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8.9 Improper Integrals

Let \( f(x) = \frac{1}{x} \)

We know \( \int_{1}^{b} \frac{1}{x} dx \) is the area under \( \frac{1}{x} \) from \( x=1 \) to \( x=b \).

\[ \int_{1}^{b} \frac{1}{x} dx = \ln|x| \Big|_{1}^{b} = \ln b - \ln 1 = \ln b \]
Graph of the function y = 1/x showing a shaded region under the curve from x=1 to x=b.

What if \( b \to \infty \)?

\[ \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x} dx = \lim_{b \to \infty} \ln b = \infty \]

\( \rightarrow \) the integral is unbounded as \( b \to \infty \).

\[ \int_{1}^{\infty} \frac{1}{x} dx \]

This is a type of improper integrals \( \rightarrow \) at least one integration limit is \( \infty \) or \( -\infty \).

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Let's try \( f(x) = \frac{1}{x^2} \)

\( \int_{1}^{b} \frac{1}{x^2} dx \)

\[ \int_{1}^{b} x^{-2} dx = -x^{-1} \Big|_{1}^{b} = -\frac{1}{x} \Big|_{1}^{b} = -\frac{1}{b} + 1 \]
Graph of y = 1/x^2 showing the area under the curve from x=1 to x=b is shaded.

As \( b \to \infty \), \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} dx = \lim_{b \to \infty} (1 - \frac{1}{b}) = 1 \)

The area under \( \frac{1}{x^2} \) is bounded even as \( b \to \infty \).

\[ \int_{1}^{\infty} \frac{1}{x^2} dx = 1 \]

If the improper integral goes to \( \infty \) or \( -\infty \), we say the integral diverges or is divergent (e.g. \( \int_{1}^{\infty} \frac{1}{x} dx \)).

If the improper integral results in a number, we say the integral converges or is convergent (e.g. \( \int_{1}^{\infty} \frac{1}{x^2} dx \)).

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Convergence of Improper Integrals

It turns out:

\[ \int_{a}^{\infty} \frac{1}{x^p} dx \begin{cases} \text{converges if } p > 1 \\ \text{diverges if } p \leq 1 \end{cases} \]

The difference is how fast \( \frac{1}{x^p} \) decreases.

The improper integral can have both integration limits being \( \infty \) and \( -\infty \).

Example

\[ \int_{-\infty}^{\infty} \frac{1}{x^2+1} dx \]
Graph of the function y = 1/(x^2+1) showing a bell-shaped curve with the area under the curve shaded.
\[ = \int_{-\infty}^{0} \frac{1}{x^2+1} dx + \int_{0}^{\infty} \frac{1}{x^2+1} dx \]\[ = \lim_{a \to -\infty} \int_{a}^{0} \frac{1}{x^2+1} dx + \lim_{a \to \infty} \int_{0}^{a} \frac{1}{x^2+1} dx \]\[ = \lim_{a \to -\infty} \tan^{-1}(x) \Big|_a^0 + \lim_{a \to \infty} \tan^{-1}(x) \Big|_0^a \]
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\[ = \lim_{a \to -\infty} (\tan^{-1}(0) - \tan^{-1}(a)) + \lim_{a \to \infty} (\tan^{-1}(a) - \tan^{-1}(0)) \]

Note: \( \tan^{-1}(0) = 0 \)

\[ = \lim_{a \to -\infty} (-\tan^{-1}(a)) + \lim_{a \to \infty} (\tan^{-1}(a)) \]

Evaluating the limits:

\[ = -(-\frac{\pi}{2}) + \frac{\pi}{2} \]\[ = \pi \]
\[ \int_{-\infty}^{\infty} \frac{1}{x^2+1} dx = \pi \text{ (convergent)} \]

\( \tan^{-1}(\infty) = ? \)
\( \tan^{-1}(-\infty) = ? \)

Graph of y = tan(x) with vertical asymptotes at x = -pi/2 and x = pi/2.
Graph of y = arctan(x) with horizontal asymptotes at y = pi/2 and y = -pi/2.
Graph of y = 1/(x^2+1) with shaded area labeled 'area approaches pi'.
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Another Type of Improper Integrals

Integration limits are finite but the integrand becomes undefined at some point on the interval.

\[ \int_{a}^{b} f(x) \, dx \]

Also improper if \( f(x) \to \infty \) or \( -\infty \) somewhere on \( a \le x \le b \).

