对数积分不等式证明求助：如何推导给定积分不等式？

Hey there! Let's break down how to prove this integral inequality—you're already halfway there with the inequalities you've got, and the key trick here is splitting the integral at $\sqrt{x}$ to match the square root terms on the right-hand side. Let's walk through it step by step.

Step 1: Split the Integral #

First, we split the left-hand side integral into two intervals separated by $\sqrt{x}$. This lets us handle each segment with a different upper bound for $\frac{1}{\log t}$:
$$\int_2^x \frac{dt}{\log t} = \int_2^{\sqrt{x}} \frac{dt}{\log t} + \int_{\sqrt{x}}^x \frac{dt}{\log t}$$
(Note: For $2 \leq x < 4$, $\sqrt{x} < 2$, so the first integral becomes negative. That's fine—since the right-hand side is positive, the inequality will still hold because a negative number plus the second integral is smaller than the positive right-hand side.)

Step 2: Bound the First Integral ($\int_2^{\sqrt{x}} \frac{dt}{\log t}$) #

On the interval $[2, \sqrt{x}]$, the natural logarithm is increasing, so $\log t \geq \log 2$ for all $t \geq 2$. Taking reciprocals (which reverses inequalities for positive numbers), we get:
$$\frac{1}{\log t} \leq \frac{1}{\log 2}$$
Integrating both sides over $[2, \sqrt{x}]$:
$$\int_2^{\sqrt{x}} \frac{dt}{\log t} \leq \frac{\sqrt{x} - 2}{\log 2}$$
Since $\sqrt{x} - 2 < \sqrt{x}$ (for $x \geq 2$), we can simplify this to:
$$\int_2^{\sqrt{x}} \frac{dt}{\log t} \leq \frac{\sqrt{x}}{\log 2}$$

Step 3: Bound the Second Integral ($\int_{\sqrt{x}}^x \frac{dt}{\log t}$) #

Now look at the interval $[\sqrt{x}, x]$. Again, $\log t$ is increasing, so the smallest value of $\log t$ here is $\log \sqrt{x}$ (at $t = \sqrt{x}$). Taking reciprocals gives:
$$\frac{1}{\log t} \leq \frac{1}{\log \sqrt{x}}$$
Integrate over $[\sqrt{x}, x]$:
$$\int_{\sqrt{x}}^x \frac{dt}{\log t} \leq \frac{x - \sqrt{x}}{\log \sqrt{x}}$$

Step 4: Combine the Bounds #

Adding the two inequalities from Steps 2 and 3 gives exactly the inequality we need to prove:
$$\int_2^x \frac{dt}{\log t} \leq \frac{\sqrt{x}}{\log 2} + \frac{x - \sqrt{x}}{\log \sqrt{x}}$$

Quick Note on the Number Theory Context #

This is a classic upper bound for the logarithmic integral $\text{li}(x) = \int_2^x \frac{dt}{\log t}$, which is closely tied to the Prime Number Theorem (PNT). The PNT tells us that $\pi(x) \sim \text{li}(x)$ (where $\pi(x)$ is the number of primes ≤ x), so simple bounds like this help in deriving rough upper estimates for $\pi(x)$ without getting into more complex error terms.

内容的提问来源于stack exchange，提问作者Dominated Convergence Theorem