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Let $n=1,2,...$ and define $f(n)$ by $$f(n)=\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$$

For some values of $n$, Matehamtica's shows that $f(n)$ is finite and seems to converge to $\infty$ as $n\rightarrow \infty$.

1 - How to find a bound for $f(n)$?

2- $f(n)\rightarrow\infty$, as $n\rightarrow\infty$?

Thank you for your time.

S L
  • 11,731
Tomás
  • 22,559

2 Answers2

6

Note that, with the substitution $x \mapsto nx$, we have

$$ \frac{f(n)}{n} = \int_{0}^{\infty} \log^2 \left( \frac{x + 1}{x + n^{-2}} \right) \, dx. $$

Now, let us denote

$$ h_n (x) = \log^2 \left( \frac{x + 1}{x + n^{-2}} \right) \quad \text{and} \quad h(x) = \log^2 \left( \frac{x + 1}{x} \right). $$

Then we easily observe that

  1. $h_n (x) \to h(x)$ pointwise and $h_n (x) \leq h(x)$. Indeed, $$ 1 \leq \frac{x+1}{x+n^{-2}} \leq \frac{x+1}{x} $$ and the claim follows by observing that $x \mapsto \log^2 x$ is non-negative increasing on $x \geq 1$.
  2. $h(x)$ is integrable. It is a direct consequence of the following estimation: \begin{align*} \int_{0}^{\infty} h(x) \, dx &= \int_{0}^{1} h(x) \, dx + \int_{1}^{\infty} h(x) \, dx \\ &= \int_{1}^{\infty} \frac{h(1/x)}{x^2} \, dx + \int_{1}^{\infty} h(x) \, dx \\ &= \int_{1}^{\infty} \frac{\log^2 (1+x)}{x^2} \, dx + \int_{1}^{\infty} \log^2 \left(1 + \frac{1}{x} \right) \, dx \\ &\leq \int_{1}^{\infty} \frac{\log^2 (1+x)}{x^2} \, dx + \int_{1}^{\infty} \frac{dx}{x^2} < \infty. \end{align*}

Then by dominated convergence we obtain

$$ \frac{f(n)}{n} = \int_{0}^{\infty} h_n(x) \, dx \xrightarrow[]{n\to\infty} \int_{0}^{\infty} h(x) \, dx. $$

(If you want to avoid the use of dominated convergence, which requires some basic measure theory, you can first show that $h_n(x) \uparrow h(x)$ as $n \to \infty$ pointwise. Then by Dini's theorem, on any compact subset of $(0, \infty)$ we have $h_n (x) \to h(x)$ uniformly. Now a suitable $\epsilon-\delta$ argument will give the same conclusion.)

This already shows that

$$ f(x) \sim cn $$

for some constant $c > 0$. To determine the constant $c$, we have to show that

$$ \int_{0}^{\infty} h(x) \, dx = \frac{\pi^2}{3}. $$

But this is also straightforward. Indeed, put $\displaystyle u = \frac{x}{x+1}$. Then $\displaystyle dx = \frac{du}{(1-u)^2}$ and therefore

\begin{align*} \int_{0}^{\infty} h(x) \, dx &= \int_{0}^{\infty} \log^2 \left( \frac{x + 1}{x} \right) \, dx = \int_{0}^{1} \frac{\log^2 u}{(1-u)^2} \, du \\ &= \left[ \frac{u}{1-u} \log^2 u \right]_{0}^{1} - 2 \int_{0}^{1} \frac{\log u}{1-u} \, du \\ &= 2 \int_{0}^{\infty} \frac{t}{e^{t}-1} \, dt \qquad (u = e^{-t}) \\ &= 2 \zeta(2) = \frac{\pi^2}{3}. \end{align*}

Sangchul Lee
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$$ \frac{x + n}{x + n^{-1}} = 1 + \frac{n-1}{nx + 1}$$

Using the expansion of $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$

We see that the resulting expansion of the integrand is $$\frac{n - 1}{nx + 1} - \frac{(n - 1)^2}{2(n x+ 1)^2} + O\left (\left(\frac{n-1}{nx + 1} \right)\right)^3$$

Squaring this again, we get $(\frac{n - 1}{nx + 1})^2 + O((\frac{n-1}{n x + 1})^3)$

The integral of the left term is just $$\left.-\frac{(n - 1)^2}{n(1 + n x)}\right|^{\infty}_0 = \frac{(n-1)^2}{n}$$

muzzlator
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  • To strictly get a lower bound might have to take a few more terms. I'm just giving general asymptotic behaviour of the biggest contributing term. – muzzlator Mar 21 '13 at 16:07