Example

\[ \int_{-2}^{3} \frac{1}{x^4} \, dx \]

Note \( \frac{1}{x^4} \to \infty \) as \( x \to 0 \) which is in \( -2 \le x \le 3 \).

So this is an improper integral.

We want to stay away from \( x = 0 \) (where \( \frac{1}{x^4} \to \infty \) (or \( -\infty \))).

\[ \int_{-2}^{0} \frac{1}{x^4} \, dx + \int_{0}^{3} \frac{1}{x^4} \, dx \]\[ = \lim_{b \to 0^-} \int_{-2}^{b} \frac{1}{x^4} \, dx + \lim_{a \to 0^+} \int_{a}^{3} \frac{1}{x^4} \, dx \]
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\[ = \lim_{b \to 0^-} \left. \left( -\frac{1}{3x^3} \right) \right|_{-2}^{b} + \lim_{a \to 0^+} \left. \left( -\frac{1}{3x^3} \right) \right|_{a}^{3} \]\[ = \lim_{b \to 0^-} \left( -\frac{1}{3b^3} - \frac{1}{24} \right) + \lim_{a \to 0^+} \left( -\frac{1}{81} + \frac{1}{3a^3} \right) \]
Note: \( b \) is a small neg. #, so \( -\frac{1}{3b^3} \) is a large pos. #
Note: \( a \) is a small pos. #, so \( \frac{1}{3a^3} \) is a large pos. #
\[ = \infty - \frac{1}{24} - \frac{1}{81} + \infty = \infty \text{ (divergent)} \]

Improper integrals of this type can be easily missed and wrong results will result.

\[ \int_{-2}^{3} \frac{1}{x^4} \, dx \quad \text{pretend we didn't realize this is improper} \]\[ = \left. -\frac{1}{3x^3} \right|_{-2}^{3} = -\frac{1}{81} - \frac{1}{24} = -\frac{35}{648} \quad \text{wrong! completely meaningless} \]
Graph of y = 1/x^4 showing vertical asymptote at x=0 and shaded area from -2 to 3.
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Comparison Test for Improper Integrals

We can compare integrals to (sometimes) quickly determine if an improper integral will converge.

For example, we found that:

\[ \int_{1}^{\infty} \frac{1}{x^2} dx = 1 \text{ (converges)} \]

Since \( 0 \le \frac{1}{x^2+1} \le \frac{1}{x^2} \) because \( \frac{1}{x^2+1} \) has a larger denominator,

\[ 0 \le \underbrace{\int_{1}^{\infty} \frac{1}{x^2+1} dx}_{\text{must also converge}} \le \underbrace{\int_{1}^{\infty} \frac{1}{x^2} dx}_{\substack{\text{convergent} \\ \text{so is finite}}} \]

Similarly, \( \int_{1}^{\infty} \frac{1}{x} dx \) diverges.

And \( \frac{1}{x-\frac{1}{2}} \ge \frac{1}{x} \) since \( \frac{1}{x-\frac{1}{2}} \) has a smaller denominator.

\[ \underbrace{\int_{1}^{\infty} \frac{1}{x-\frac{1}{2}} dx}_{\text{must be a bigger } \infty} \ge \underbrace{\int_{1}^{\infty} \frac{1}{x} dx}_{\infty} \]

So \( \int_{1}^{\infty} \frac{1}{x-\frac{1}{2}} dx \) diverges.