91Ó°ÊÓ

An improper integral is typically used to describe integrals that have one or more infinite limits or integrands that approach infinity at some points within the integration range. They often appear in calculus when integrating functions over unbounded intervals or in cases where the integrand becomes unbounded within the interval of integration.
In our problem, the integral \(\int_{2}^{x} \frac{1}{\log t} \; dt\) is improper because we are considering the upper limit of the integral going to \(\infty\). Understanding how to evaluate improper integrals is crucial, as it tells us whether the integral converges to a finite value or diverges to infinity.
To handle these kinds of problems, we usually convert the infinite limits to a finite limit with a variable, and then take the limit of the integral as this variable approaches infinity. In this case, we need to evaluate the behavior of: \[ \int_{2}^{b} \frac{1}{\log t} \; dt \] and then take the limit as \(b \rightarrow \infty\).
Since \( \frac{1}{\log t}\) decreases very slowly and never actually becomes zero, this improper integral will increase without bound.

The logarithmic function, denoted as \( \log t\), is a fundamental mathematical function that grows very slowly compared to polynomial and exponential functions.
As \( t \rightarrow \infty\), the value of \( \log t \rightarrow \infty\). Even though \( \log t\) increases without bound, it does so at a much slower rate. This characteristic is essential in our problem because it affects how \( \frac{1}{\log t}\) behaves for large values of \( t\).
Specifically, as \( t \rightarrow \infty\), \( \frac{1}{\log t}\) decreases, but at a very slow rate, which contributes to the integral's divergence over an infinite interval. This slow decrease counterbalances but does not extinguish the unbounded nature of the integral.

In the context of integrals, divergence refers to the integral not converging to a finite value. Instead, it grows without bound. We are asked to show that the function \( F(x) = \int_{2}^{x} \frac{1}{\log t} \; dt\) is unbounded, which means it diverges as \( x \rightarrow \infty\).
To show this, we analyzed the behavior of the integrand \(\frac{1}{\log t}\) and how it influences the integral's growth. Given that \(\log t\) approaches \(\infty\) slowly, \(\frac{1}{\log t}\) decreases slowly. This slow decrease means the integrand maintains a non-negligible value over a long range, resulting in the integral accumulating significant area, thereby diverging.
This formalizes to: \[ \lim_{x \rightarrow \infty} \int_{2}^{x} \frac{1}{\log t} \; dt = \infty \] indicating that the integral does not converge to a finite number but instead grows without bound. The divergence of the integral \( F(x)\) signifies its unbounded nature.