(AI9) Improper Integrals (p-forms)#

In this lesson we are going to investigate the convergence and divergence of a specific class of improper integrals: the \(p\)-forms. We will be using these pretty extenstively when we talk about sequences and series.

By the end of this lesson, you will be able to determine whether a \(p\)-form improper integral is convergent or divergent, and if convergent, be able to evaluate the integral.

Lecture Videos#

Convergence and Divergence#

Improper Integrals (\(p\)-forms)

The convergence of the \(p\)-form improper integral depends on the value of \(p\):

\[ \int_1^{\infty} \dfrac{1}{x^p}\, dx \]

Specifically, this improper integral is:

  • convergent when \(p>1\)

  • divergent when \(p\leq 1\)

Antiderivative#

The function in these integrals

\[ f(x)=\dfrac{1}{x^p} \]

is a power function and in general there are two separate cases when we integrate a power function:

Proof#

We break the proof of this result into 3 cases:

Case \(p=1\qquad\) (Click to see the steps.)
Case \(p>1\qquad\) (Click to see the steps.)
Case \(p<1\qquad\) (Click to see the steps.)

Example 1#

Determine if the following improper integral is convergent or divergent.

\[ \int_1^{\infty} \dfrac{1}{x^2}\, dx \]
Solution (Click to see the steps.)

This is a \(p\)-form improper integral with \(p=2\) which is greater than 1.

This integral is therefore convergent.

Example 2#

Determine if the following improper integral is convergent or divergent.

\[ \int_1^{\infty} \dfrac{1}{\sqrt{x}}\, dx \]
Solution (Click to see the steps.)

This is a \(p\)-form improper integral with \(p=\tfrac{1}{2}\) which is less than 1.

This integral is therefore divergent.

Note

We can relax the starting point of our integration and the general result is still true. For any number \(a>0\), the improper integral

\[ \int_a^{\infty} \dfrac{1}{x^p}\, dx \]

is convergent when \(p>1\) and divergent when \(p\leq 1\).

Example 3#

Determine if the following improper integral is convergent or divergent.

\[ \int_5^{\infty} \dfrac{1}{x}\, dx \]
Solution (Click to see the steps.)

This is a \(p\)-form improper integral with \(p=1\).

This integral is therefore divergent.