不含x=exp(t)替换时，研究积分∫₁^+∞(logx/x)^p dx（p≥1）的敛散性

Hey folks! Let's break down how to figure out if this integral converges or diverges without relying on the ( x = e^t ) substitution. We'll use standard comparison tests and analyze the integrand's behavior near its two key "trouble spots": ( x=1 ) (the lower limit) and ( x \to +\infty ) (the upper limit).

Step 1: Split the Integral #

First, split the integral into two separate parts to isolate each problem area:
$$\int_1^{+\infty} \left( \frac{\log x}{x} \right)^p dx = \int_1^e \left( \frac{\log x}{x} \right)^p dx + \int_e^{+\infty} \left( \frac{\log x}{x} \right)^p dx$$
We'll analyze each part independently.

Step 2: Check Convergence Near ( x=1 ) #

When ( x \to 1^+ ), we can use the Taylor approximation for ( \log x ): ( \log x \approx x - 1 ) (since ( \log(1+h) \approx h ) for small ( h = x-1 )). Also, ( x \approx 1 ) here, so ( x^p \approx 1 ). That simplifies the integrand to:
$$\left( \frac{\log x}{x} \right)^p \approx (x-1)^p$$
Recall that for an integral ( \int_a^b (x-a)^q dx ), it converges if ( q > -1 ). Here, ( q = p \geq 1 ), which is way greater than -1. So the integral ( \int_1^e \left( \frac{\log x}{x} \right)^p dx ) converges for all ( p \geq 1 ).

Step 3: Check Convergence as ( x \to +\infty ) #

This is the critical part. First, note that for ( x \geq e ), the integrand ( f(x) = \left( \frac{\log x}{x} \right)^p ) is strictly decreasing (take the derivative: ( f'(x) = p(\log x)^{p-1}x{-p-1}(1 - \log x) ), which is negative when ( \log x > 1 ), i.e., ( x > e )). That lets us use comparison tests confidently.

Case 1: ( p > 1 ) #

We need to compare ( f(x) ) to a function we know converges. Let's pick a small ( \delta > 0 ) such that ( p - \delta > 1 ) (for example, ( \delta = \frac{p-1}{2} ), which works since ( p > 1 )). Now compute the limit:
$$\lim_{x \to +\infty} x^{p-\delta} \cdot f(x) = \lim_{x \to +\infty} \frac{(\log x)^p}{x\delta}$$
Since ( \delta > 0 ), ( x^\delta ) grows much faster than any power of ( \log x ), so this limit equals 0.

By the limit comparison test, since ( \int_e^{+\infty} \frac{1}{x^{p-\delta}} dx ) converges (because ( p-\delta > 1 )), the integral ( \int_e^{+\infty} f(x) dx ) converges when ( p > 1 ).

Case 2: ( p = 1 ) #

When ( p=1 ), the integrand simplifies to ( \frac{\log x}{x} ). We can compute its antiderivative directly:
$$\int \frac{\log x}{x} dx = \frac{1}{2} (\log x)^2 + C$$
As ( x \to +\infty ), ( (\log x)^2 ) blows up to infinity, so the integral ( \int_e^{+\infty} \frac{\log x}{x} dx ) diverges.

Final Conclusion #

• If ( p > 1 ): The entire integral ( \int_1^{+\infty} \left( \frac{\log x}{x} \right)^p dx ) converges (both parts converge).
• If ( p = 1 ): The integral diverges (the infinite part blows up, even though the near-x=1 part converges).

内容的提问来源于stack exchange，提问作者Nicola M.