Range of 2 parameters and Convergency of Improper Integral

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Range of 2 parameters and Convergency of Improper Integral

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Source: 2012 Kyoto University Master Course in Mathematics

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Aug 21, 2012, 5:38 PM • 1 Y

Y by Adventure10

Let $\alpha,\ \beta$ be real numbers. Find the ranges of $\alpha,\ \beta$ such that the improper integral $\int_1^{\infty} \frac{x^{\alpha}\ln x}{(1+x)^{\beta}}$ converges.

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#2 h Apr 27, 2025, 1:18 PM

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Upon inspection, $\int_{1}^{\infty} \frac{x^{\alpha}\ln(x)}{(1+x)^{\beta}} dx$ , we necessitate $\beta-\alpha > 1$ for convergence. Let us utilize the Comparison Test for this improper integral according to:

$\int_{1}^{\infty}\frac{x^{\alpha}\ln(x)}{(1+x)^{\beta}} dx \le \int_{1}^{\infty} \frac{x^{\alpha}x^{p}}{(1+x)^{\beta}} dx$

of which the RHS integral converges $\Leftrightarrow \beta -(\alpha + p) > 1$ . Suppose that $\ln(x) \le x^{p}$ for all $x\ge 1$ . If we solve for $p$ in terms of $x$ , we obtain the function $p=f(x)=\frac{\ln[\ln(x)]}{\ln(x)}$ where $f'(x)=0$ yields the critical point:

$f'(x) = \frac{1 - \ln[\ln(x)]}{x \ln^{2}(x)} = 0 \Rightarrow x = e^{e}$ ;

and $f''(x)$ at this critical point produces:

$f''(x) = \frac{-3 + \ln(x)[-1 + \ln[\ln(x)]] + 2\ln[\ln(x)]}{x^2 \ln^{3}(x)} \Rightarrow f''(e^{e}) = -\frac{1}{e^{2e+3}} < 0$ (a maximum).

This gives us $p=f(e^{e}) = \frac{1}{e}$ , which means that $x^{1/e}$ is the supremum of $\ln(x)$ such that $\ln(x) \le x^{p}$ for all $x \in [1, \infty)$ . Returning to the RHS integral of the original inequality above, we require:

$\beta - (\alpha+p) > 1 \Rightarrow \beta > \alpha + 1+\frac{1}{e}$ ;

which gives us the ranges:

$\textcolor{red}{\alpha \in \mathbb{R}, \beta \in (\alpha + 1+1/e, \infty)}$ .

This post has been edited 13 times. Last edited by Mathzeus1024, Apr 27, 2025, 5:59 PM

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#3 h Apr 30, 2025, 9:37 PM

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Mathzeus1024 wrote:

$x^{1/e}$ is the supremum of $\ln(x)$ such that $\ln(x) \le x^{p}$ for all $x \in [1, \infty)$ .

sorry an inequality is a sufficient condition but not necessary (too strong)
you just have to consider $x\longrightarrow \infty$ , (for any $p>0$ , )
then the positive integrand is equivalent to $x^{\alpha-\beta }\ln x$ ,
now put $x=e^u$ to see that answer's $\beta>\alpha +1$ .

This post has been edited 11 times. Last edited by Hello_Kitty, Apr 30, 2025, 9:55 PM

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#4 h May 1, 2025, 11:37 AM

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Hello_Kitty wrote:

Mathzeus1024 wrote:

$x^{1/e}$ is the supremum of $\ln(x)$ such that $\ln(x) \le x^{p}$ for all $x \in [1, \infty)$ .

sorry an inequality is a sufficient condition but not necessary (too strong)
you just have to consider $x\longrightarrow \infty$ , (for any $p>0$ , )
then the positive integrand is equivalent to $x^{\alpha-\beta }\ln x$ ,
now put $x=e^u$ to see that answer's $\beta>\alpha +1$ .

Interesting, thanks.

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