$\lim_{n \to \infty} n \int_{1}^{\infty} \frac{dx}{1+x^n}$, solution critique needed

Question

I am particularly curious about exchanging the sum and integral in the third equation, if this is unjustified then please let me know.

We have that

$$\lim_{n \to \infty} n \int_{1}^{\infty}\frac{dx}{1+x^n} =\lim_{a \to \infty}\lim_{n \to \infty} n \int_1^{a}\frac{dx}{1+x^n} $$

Also

$$\frac{1}{1+x^n} = \frac{1}{x^n}\frac{1}{1+\frac{1}{x^n}} = \sum_{i=0}^{\infty}\bigg(\frac{-1}{x^n}\bigg)^i\frac{1}{x^n} =\sum_{i=0}^{\infty}(-1)^i \frac{1}{x^{(i+1)n}}$$

Since we are integrating for $x \geq 1$, the above series converges for every $x$ so we may integrate term-by-term to obtain

$$n \int_1^{a}\frac{1}{1+x^n}dx = n\sum_{i=0}^{\infty}\int_1^{a}\frac{1}{x^{(i+1)n}} \, dx$$

$$= n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-n\left[\frac{a^{1-n}}{n-1}-\frac{a^{1-2n}}{2n-1}+\cdots\right] $$

where the second term goes to zero as $n$ goes to infinity. Further, we have that

$$\lim_{n \to \infty}\frac{n}{nk-1} = \frac{1}{k}$$

so the first term becomes $1 -\frac{1}{2}+\frac{1}{3}-... = \ln(2)$. In conclusion we get

$$\lim_{a \to \infty}\lim_{n \to \infty} n \int_1^{a}\frac{dx}{1+x^n} =\lim_{a \to\infty} \ln(2) = \ln(2) $$

independently of $a$.

Answer 1

The interchange of the order of summation and integration needs to be justified. Then the interchange of the order of limit and summation needs to be justified. Here is how I would proceed:

$$ \begin{align} \lim_{n\to\infty}n\int_1^\infty\frac1{1+x^n}\,\mathrm{d}x &=\lim_{n\to\infty}n\int_1^\infty\sum_{k=1}^\infty(-1)^{k-1}x^{-kn}\,\mathrm{d}x\tag1\\ &=\lim_{n\to\infty}n\int_1^\infty\sum_{k=1}^\infty\left(x^{-(2k-1)n}-x^{-2kn}\right)\,\mathrm{d}x\tag2\\ &=\lim_{n\to\infty}n\sum_{k=1}^\infty\left(\frac1{(2k-1)n-1}-\frac1{2kn-1}\right)\tag3\\ &=\lim_{n\to\infty}\sum_{k=1}^\infty\frac{n^2}{((2k-1)n-1)(2kn-1)}\tag4\\ &=\sum_{k=1}^\infty\frac1{(2k-1)2k}\tag5\\ &=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)\tag6\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\tag7\\[6pt] &=\log(2)\tag8 \end{align} $$ Explanation:
$(1)$: write the integrand as a geometric series
$(2)$: group terms by twos
$(3)$: apply Fubini's Theorem
$(4)$: rewrite the terms
$(5)$: Dominated Convergence (see this answer)
$(6)$: rewrite terms
$(7)$: write as an alternating series
$(8)$: evaluate the series

Answer 2

I thought it might be instructive to present a way forward that does not require applying the Dominated Convergence Theorem, Uniform Convergence, or Fubini's Theorem. Rather, it relies on tools of elementary calculus only, including substitution, integration by parts, and a basic inequality of the logarithm function. To that end, we now proceed.

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Enforcing the substitution $x\mapsto 1/x$, we find that for $n\ge 2$

$$\int_1^\infty \frac{n}{1+x^n}\,dx=\int_0^1 \frac{(nx^{n-1})}{x(x^n+1)}\,dx$$

Integrating by parts with $u=x^{-1}$ and $v=\log(1+x^n)$ reveals

$$\int_1^\infty \frac{n}{1+x^n}\,dx=\log(2)-\int_0^1 \frac{\log(1+x^n)}{x^2}\,dx$$

Using $\log(1+x^n)\le x^n$, we see for $n>2$

$$\begin{align} \left|\int_0^1 \frac{\log(1+x^n)}{x^2}\,dx\right|&\le \int_0^1 x^{n-2}\,dx\\\\ &=\frac{1}{n-1} \end{align}$$

whence letting $n\to \infty$ we find the coveted limit

$$\begin{align} \lim_{n\to \infty}\int_1^\infty \frac{n}{1+x^n}\,dx&=\log(2)-\lim_{n\to \infty}\int_0^1 \frac{\log(1+x^n)}{x^2}\,dx\\\\ &=\log(2) \end{align}$$

as expected!

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Tools Used: Integration by substitution, integration by parts, an the inequality $\log(1+x)\le x$.

Answer 3

Let $x= y^{1/n}$ in the given integral. The expression becomes

$$\int_1^\infty \frac{y^{1/n}}{y(1+y)}\,dy.$$

Now for any $y\in [1,\infty),$ $y^{1/n} \to 1$ as $n\to \infty.$ Furthermore, for $n>1,$

$$\frac{y^{1/n}}{y(1+y)} \le \frac{y^{1/2}}{y(1+y)} < \frac{1}{y^{3/2}} \in L^1[1,\infty).$$

By the dominated convergent theorem, the desired limit is

$$\int_1^\infty \frac{1}{y(1+y)}\,dy = \ln 2.$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\pars{n\int_{1}^{\infty}{\dd x \over 1 + x^{n}}} & = \lim_{n \to \infty}\pars{n\int_{0}^{1}{x^{n - 2} \over 1 + x^{n}}\,\dd x} = \lim_{n \to \infty}\int_{0}^{1}{x^{-1/n} \over 1 + x}\,\dd x \end{align}

However, $\ds{\exists\ c \mid 0 < c < 1\ \mbox{which satisfies}\ \int_{0}^{1}{x^{-1/n} \over 1 + x}\,\dd x = c^{-1/n}\int_{0}^{1}{\dd x \over 1 + x} = c^{-1/n}\ln\pars{2}}$ ( First Mean Value Theorem for Definite integrals ).

Then, $$ \bbx{\lim_{n \to \infty}\pars{n\int_{1}^{\infty}{\dd x \over 1 + x^{n}}} = \ln\pars{2}} $$