Asymptotic expansion of $I_n=\int_{0}^{1}\frac{1}{1+x^n} dx$

Question

The problem is to find the asymptotic expansion of $I_n=\int_{0}^{1}\frac{1}{1+x^n} dx$ (at least the first 3 terms).

By using some simple bounding, I first showed that $I_n$ tends to $1$. Then I calculated the limit $n(1-I_n)$ using integration by parts and some more bounding. At the end I found:

$I_n=1-\frac{\ln(2)}{n}+o(\frac{1}{n})$

I guess there is another way to find this result by invoking the expansion of $\frac{1}{1+x^n}$. This is more annoying, because you have to justify that you can switch the integral inside and the infinite sum, and at the end you will get the following sum

$I_n\stackrel{?}{=}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{kn+1}=1-\frac{1}{n}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k+\frac{1}{n}}$

In this case, one has to justify that the little $\frac{1}{n}$ doesn't create any problems and that we have in fact:

$\lim_{n \to \infty} \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k+\frac{1}{n}}\stackrel{?}{=}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}=\ln(2)$

So my question is if the second approach can be made rigorous, and how can we find the third term in the asymptotic expansion of $I_n$? Also, is there a standard method to solve these types of problems?

Answer 1

The second approach can be made rigorous: For every partial sum we have

$$0 \leqslant \sum_{k = 0}^m (-1)^k x^{kn} \leqslant 1,$$

since $0 \leqslant x^{(k+1)n} \leqslant x^{kn} \leqslant 1$ for $x\in [0,1]$ and all $k\in \mathbb{N}$ if $n > 0$. If you can use it, just mumble "dominated convergence", and if you're dealing with Riemann integrals, combine the global boundedness with the uniform convergence on $[0,1-\varepsilon]$ for every $\varepsilon > 0$ to justify

$$I_n = \sum_{k = 0}^{\infty} \frac{(-1)^k}{kn+1} = 1 - \sum_{k = 1}^{\infty} \frac{(-1)^{k+1}}{kn+1}.$$

Now we can approximate the $\frac{1}{kn+1}$ with the simpler to handle $\frac{1}{kn}$:

$$I_n = 1 - \sum_{k = 1}^{\infty} \frac{(-1)^{k+1}}{kn} + \sum_{k = 1}^{\infty} (-1)^{k+1}\biggl(\frac{1}{kn} - \frac{1}{kn+1}\biggr).$$

We know that the first series evaluates to $\frac{\ln 2}{n}$, and the remaining sum can be handled similarly:

$$\frac{1}{kn} - \frac{1}{kn+1} = \frac{1}{kn(kn+1)} = \frac{1}{(kn)^2} - \frac{1}{(kn)^2(kn+1)},$$

and

$$\sum_{k = 1}^{\infty} \frac{(-1)^{k+1}}{(kn)^2} = \frac{\pi^2}{12n^2}$$

gives the next term in the expansion.

This can be continued as far as one desires, leading to

$$I:n = 1 - \frac{\ln 2}{n} + \sum_{m = 2}^{\infty} (-1)^m\bigl(1-2^{1-m}\bigr)\frac{\zeta(m)}{n^m}.$$

Answer 2

$I_n =\int_{0}^{1}\frac{1}{1+x^n} dx $

Since, for any integer $m$, $\frac{1+x^{2m+1}}{1+x} =\sum_{k=0}^{2m} (-1)^k x^k $, $\frac{1}{1+x} =\sum_{k=0}^{2m} (-1)^k x^k-\frac{x^{2m+1}}{1+x} $, so $\frac{1}{1+x^n} =\sum_{k=0}^{2m} (-1)^k x^{nk}-\frac{x^{n(2m+1)}}{1+x} $.

Integrating,

$\begin{array}\\ I_n &=\int_{0}^{1}\frac{1}{1+x^n} dx\\ &=\int_{0}^{1}\left(\sum_{k=0}^{2m} (-1)^k x^{nk}-\frac{x^{n(2m+1)}}{1+x}\right) dx\\ &=\sum_{k=0}^{2m} (-1)^k \int_{0}^{1} x^{nk}dx-\int_{0}^{1}\frac{x^{n(2m+1)}dx}{1+x}\\ &=\sum_{k=0}^{2m} (-1)^k \frac{x^{nk+1}}{nk+1}|_{0}^{1}-J_{n(2m+1)} \qquad\text{where } J_{n(2m+1)} =\int_{0}^{1}\frac{x^{n(2m+1)}dx}{1+x}\\ &=\sum_{k=0}^{2m} \frac{(-1)^k}{nk+1}-J_{n(2m+1)}\\ \end{array} $

Since $1 \le 1+x \le 2$, $J_{n(2m+1)} =\int_{0}^{1}\frac{x^{n(2m+1)}dx}{1+x} =c\int_{0}^{1}x^{n(2m+1)}dx =\frac{c}{n(2m+1)+1} $ where $\frac12 \le c \le 1$.

Therefore the series converges, so $I_n =\sum_{k=0}^{\infty} \frac{(-1)^k}{nk+1} $.

To get an expression for this sum, note that it is a multisection of $\sum_{k=1}^{\infty} \frac{(-1)^k}{k} $.

Answer 3

Note that we can write

$$\begin{align} I_n&=\int_0^1\frac{1}{1+x^n}\,dx\\\\ &=1-\frac1n\int_0^1\frac{x^{1/n}}{1+x}\,dx\tag 1 \end{align}$$

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The first methodology in the OP is quite a tenable way forward. Here, we use successive integration by parts to obtain the full asymptotic series in reciprocal powers of $n$.

