mathcal{O}\left(\frac{1}{n^3} \right)$. – Olivier Oloa Dec 26 '15 at 19:57
• You're welcome! Olivier I consider you one of the "good guys" here on MSE. I was suggesting that some readers might really benefit by your adding this explanation. In any case +1 for the solid post! – Mark Viola Dec 26 '15 at 20:12
• @Dr.MV Let me return the compliment by saying that I appreciate many of your interventions here. – Olivier Oloa Dec 26 '15 at 21:06

First make the change of variables $x=e^{-s/n}$, so that $$I(n)=n\int_0^1\frac{x^n dx}{1+x}=\int_0^{\infty}\frac{e^{-s}ds}{e^{s/n}+1}.$$ Now it suffices to Taylor expand in $\frac1n$ to any desired order. All the resulting integrals will be of the form $\alpha_k=\int_0^{\infty}s^{k}e^{-s}ds=k!$ and therefore are easily computable. In particular, $$I(n)=\frac12-\frac1{4n}+\frac1{8n^3}+O\left(n^{-5}\right).$$ To write complete asymptotic expansion, it is helpful to notice that $$\frac{1}{e^{s/n}+1}=\frac12\left(1-\tanh\frac s{2n}\right)=\frac12-\frac12\sum_{k=1}^{\infty}{\frac{2^{2k}(2^{2k}-1)B_{2k}}{(2k)!}}\left(\frac s{2n}\right)^{2k-1},$$ which implies that (here $B_k$ denote Bernoulli numbers) $$I(n)=\frac12-\sum_{k=1}^{\infty}{\frac{(2^{2k}-1)B_{2k}}{2k}}n^{1-2k}.$$

Consider $\,t:=x^n\,$ then the integral becomes \begin{align} I(n)&:=\int_0^1 \frac{x^n}{1+x} \, dx\\ &=\int_0^1 \frac{t}{1+t^{1/n}} \frac{t^{1/n-1}}n\, dt\\ &=\frac 1n \int_0^1 \frac{t^{1/n}}{1+t^{1/n}}\, dt\\ \end{align} You may expand $\,\dfrac{t^{1/n}}{n\;(1+t^{1/n})}\,$ in series as $n\to \infty$ (i.e. expand $e^{\log(t)/n}$) (faster with a CAS) : $$\frac 1{2n}+\frac{\log(x)}{4n^2}-\frac{\log(x)^3}{48 n^4}+\cdots$$ and integration between $0$ and $1$ should return you : $$\frac 1{2n}-\frac{1}{4n^2}+\frac{1}{8 n^4}-\cdots$$ (the integral of $\log(x)^n$ was given not long ago as $(-1)^n\,n!$)

Explicit solution: The coefficients may be obtained from the generating function of the Bernoulli numbers so that the final result is simply : $$I(n)\sim \sum_{k>0} \frac {1-2^{k}}{k}\frac{B_{k}}{n^k}$$

• Hi dear @Startwearingpurple! (you merit a +1btw since a little earlier! :-)) In fact it should be the expansion of $\,\dfrac 1n\dfrac{t^{1/n}}{1+t^{1/n}}\,$. Ok got it the power of $3$ was wrong, thanks for that! – Raymond Manzoni Dec 26 '15 at 19:25
• Yes, that was it. It is not immediately obvious but the asymptotic expansion does not contain even powers of $1/n$ except for the very first term $1/2$. – Start wearing purple Dec 26 '15 at 19:32
• @Startwearingpurple: as in the Bernoulli numbers! In fact the numerators are given by OEIS A002425 so no much doubts about this! – Raymond Manzoni Dec 26 '15 at 19:37