Explore Integrals

# Improper integrals

The definite integral with limits $$a$$ and $$b$$ represents a signed area of the region under the graph of the function, between the vertical lines $$x=a$$ and $$x=b$$ and the $$x$$-axis. But what if one of the limits is infinity? Or what if the function has a discontinuity between $$a$$ and $$b$$? That’s where improper integrals come into the picture!

## What is an improper integral?

If $$f$$ is continuous on $$[a,b\rangle$$ and discontinuous in $$b$$, then the integral of $$f$$ over $$[a,b\rangle$$ is improper and by definition calculated by using a one-sided limit:

$$\int_a^{b}f(x)dx=\lim_{c\to {b}^{-}}\int_a^{c}f(x)dx$$

If $$b$$ is $$\infty$$, then the mentioned integral is also improper and:

$$\int_a^{\infty}f(x)dx=\lim_{c\to \infty}\int_a^{c}f(x)dx$$

Similar definitions apply when $$f$$ is continuous on $$\langle a,b]$$ and discontinuous at $$a$$ or when $$f$$ is continuous on $$\langle -\infty,b]$$:

$$\int_a^{b}f(x)dx=\lim_{c\to a^{+}}\int_c^{b}f(x)dx$$
$$\int_{-\infty}^{b}f(x)dx=\lim_{c\to -\infty}\int_c^{b}f(x)dx$$

However, when the improper integral is written as a limit, we still need to use the integral rules to solve it. But what are the properties of integrals?

Here they are:

### Why are improper integrals so useful?

Improper integrals are like definite integrals for which at least one of the limits is infinity, or there’s a discontinuity on the interval that is defined by the limits.

This can be used in calculating the energy of the objects in a gravitational field $$\int_r^{+\infty}F(x)dx$$, where $$r$$ is the inital actual distance between objects and $$F(x)$$ is the gravitational force between them at distance $$x$$. We can also use it in statistics and probability, in which some quantity is modeled by normal distribution.

See? Improper integrals can be endless fun! (Get it? Sorry.)

## How to solve an improper integral

Time to do some math! Let’s walk through a couple example problems together.

### Example 1

Find the integral:

$$\int_{-1}^{1}\frac{1}{\sqrt{x^2}}dx$$

To integrate the function on the interval $$[-1,1]$$, we first need to see if the integrand function has discontinuities on that interval. To do that, we’ll determine the domain of $$\frac{1}{\sqrt{x^2}}$$. The root is odd, which means it’s defined for every real number, but we still need to determine the zeroes of the denominator (because the denominator cannot be zero!). To determine the zeroes of the denominator, set $$\sqrt{x^2}$$ equal to zero:

$$\sqrt{x^2}=0$$

The only way a root could be zero is when the radicand equals zero, so equate the radicand with $$0$$:

$${x^2}=0$$

The only way a power can be zero is when the base is equal to $$0$$, so equate the base of the power with $$0$$:

$$x=0$$

Hence, the function $$\frac{1}{\sqrt{x^2}}$$ is not defined for $$x=0$$ in the interval $$[-1,1]$$. So, let’s use $$\int_a^c f(x)dx=\int_a^b f(x)dx+\int_b^c f(x)dx$$ and the point of discontinuity $$b=0$$ to transform the expression:

$$\int_{-1}^{0}\frac{1}{\sqrt{x^2}}dx+\int_{0}^{1}\frac{1}{\sqrt{x^2}}dx$$

Let’s focus on the first integral. The function has an infinite discontinuity at the upper limit $$b=0$$. To evaluate the improper integral, by definition, we’ll rewrite it using a one-sided limit and a definite integral.

$$\lim_{a\rightarrow0^{-}}\int_{-1}^{a}\frac{1}{\sqrt{x^2}}dx+\int_{0}^{1}\frac{1}{\sqrt{x^2}}dx$$

Now let’s focus on the second integral. The function has an infinite discontinuity at the lower limit $$b=0$$. To evaluate the improper integral, by definition, we have to rewrite it using a one-sided limit and a definite integral.

$$\lim_{a\rightarrow0^{-}}\int_{-1}^{a}\frac{1}{\sqrt{x^2}}dx+\lim_{a\rightarrow0^{+}}\int_{a}^{1}\frac{1}{\sqrt{x^2}}dx$$

We need to evaluate the definite integrals, but to do that, we evaluate the indefinite integral first (notice that the integrand of the first and the second integral are the same, so we need to determine one indefinite integral):

$$\int \frac{1}{\sqrt{x^2}}dx$$

Use $$\sqrt[n]{a^m}=a^{\frac{m}{n}}$$ to transform the expression in the denominator:

$$\int \frac{1}{x^{\frac{2}{3}}}dx$$

Now use $$\int{\frac1{x^n}}dx=-\frac1{(n-1)\times x^{n-1}}, n\not =1$$ to evaluate the integral:

$$-\frac1{(\frac23-1)\times x^{\frac23-1}}$$

Find the difference in the denominator:

$$-\frac1{(-\frac13)\times x^{-\frac13}}$$

Simplify the expression:

$$3\times x^{\frac13}$$

Use $$a^{\frac{m}{n}}=\sqrt[n]{a^m}$$ to transform the expression:

$$3\times \sqrt{x}$$

Now, we can substitute the expression back into our definite integrals:

$$\lim_{a\rightarrow0^{-}}(3\times \sqrt{x})|{-1}^a+\lim{a\rightarrow0^{+}}(3\times \sqrt{x})|_a^1$$

