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# Improper Integrals

Definite integrals have a wide variety of applications. Definite integrals are used to find the area below a curve, to find the displacement of an accelerating car, and even to measure the amount fuel an airplane uses during flight!

But what happens if the interval of integration is infinite? Or if the function is discontinuous in the interval of integration? These integrals are known as improper integrals, and we have to be careful when working with them. Let's see how this is done.

## Types of Improper Integrals

Consider a definite integral. Depending on the interval of integration, the following cases might show up:

• the integral is defined over an infinite interval; or

• the function you are integrating has a discontinuity in the interval of integration.

If either of the above circumstances is met, the integral is known as an Improper Integral.

An Improper Integral is a definite integral that is either defined over an infinite interval, or that the function of its integrand contains a discontinuity in the interval.

## Evaluating Improper Integrals

Once that you identify an improper integral, the next step is to evaluate it. The evaluation of both cases involve limits. Let's see how to deal with each one.

### Over an Infinite Interval

Consider the following definite integral:

$$\int_{0}^{2}3e^{\text{-}x}\,\mathrm{d}x .$$

The above integral can be seen as the area between $$y=3e^{\text{-}x}$$, the vertical lines $$x=0$$ and $$x=2$$, and the $$x$$-axis.

Area below an exponential curve - StudySmarter Originals

But what if you keep going further to the right?

Further area below the exponential curve - StudySmarter Originals

You can keep going even further, out to infinity, as in the example:

$$\int_{0}^{\infty}3e^{\text{-}x}\,\mathrm{d}x.$$

In order to evaluate an integral like this, you need to take limits.

Let $$f(x)$$ be a function that is continuous for $$a\leq x$$. If the limit exists, the improper integral $$\int_{a}^{\infty}f(x)\,\mathrm{d}x$$ is defined as follows:

$$\int_{a}^{\infty}f(x)\,\mathrm{d}x=\lim_{b\rightarrow \infty}\int_{a}^{b}f(x)\,\mathrm{d}x .$$

The other types of improper integrals involving infinite intervals are defined similarly.

Let $$f(x)$$ be a function that is continuous for $$x\leq b$$. If the limit exists, the improper integral $$\int_{\text{-}\infty}^{b}f(x)\,\mathrm{d}x$$ is defined as follows:

$$\int_{\text{-}\infty}^{b}f(x)\,\mathrm{d}x=\lim_{a\rightarrow \infty}\int_{a}^{b}f(x)\,\mathrm{d}x .$$

If the integral is to be done over all real numbers you need split it into two integrals, each of which involves the above definitions.

Let $$f(x)$$ be a function that is continuous everywhere. If both limits exist, the improper integral $$\int_{\text{-}\infty}^{\infty}f(x)\,\mathrm{d}x$$ is defined as follows:

$$\int_{\text{-}\infty}^{\infty}f(x)\,\mathrm{d}x=\int_{\text{-}\infty}^{0}f(x)\,\mathrm{d}x+\int_{0}^{\infty}f(x)\,\mathrm{d}x .$$

For the evaluation of these types of improper integrals we follow these steps:

1. Write the improper integral using an appropriate limit.
2. Evaluate the definite integral. This is usually done with the Fundamental Theorem of Calculus.
3. Evaluate the resulting limit.

Let's see an example of an integral over an infinite interval.

Evaluate the following improper integral:

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x.$$

Answer:

You should begin by noticing that, despite the function being discontinuous at $$x=0$$, the discontinuity is not included in your interval of integration, so you are good to go.

The discontinuity is not included in the integration interval - StudySmarter Originals

You can now follow the steps for evaluating improper integrals.

1. Write the improper integral as a limit.

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x = \lim_{b\rightarrow\infty} \int_{1}^{b} \frac{1}{x^2} \mathrm{d}x.$$

2. Evaluate the definite integral.

For this step you can find the antiderivative of the function with the Power Rule,

$$\int \frac{1}{x^2} \mathrm{d}x = -\frac{1}{x}+C,$$

and then use the Fundamental Theorem of Calculus, so

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x=\lim_{b\rightarrow\infty} \left[ \left( -\frac{1}{b} \right) - \left( -\frac{1}{1} \right) \right].$$

3. Evaluate the limit and simplify.

\begin{align} \int_{1}^{\infty}\frac{1}{x^2} \mathrm{d}x &= 0 -(-1) \\ &= 1. \end{align}

### Over an Interval Where the Function Is Not Continuous

It is also possible to have an integral whose integrand is a function that has a discontinuity in the given interval. In such cases you are not allowed to include the discontinuity on the integral, but you can get close enough to it.

