3.1: Improper Integrals

Definitions

• Improper Integral
• Convergence and Divergence

Definition: Improper Integral

1. Let \(f(x)\) be continuous over an interval of the form \([a,\infty )\). Then \( \displaystyle \int_a^{\infty} f(x) \, dx \) is called an improper integral, and \[\int ^{\infty }_a f(x)\,dx \equiv \lim_{t \to \infty }\int ^t_a f(x)\,dx, \label{improper1} \]provided this limit exists.
2. Let \(f(x)\) be continuous over an interval of the form \((− \infty ,b]\). Then \( \displaystyle \int_{-\infty}^b f(x) \, dx \) is also called an improper integral, and \[\int ^b_{− \infty } f(x)\,dx \equiv \lim_{t \to − \infty }\int ^b_tf(x)\,dx, \label{improper2} \]provided this limit exists.

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.

3. Let \(f(x)\) be continuous over \((− \infty ,\infty )\). Then \( \displaystyle \int_{-\infty}^{\infty} f(x) \, dx \) is also called an improper integral, and \[\int ^{\infty }_{− \infty }f(x)\,dx \equiv \int ^a_{− \infty }f(x)\,dx+\int ^{\infty }_a f(x)\,dx \label{improper3} \]for any value of \( a \), provided that \(\displaystyle \int ^a_{− \infty }f(x)\,dx\) and \(\displaystyle \int ^{\infty }_a f(x)\,dx\) both converge.
   If either of these two integrals diverge, then \(\displaystyle \int ^{\infty }_{− \infty }f(x)\,dx\) diverges.

Definition: Converging and Diverging Improper Integral

1. Let \(f(x)\) be continuous over \([a,b)\). Then, \[\int ^b_af(x)\,dx \equiv \lim_{t \to b^−}\int ^t_af(x)\,dx, \label{improperundefb} \]provided this limit exists.
2. Let \(f(x)\) be continuous over \((a,b]\). Then, \[\int ^b_af(x)\,dx=\lim_{t \to a^+}\int ^b_tf(x)\,dx, \label{improperundefa} \]provided this limit exists.
3. If \(f(x)\) is continuous over \([a,b]\) except at a point \(c\) in \((a,b)\), then \[\int ^b_af(x)\,dx \equiv \int ^c_af(x)\,dx+\int ^b_cf(x)\,dx,\label{improperundefc} \] provided both \(\displaystyle \int ^c_af(x)\,dx\) and \(\displaystyle \int ^b_cf(x)\,dx\) converge. If either of these integrals diverges, then \(\displaystyle \int ^b_af(x)\,dx\) diverges.

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.

Theorems

• \( p \)-Integral Test (prove)
• Comparison Test for Improper Integrals (justify)

Theorem: \( p \)-Integral Test

The improper integral\[ \int ^{\infty }_a \dfrac{1}{x^p}\,dx \nonumber \]where \( a \gt 0 \), converges for \( p \gt 1 \) and diverges otherwise.

Proof

Assuming for now that \( p \neq 1 \),\[ \begin{array}{rcl}
\displaystyle \int_a^{\infty} \dfrac{1}{x^p} \, dx & = & \displaystyle \lim_{t \to \infty} \int_a^t \dfrac{1}{x^p} \, dx \\[16pt]
& = & \displaystyle \lim_{t \to \infty} \left( \dfrac{x^{-p + 1}}{-p + 1} \right) \bigg|_a^t \\[16pt]
& = & \dfrac{1}{1 - p} \displaystyle \lim_{t \to \infty} \left( t^{1 - p} - a^{1 - p} \right) \\[16pt]
\end{array} \nonumber \]If \( p \gt 1 \), then \( 0 \gt 1 - p \) and so \( \displaystyle \lim_{t \to \infty} t^{1 - p} = 0 \). Hence, the improper integral converges (specifically, it converges to \( \frac{a^{1 - p}}{p - 1} \)).

On the other hand, if \( p \lt 1 \), then \( 0 \lt 1 - p \) and so \( \displaystyle \lim_{t \to \infty} t^{1 - p} = \infty \). Hence, the improper integral diverges.

Finally, if \( p = 1 \), then\[ \begin{array}{rcl}
\displaystyle \int_a^{\infty} \dfrac{1}{x^p} \, dx & = & \displaystyle \lim_{t \to \infty} \int_a^t \dfrac{1}{x} \, dx \\[16pt]
& = & \displaystyle \lim_{t \to \infty} \left( \ln(x) \right) \bigg|_a^t \\[16pt]
& = & \displaystyle \lim_{t \to \infty} \left( \ln(t) - \ln(a) \right) \\[16pt]
\end{array} \nonumber \]However, \( \displaystyle \lim_{t \to \infty} \ln(t) = \infty \). Hence, the improper integral diverges.