Integrating by parts the integral on the right-hand side of $(1)$ with $u=x^{1/n}$ and $v=\log(1+x)$, we obtain

$$I_n=1-\frac{\log(2)}{n}+\frac1{n^2}\int_0^1 x^{1/n}\frac{\log(1+x)}{x}\,dx \tag2 $$

Then, integrating by parts the integral on the right-hand side of $(2)$, we obtain

$$\begin{align} I_n&=1-\frac{\log(2)}{n}+\frac1{n^2}\underbrace{\int_0^1 \frac{\log(1+x)}{x}\,dx}_{\eta(2)=-\text{Li}_2(-1)=\pi^2/12}-\frac1{n^3}\int_0^1 x^{-1+1/n}\underbrace{\int_0^x \frac{\log(1+x')}{x'}\,dx'}_{\log(1+x) \le -\text{Li}_2(-x)\le x}\,dx\\\\ &=1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac{1}{n^3}\right) \end{align}$$

One can continue to integrate by parts by introducing higher order polylogarithm functions $-\text{Li}_\ell(-1)=\eta(\ell)$, in terms of the Dirichlet Eta function. The full result is

$$I_n=1-\frac{\log(2)}{n}+\sum_{k=2}^\infty \frac{(-1)^{k}\eta(k)}{n^k}$$

which agrees with the result reported by @DanielFischer.

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PART $2$: I thought it would be a useful complement to Daniel Fischer's post to present the details regarding the interchange of operations. To that end, we proceed.

Note that we have for $n\ge 1$

$$\begin{align} I_n&=\int_0^1 \frac{1}{1+x^n}\,dx\\\\ &\int_0^1 \sum_{k=0}^\infty (-1)^kx^{nk}\,dx\\\\ &=\lim_{\delta\to 0^+}\int_0^{1-\delta}\sum_{k=0}^\infty (-1)^kx^{nk}\,dx\tag3 \end{align}$$

Since the series in $(3)$ converges uniformly on $[0,1-\delta]$ for $0<\delta<1]$, we can interchange the order of integration with the series to obtain

$$\begin{align} I_n&=\lim_{\delta\to 0^+}\sum_{k=0}^\infty (-1)^k\int_0^{1-\delta}x^{nk}\,dx\\\\ &=\lim_{\delta\to 0^+}\sum_{k=0}^\infty \frac{(-1)^k(1-\delta)^{nk+1}}{1+nk} \tag 4 \end{align}$$

For each $n$, we have $\left|\sum_{k=1}^N (-1)^k\right|\le 1$ for all $N$, $\frac{(1-\delta)^{1+nk}}{1+nk}$ is monotonic for each $\delta \in [0,1]$, and $\frac{(1-\delta)^{1+nk}}{1+nk}\to 0$ uniformly for $\delta\in[0,1]$.

Applying Dirichlet's Test for Uniform Convergence, we find that the series in $(4)$ converges uniformly for $\delta\in [0,1]$. And inasmuch as $\frac{(-1)^k(1-\delta)^{1+nk}}{1+nk}$ is a continuous function of $\delta$, then the series is also continuous and we may interchange the limit and the series to arrive at the coveted result

$$I_n=\sum_{k=0}^\infty\frac{(-1)^k}{1+nk}$$

Answer 4

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The integral can be expressed in closed form in terms of Digamma Functions $\Psi$. Namely,

\begin{align} I_{n} & \equiv \int_{0}^{1}{\dd x \over 1 + x^{n}} = \int_{0}^{1}{1 - x^{n} \over 1 - x^{2n}}\,\dd x \,\,\,\stackrel{x^{2n}\,\,\, \mapsto\ x}{=}\,\,\, {1 \over 2n} \int_{0}^{1}{x^{1/\pars{2n} - 1}\ -\ x^{1/\pars{2n} - 1/2} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 2n}\bracks{\int_{0}^{1}{1\ -\ x^{1/\pars{2n} - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{1/\pars{2n} - 1} \over 1 - x}\,\dd x} \\[5mm] & = {1 \over 2n}\bracks{\Psi\pars{{1 \over 2} + {1 \over 2n}} - \Psi\pars{1 \over 2n}} = {1 \over 2n}\bracks{\Psi\pars{{1 \over 2} + {1 \over 2n}} - \Psi\pars{1 + {1 \over 2n}} + {1 \over 1/\pars{2n}}} \\[5mm] & = 1 + {1 \over 2n}\bracks{\Psi\pars{{1 \over 2} + {1 \over 2n}} - \Psi\pars{1 + {1 \over 2n}}} \\[1cm] & = 1 + \underbrace{\Psi\pars{1/2} - \Psi\pars{1} \over 2}_{\ds{-\ln\pars{2}}}\ \,{1 \over n} + \underbrace{\Psi\,'\pars{1/2} - \Psi\,'\pars{1} \over 4} _{\ds{\pi^{2} \over 12}}\,{1 \over n^{2}} + \mrm{O}\pars{1 \over n^{3}} \end{align} $$ \bbx{\ds{% \int_{0}^{1}{\dd x \over 1 + x^{n}} \sim 1 - {\ln\pars{2} \over n} + {\pi^{2}/12 \over n^{2}} \quad\mbox{as}\ n \to \infty}} $$

$\ds{\Psi\ \mbox{and}\ \Psi\,'}$ values for arguments $\ds{1/2}$ and $1$ are given at the above mentioned link.