The rational function is discontinuous at the origin - StudySmarter Originals

You also need to take limits in order to evaluate integrals involving discontinuities. Let's see how.

Let $$f(x)$$ be a function that is continuous on the interval $$a\leq x < b$$ but discontinuous at $$x=b$$. Its improper integral is defined as:

$$\int_{a}^{b} f(x) \, \mathrm{d}x = \lim_{t\rightarrow b^{-}} \int_{a}^{t} f(x) \, \mathrm{d}x.$$

In the above case the discontinuity was on the upper limit of integration. Hence, you need to take the limit as you approach the discontinuity from the left.

Approaching a discontinuity from the left - StudySmarter Originals.

Let's now see what to do if the discontinuity is in the lower limit of integration.

Let $$f(x)$$ be a function that is continuous for $$a<x\leq b$$ but discontinuous at $$x=a$$. Its improper integral is defined as:

$$\int_{a}^{b} f(x) \, \mathrm{d}x = \lim_{t\rightarrow a^{+}} \int_{t}^{b} f(x) \, \mathrm{d}x.$$

For a reminder about limits from the left and right, see One-Sided Limits.

If the discontinuity is on the lower limit of integration you take the limit as you approach it from the right.

Approaching a discontinuity from the right - StudySmarter Originals.

And what if the discontinuity is inside the interval of integration? Then you need a limit from the left and a limit from the right!

Let $$f(x)$$ be a function that is continuous for $$a\leq x\leq b$$ except at a point $$c$$ such that $$a<c<b$$. Its improper integral is defined as:

$$\int_{a}^{b} f(x) \, \mathrm{d}x = \int_{a}^{c} f(x) \, \mathrm{d}x + \int_{c}^{b} f(x) \, \mathrm{d}x.$$

The steps involved for evaluating these types of improper integrals are pretty much the same as the other case, you just need to pay attention at the discontinuity. Here are the steps:

1. Write the integral using a limit that approaches the discontinuity. If the discontinuity is inside the interval of integration you will have to split the integral and write two limits.
2. Evaluate the resulting integral(s). This is usually done with the Fundamental Theorem of Calculus.
3. Evaluate the resulting limit(s).

Let's see an example of an improper integral involving a discontinuity.

Evaluate the following integral:

$$\int_{-3}^{3}\frac{1}{x^3} \mathrm{d}x.$$

Answer:

1. Write both integrals using appropriate limits that approach the discontinuity.

Note that the function is discontinuous at $$x=0$$, so you have to split it into two integrals

$$\int_{-3}^{3} \frac{1}{x^3} \mathrm{d}x = \int_{-3}^{0} \frac{1}{x^3} \mathrm{d}x + \int_{0}^{3} \frac{1}{x^3} \mathrm{d}x.$$

2. Evaluate the resulting integral(s).

Next, you need to evaluate each integral. This can be done by finding the antiderivative of the integrand (with the help of The Power Rule)

$$\int \frac{1}{x^3} \mathrm{d}x = -\frac{1}{2} \cdot \frac{1}{x^2} +C,$$

and then using The Fundamental Theorem of Calculus for each pair of integration limits, so

$$\int_{-3}^{0} \frac{1}{x^3} \mathrm{d}x=\lim_{t\rightarrow 0^{-}} \left[ \left( -\frac{1}{2} \cdot \frac{1}{t^2} \right) -\left(-\frac{1}{2} \cdot \frac{1}{(-3)^2}\right) \right],$$

and

$$\int_{0}^{3} \frac{1}{x^3} \mathrm{d}x=\lim_{t\rightarrow 0^{+}} \left[ \left( -\frac{1}{2} \cdot \frac{1}{3^2} \right) -\left(-\frac{1}{2} \cdot \frac{1}{t^2}\right) \right].$$

3. Evaluate the resulting limits.

You can note that neither of the required limits exist. This is enough information to conclude the integral diverges!

In fact, if you find that any of the limits does not exist, you can stop there and conclude the integral divertes. There is no need to evaluate the other limit!

In the above example you found that the value of the integral does not exist. But what does this means? We are talking about the convergence of the integral.

## Convergence of Improper Integrals

You have seen under which circumstances you can find the values of improper integrals. In such cases we say the integrals converge.

Suppose we have an improper integral. If all the limits involved in its evaluation exist, then the integral is said to converge. If any of the limits does not exist the value of the integral does not exist, and it is said to diverge.

Typically, we will say that the integral either diverges, or that it converges to a certain value.

From our previous examples you found that

$$\int_{1}^{\infty} \frac{1}{x^2} \mathrm{d}x$$

converges, and its value is $$1$$.

You also found that the value of

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x$$

does not exist because the limit involved in its evaluation does not exist, hence, you can say that it diverges.