Therefore, \[ \int ^{\infty }_a \dfrac{1}{x^p}\,dx \nonumber \]where \( a \gt 0 \), converges for \( p \gt 1 \) and diverges otherwise.

Theorem: Comparison Test for Improper Integrals

Let \(f(x)\) and \(g(x)\) be continuous over \([a,\infty ).\) Assume that \(0 \leq f(x) \leq g(x)\) for \(x \geq a\).

1. If \[\int ^{\infty }_af(x)\,dx=\lim_{t \to \infty }\int ^t_af(x)\,dx=\infty , \nonumber \]then \[\int ^{\infty }_ag(x)\,dx=\lim_{t \to \infty }\int ^t_ag(x)\,dx=\infty . \nonumber \]That is, if the area between \( f(x) \) and the \( x \)-axis is divergent (in this case, infinite), then so is the area between \( g(x) \) and the \( x \)-axis.
2. If \[\int ^{\infty }_ag(x)\,dx=\lim_{t \to \infty }\int ^t_ag(x)\,dx=L, \nonumber \]where \(L\) is a real number, then \[\int ^{\infty }_af(x)\,dx=\lim_{t \to \infty }\int ^t_af(x)\,dx=M \nonumber \]for some real number \(M \leq L\). That is, if the area between \( g(x) \) and the \( x \)-axis is convergent (i.e., finite), then so is the area between \( f(x) \) and the \( x \)-axis.

Cautions

Caution

When splitting an improper integral into two improper integrals, if either of the splits diverges, the entire improper integral diverges.

Caution

In general, symmetry is not useful when dealing with improper integrals.

Examples

Lecture Example \(\PageIndex{1}\)

Determine if the following integral converges or diverges. If it converges, determine its value.\[ \int_{1}^{\infty} \dfrac{1}{x^4} \, dx. \nonumber \]

Lecture Example \(\PageIndex{2}\)

For what values of \( p \) is\[ \int_{1}^{\infty} \dfrac{1}{x^p} \, dx \nonumber \]convergent?

Lecture Example \(\PageIndex{3}\)

Determine if the following integral converges or diverges. If it converges, determine its value.\[ \int_{-\infty}^{0} \dfrac{1}{\sqrt{3 - x}} \, dx. \nonumber \]

Lecture Example \(\PageIndex{4}\)

Determine if the following integral converges or diverges. If it converges, determine its value.\[ \int_{-\infty}^{\infty} x e^{-x^2} \, dx. \nonumber \]

Lecture Example \(\PageIndex{5}\)

Determine if the following integral converges or diverges. If it converges, determine its value.\[ \int_{0}^{\sqrt{2}} \dfrac{1}{\sqrt{2 - x^2}} \, dx. \nonumber \]

Something to Mention

What is the derivative of the arcsine?

Lecture Example \(\PageIndex{6}\)

Determine if the following integral converges or diverges. If it converges, determine its value.\[ \int_{1}^{3} \dfrac{dx}{x^2 - 2}.  \nonumber \]

Something to Mention

The sub, \( x = \sqrt{2} \sec(\theta) \), will not work. Why? PFD to the rescue!!!

Lecture Example \(\PageIndex{7}\)

Determine if the integral converges or diverges (do not find its value).

1.    \[ \int_{1}^{\infty} \dfrac{dx}{x^3 + 7x^2 + 2x + 1} \nonumber \]
2.    \[ \int_{3}^{\infty} \dfrac{1}{x + e^x} \, dx \nonumber \]
3.    \[ \int_{1}^{\infty} \dfrac{1 + 3\sin^4(2x)}{\sqrt{x}} \, dx \nonumber \]

Online Lecture Examples \(\PageIndex{8}\)

This is a few extra examples, if you need them.

Lecture Example \(\PageIndex{9}\)

We have already shown that the curve \( y = \frac{1}{x} \) to the right of \( x = 1 \) encloses an infinite area. The infinitely long "horn" obtained by rotating this region about the \(x\)-axis is known as Gabriel’s Horn. While most people would assume the volume of Gabriel’s Horn must be infinite, let’s see if it is confirmed by evaluating the improper integral.