Improper Integrals

Improper integrals are a subset of integrals that you cannot compute directly and thus must use limits to calculate. Improper integrals can either diverge or converge, depending on whether their value is infinite or finite.

The Essentials

There are two main cases of improper integrals:

  1. The bounds involve infinity: \( \int_0^\infty xdx\), \( \int_{-\infty}^5 xdx \)
  2. The integrand is undefined at one of the bounds: \( \int_0^1 \frac{1}{x}dx \), \( \int_{-2}^3 {\frac{1}{3-x}}dx \)

We handle both of these cases similarly.

  1. Set your "problem" bound(s) equal to some letter variable: \( a=\infty \), \( b=3 \)
  2. Substitute your chosen variable into your integral bounds: \( \int_{0}^3 {\frac{1}{3-x}}dx \longrightarrow \int_{0}^b {\frac{1}{3-x}}dx \)
  3. Solve your integral as usual and evaluate with your new bounds. You should end up with an answer in terms of your chosen letter variable: \( \int_{0}^b {\frac{1}{3-x}}dx \longrightarrow -ln|3-b| \)
  4. Take the limit of your answer as your variable approaches its original value: \( \lim\limits_{b\to 3} -ln|3-b| = \infty \)
  5. If you end up with a number, then your integral converges and is equal to that number. If you end up with \( \pm \infty \), then your integral diverges.

Examples

To evaluate the integral \( \int_0^\infty xdx \), we can follow the steps above:

\[ \infty = a \longrightarrow \int_0^\infty xdx = \int_0^a xdx \]
\[ \int_0^a xdx = \frac{x^2}{2}\Big|_0^a = \frac{a^2}{2} - \frac{0^2}{2} = \frac{a^2}{2} \]

We now have a solution to our integral in terms of \( a \). However, \( a \) is equal to infinity. In order to figure out our true solution, we will have to take the limit as \( a \) approaches infinity:

\[ \lim\limits_{a\to \infty} \frac{a^2}{2} = \infty \]

In this case, as \( a \) goes to infinity, our solution also goes to infinity. Thus, we say that the integral diverges.

This makes sense if we consider the graph of the function \( y=x \):

y equals x

As our x-values increase, this function gets bigger and bigger. If we are integrating this function from 0 to infinity, that translates to computing the area under the curve from 0 to infinity, and we can see that this area will be infinitely large.

If we are evaluating the integral \( \int_0^\infty \frac{1}{x^2}dx \), we can go through the same steps as above. In this case, both of our bounds are "problem" bounds -- the integrand is undefined at both \( x=0 \) and \( x=\infty \). Thus, we will have to make substitutions for both of these.

\[ a=0, b=\infty \longrightarrow \int_0^\infty \frac{1}{x^2}dx = \int_a^b \frac{1}{x^2}dx \]
\[ \int_a^b \frac{1}{x^2}dx = -\frac{1}{a} - (-\frac{1}{b}) \]
\[ \lim\limits_{a\to 0}\lim\limits_{b\to \infty} -\frac{1}{a} + \frac{1}{b} \longrightarrow -\infty + 0 = -\infty \]

Thus, this integral diverges. This is less intuitive, since looking at the graph of \( \frac{1}{x^2} \) might suggest convergence:

one over x squared

However, this function goes to infinity as x approaches 0. This is what gives us an infinite area under the curve.

Practice

Evaluate these integrals:

  1. \( \int_0^e ln(x)dx \)
  2. \( \int_0^\infty e^x dx \)
  3. \( \int_0^9 \frac{1}{\sqrt{9-x}}dx \)

Solution:

  1. 0
  2. diverges
  3. 6

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