## Examples of Evaluating Improper Integrals

Let's take a look at more examples of improper integrals.

Evaluate the following integral:

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x.$$

Answer:

Begin by noticing that the function $$f(x)=\frac{1}{x}$$ is not defined when $$x=0$$, so you are dealing with an improper integral. 1. Write the improper integral as a limit.You should start by writing the improper integral as a limit, so

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x = \lim_{t\rightarrow 0^{+}} \int_{t}^{2} \frac{1}{x} \mathrm{d}x.$$

2. Evaluate the resulting integral.

Next, find the antiderivative of the function of the integrand (you can take a look at our Integrals Involving Logarithmic Functions for a refresher)

$$\int \frac{1}{x} \mathrm{d}x = \ln{x},$$

and finally, use The Fundamental Theorem of Calculus to evaluate the definite integral

$$\int_{0}^{2} \frac{1}{x} \mathrm{d}x = \ln{2} - \lim_{t\rightarrow 0^{+}} \ln{t}.$$

3. Evaluate the resulting limit.

Unfortunately, the last limit does not exist, so the integral does not exist either.

The next integral appears quite frequently in Physics and Statistics. It is used to model the limits of exponential behavior.

Evaluate the following integral:

$$\int_{0}^{\infty} e^{-x} \, \mathrm{d}x.$$

Answer:

1. Write the improper integral as a limit.

As usual, begin by writing the integral as a limit, so

$$\int_{0}^{\infty} e^{-x} \,\mathrm{d}x=\lim_{b\rightarrow \infty} \int_{0}^{b} e^{-x} \, \mathrm{d}x.$$

2. Evaluate the resulting integral.

Next, find the antiderivative of the exponential function

$$\int e^{-x} \,\mathrm{d}x = -e^{-x} + C.$$

Finally, you can use The Fundamental Theorem of Calculus to evaluate the definite integral

$$\int_{0}^{\infty} e^{-x} \,\mathrm{d}x = \lim_{b\rightarrow \infty} \left[ \left(-e^{-b} \right) - \left(-e^{0}\right) \right].$$

3. Evaluate the resulting limit.

The above limit of the exponential is equal to $$0$$. Furthermore, any non-zero real number raised to the power of $$0$$ is equal to $$1$$. Therefore

\begin{align} \int_{0}^{\infty} e^{-x} \, \mathrm{d}x &= 0 - (-1) \\ &=1. \end{align}

Need to remember how to evaluate limits like the one shown in the above example? Check out our Limits article!

## Improper Integral Formulas

Unfortunately, there is no formula for improper integrals. You need to identify which type of improper integral you are dealing with, and then using the corresponding method, as seen above in the article.

## Improper Integrals - Key takeaways

• An Improper Integral is a definite integral that is either defined over an infinite interval, or one where the function you are integrating has a discontinuity in the interval.
• There is no formula for evaluating improper integrals. Instead, there are methods depending on the type of improper integral you are dealing with.
• If the improper integral is defined over an infinite interval:1. Write the improper integral using an appropriate limit.2. Evaluate the definite integral. This is usually done with the Fundamental Theorem of Calculus.3. Evaluate the resulting limit.
• If the improper integral has a discontinuity in the interval of integration:1. Write the integral using a limit that approaches the discontinuity. If the discontinuity is inside the interval of integration you will have to split the integral and write two limits. 2. Evaluate the resulting integral(s). This is usually done with the Fundamental Theorem of Calculus.3. Evaluate the resulting limit(s).
• If all the limits involved in the evaluation of the improper integral exist, we say that such integral converges to a value.
• If any of the limits involved in the evaluation of the improper integral does not exist, we say that such integral diverges.

## Frequently Asked Questions about Improper Integrals

An improper integral is a definite integral that is either defined over an infinite interval, or that the function of its integrand contains a discontinuity in the interval.

You can solve improper integrals by taking an appropriate limit, depending on which kind of improper integral you are working with.

An example of an improper integral is integrating the exponential function from 1 to infinity. Integrating the function 1/x from -1 to 1 is another example because the function is not defined at x=0.

To know if an improper integral converges or diverges you need to rewrite it as a limit and evaluate it. If all the limits involved in its calculation exist, then the integral converges. If any of the limits involved in the calculation does not exist, the integral diverges.

There are two types of improper integrals:

• An integral that is defined over an infinite interval.
• An integral that has a function with a discontinuity in the interval.

## Final Improper Integrals Quiz

Question

What is an improper integral?

Show answer

Answer

An improper integral is a definite integral that is either defined over an infinite interval, or that the function of its integrand contains a discontinuity in the interval.

Show question

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