Use $$F(x)|_a^b=F(b)-F(a)$$ to evaluate the expressions:

$$\lim_{a\rightarrow0^{-}}(3\times \sqrt{a}-3\times \sqrt{-1})+\lim_{a\rightarrow0^{+}}(3\times \sqrt{1}-3\times \sqrt{a})$$

Remember that $$\sqrt{-1}=-1$$ and $$\sqrt{1}=1$$ and substitute that into the limits:

$$\lim_{a\rightarrow0^{-}}(3\times \sqrt{a}-3\times (-1))+\lim_{a\rightarrow0^{+}}(3\times 1-3\times \sqrt{a})$$

Simplify the expressions:

$$\lim_{a\rightarrow0^{-}}(3 \sqrt{a}+3)+\lim_{a\rightarrow0^{+}}(3-3 \sqrt{a})$$

Evaluate the limits by substituting $$0$$ for $$a$$:

$$(3 \sqrt{0}+3)+(3-3 \sqrt{0})$$

Remember that $$\sqrt{0}=0$$:

$$(3 \times0+3)+(3-3 \times0)$$

Evaluate the expression:

$$6$$

There we have it! The improper integral $$\int_{-1}^{1}\frac{1}{\sqrt{x^2}}dx$$ is equal to:

$$6$$

Deep breaths — it’s time for another one!

### Example 2

Find the integral:

$$\int_0^{+\infty}\frac{1}{x^2+1}dx$$

We can see that one of the limits of the integral is infinity, so we know we have an improper integral! To evaluate the improper integral, by definition, we need to rewrite it using a limit and a definite integral:

$$\lim_{a\rightarrow +\infty}\left(\int_0^a \frac1{x^2+1}dx\right)$$

As always, to evaluate the definite integral, we first evaluate the indefinite integral:

$$\int\frac1{x^2+1}dx$$

Use $$\int \frac1{x^2+a^2}dx=\frac1a \times \arctan{(\frac{x}{a})}$$ to evaluate the integral:

$$\frac11 \times \arctan{(\frac{x}{1})}$$

Simplify the expression:

$$\arctan{(x)}$$

To evaluate the definite integral, return the limits of integration:

$$\arctan{(x)}|_0^a$$

Use $$F(x)|_a^b=F(b)-F(a)$$ to evaluate the expression:

$$\arctan{(a)}-\arctan{(0)}$$

Since $$\arctan{(0)}=0$$, simplify the expression:

$$\arctan{(a)}$$

Now substitute $$\arctan{(a)}$$ for $$\int_0^a \frac1{x^2+1}dx$$ into the limit $$\lim_{a\rightarrow +\infty}\left(\int_0^a \frac1{x^2+1}dx\right)$$:

$$\lim_{a\rightarrow +\infty}\left(\arctan{(a)}\right)$$

Remember that when $$x$$approaches $$+\infty$$, the function $$\arctan(x)$$ approaches $$\frac{\pi}2$$, so evaluate the limit:

$$\frac{\pi}2$$

So, the improper integral $$\int_0^{+\infty}\frac{1}{x^2+1}dx$$ is equal to:

$$\frac{\pi}2$$

Nice job! Feeling good? Feeling ready to try some other problems? Just keep this process in mind:

## Study summary

1. To evaluate the improper integral, by definition, rewrite it using a limit and a definite integral.
2. Evaluate the definite integral.
3. Evaluate the limit.

## Do it yourself!

You might not think practicing math is the most glamorous way to spend your time, but it really does help in the long run. When you’ve got some time, try these practice problems and see how you do!

Find the integral:

1. $$\int_0^3 \frac{1}{\sqrt{3-x}}dx$$
2. $$\int_0^{\infty} \frac{x}{(1+x^2)^2}dx$$
3. $$\int_{-\infty}^{-1} \frac{1}{x^{3}} dx$$
4. $$\int_{-1}^1 \frac{1}{\sqrt{1-x^2}} dx$$

Solutions:

1. $$2\sqrt{3}$$
2. $$\frac12$$
3. $$-\frac12$$
4. $$\pi$$

If you’re struggling through the solving process, that’s totally okay! Stumbling a few times is actually good for learning. If you get too stuck or lost, scan the problem using your Photomath app, and we’ll walk you through to the other side!

Here’s a sneak peek of what you’ll see:

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