To be precise, there will be some X so that for all x > X, the integrand is > 1/x. (That is for the most part).tjkubo said:Whoops, I meant to ask whether it diverges for p<1.
Mathman, the integrand is not always > 1/x on (2,∞) for all p<1. For p=0.9, for example, it's < 1/x for the most part.
An Improper Riemann Integral is a type of integral used in calculus to evaluate the area under a curve that does not have a finite limit at one or both of its bounds. It is used when the integrand is either unbounded or undefined at one or both of the bounds of integration.
An Improper Riemann Integral differs from a regular Riemann Integral in that it allows for the evaluation of integrals that would otherwise be impossible using the traditional Riemann Integral. It takes into account the behavior of the integrand at the bounds of integration, rather than just the values of the integrand at each point within the interval.
Functions that are discontinuous or unbounded at one or both of the bounds of integration, such as the natural logarithm or the inverse tangent, often require the use of an Improper Riemann Integral for evaluation. Other examples include functions with infinite limits, such as the Gaussian function or the exponential function.
The evaluation of an Improper Riemann Integral involves taking the limit of a regular Riemann Integral as one or both of the bounds of integration approach infinity or negative infinity. This limit is then evaluated using standard techniques of integration, such as substitution or integration by parts.
The Improper Riemann Integral has applications in various fields of science and engineering, such as physics, economics, and electrical engineering. It is used to solve problems involving infinite series, probability distributions, and differential equations, among others.