# Series expansion of $\int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx$

I have proved $$\int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx \sim \frac{\ln(2)}{n}$$

How can I get further and find $a$ such that:

$$\int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx=\frac{\ln(2)}{n}+\frac{a}{n^2}+o\left(\frac{1}{n^2}\right)?$$

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It may be easier to write the integrand as $1 - \frac{1}{1+x^n}.$ Then integrate the usual power series for $\frac{1}{1+x^n}$ term by term. – Geoff Robinson Jan 15 '12 at 11:06
You are looking at an expansion for the logarithmic derivative of the beta function. – Eric Naslund Jan 15 '12 at 12:15
Could you explain calculations between the 6th line and the 7th line? – Chon Jan 15 '12 at 16:10

Put $I-n:=\int_0^1\frac{x^n}{1+x^n}$, and make the substitution $t=x^n$ to get \begin{align*} I_n&=\frac 1n\int_0^1\frac{t^{\frac 1n}}{1+t}dt\\ &=\frac 1n\left[\ln(1+t)t^{\frac 1n}\right]_0^1-\frac 1n\int_0^1\ln(1+t)\frac 1nt^{\frac 1n-1}dt\\ &=\frac{\ln 2}n-\frac 1{n^2}\int_0^1\frac{\ln(1+t)}tdt+\frac 1{n^2}\int_0^1\frac{\ln(1+t)}t(1-t^{\frac 1n})dt. \end{align*} Since $|1-t^{\frac 1n}|=\left|\int_0^{\frac{\ln t}n}e^t dt\right|\leq \frac 1n|\ln t|$, we have $\lim_{n\to\infty}\int_0^1\frac{\ln(1+t)}t(1-t^{\frac 1n})dt=0$ hence $a=-\int_0^1\frac{\ln(1+t)}tdt$. We can compute this integral using power series.
$$\begin{eqnarray} \int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx &=& -\int_{0}^{1} \sum_{k=1}^\infty(-x^n)^k \mathrm dx \\ &=& -\sum_{k=1}^\infty\int_{0}^{1}(-x^n)^k \mathrm dx \\ &=& -\sum_{k=1}^\infty\frac{(-1)^k}{kn+1} \\ &=& -\sum_{k=1}^\infty\frac1{kn}\frac{(-1)^k}{1+\frac1{kn}} \\ &=& \sum_{k=1}^\infty\sum_{l=0}^\infty(-1)^k(-kn)^{-(l+1)} \\ &=& -\sum_{l=0}^\infty (-1)^ln^{-(l+1)}\sum_{k=1}^\infty(-1)^kk^{-(l+1)} \\ &=& -\frac1n\sum_{k=1}^\infty\frac{(-1)^k}k+\sum_{l=1}^\infty(-1)^ln^{-(l+1)}\left(1-2^{-l}\right)\zeta(l+1) \\ &=& \frac{\log2}n-\frac{\pi^2}{12}\frac1{n^2}+o\left(\frac1{n^2}\right)\;. \end{eqnarray}$$