## 17Calculus - Improper Integrals - Direct Comparison Test

##### 17Calculus

On this page we discuss how to determine if an improper integral converges or diverges using the Direct Comparison Test.

As you have learned, evaluating integrals is not always possible. But sometimes we may not really care exactly what an integral evaluate to, i.e. we may not be interested in the exact number. We may only care if an integral converges or diverges. We have a special test that can sometimes be used to determine convergence or divergence, called the Direct Comparison Test.
Note - If you have studied infinite series and learned the Direct Comparison Test for series, this test for improper integrals works exactly the same way. If you have haven't studied infinite series yet, no worries. We will explain this thoroughly so you do not need that background.

Here is the idea.

Direct Comparison Test for Improper Integrals

Case 1: If $$f(x) \geq 0$$ on the interval $$[a,\infty]$$

and $$\displaystyle{ \int_a^{\infty}{ f(x) ~ dx } }$$ converges

and if we have a second function $$g(x)$$ which is also $$g(x) \geq 0$$ on $$[a,\infty]$$

and we know that $$g(x) \leq f(x)$$ for all $$x$$ in the interval $$[a,\infty]$$,

then we can conclude that $$\displaystyle{ \int_a^{\infty}{ g(x) ~ dx } }$$ also converges.

Case 2: If $$f(x) \geq 0$$ on the interval $$[a,\infty]$$

and $$\displaystyle{ \int_a^{\infty}{ f(x) ~ dx } }$$ diverges

and if we have a second function $$g(x)$$ which is also $$g(x) \geq 0$$ on $$[a,\infty]$$

and we know that $$g(x) \geq f(x)$$ for all $$x$$ in the interval $$[a,\infty]$$,

then we can conclude that $$\displaystyle{ \int_a^{\infty}{ g(x) ~ dx } }$$ also diverges.

A way to summarize this test is to say that a convergent integral holds down a smaller integral and keeps the smaller one from diverging. Similarly, a divergent integral blocks a larger integral, pushing it out to infinity, essentially keeping it from converging.

This video clip has a great explanation of this test including graphs.

### PatrickJMT - Direct Comparison Test for (Improper) Integrals [2min-50secs]

video by PatrickJMT

If you have an improper integral that cannot be evaluated you need to do three things.
1. First, you need to choose a test integral that you know converges or diverges and can will help you show convergence or divergence. This is quite tricky since you need to kind of have a feel for the one you do not know about in order to know what to choose.
3. Secondly, you need to set up the inequality. This may seem simple but it is not. The direction of the inequality is based on whether you think the improper integral converges or diverges.
3. Third, you need to show that the inequality holds. Again, this can be tricky without some direction and experience.

On the first page discussing the concepts of improper integrals, we show this special integral but don't say much about it. This integral will come in handy when using this comparison test.

Special Improper Integral

$$\displaystyle{ \int_{1}^{\infty}{\frac{dx}{x^p}} = \left\{ \begin{array}{llr} \text{converges to} \displaystyle{\frac{1}{p-1}} & & p \gt 1 \\ \text{diverges} & & p \leq 1 \end{array} \right. }$$

This special integral is one that you know when it converges and when it diverges and you can use it as your test series for the comparison test.

How To Choose Test Integral

As mentioned above, the first key to choose a test integral. We will call this integral $$\displaystyle{ \int_{a}^{\infty}{ t(x) ~ dx } }$$ with integrand $$t(x)$$.

Let's say we have an improper integral that looks like $$\displaystyle{ \int_{a}^{\infty}{ f(x) ~ dx } }$$. One technique to start with is to think about what happens to $$f(x)$$ as $$x$$ gets very large, i.e. $$x \to \infty$$. For example, if we have $$g(x) = x^2 + 3x + 1$$. When $$x$$ is very, very large, the $$1$$ has a negligible effect on the value of $$g(x)$$. So does $$3x$$. So we can say for very, very large $$x$$, $$g(x) \approx x^2$$. The $$x^2$$ term dominates significantly.

One way to use this idea is that if you have an integrand like $$\displaystyle{ \frac{1}{x^2 + 3x + 1} }$$, then when $$x$$ is very large, this integrand is going to be very near $$1/x^2$$. To get this we drop the negligible terms in the denominator. Now we can use the above special integral for $$p=2$$ and we know that the test integral converges.

Setting Up The Inequality

Once we have a test series, we have to prove one of the inequalities depending on whether we think the improper integral converges or diverges. Setting up the inequality correctly is critical.
If we think the original integral converges, the inequality is $$f(x) \leq t(x)$$.
If we think the original integral diverges, the inequality is $$f(x) \geq t(x)$$

Proving The Inequality Holds

Again, this can be tricky for some functions. There are three main techniques to use here.
1. Directly
2. Prove an inequality is always positive
3. Using Slope

Technique 1 - Directly
First, directly. In this case, set up the inequality and perform algebraic operations until you get an inequality that always holds. For example, we can show that $$n \leq n+2$$ by subtracting n from both sides to get $$0 \leq 2$$. This last inequality is always true.

Technique 2 - Prove an inequality is always positive
If the direct way is not possible, try moving all the terms to one side leaving zero on the other side and then explain how the expression is always positive. For example, if we can get something like $$0 \leq \displaystyle{ \frac{n+5}{n^3} }$$ we can argue that, since n is always positive, both the numerator and denominator are positive, resulting in the right side always being greater than zero. The key to remember here is that n starts at zero or one and is always positive after that.

This technique also works if you can find a value N such that the expression holds for all $$n \geq N$$. Similar to the last example, you can use this argument for the inequality $$0 \leq \displaystyle{ \frac{n-5}{n^3} }$$by saying that for $$n \geq 6$$, the inequality holds.

Technique 3 - Using Slope
The third technique is to use the concept of slope and is best demonstrated with an example. Let's show that $$n \geq \ln(n)$$ using slope.

The first thing to remember about slope is, to find the slope, you take the derivative and the derivative is defined only on continuous functions. In our case $$n$$ is discrete ($$n$$ takes on the discrete values 1, 2, 3, 4, ... but no value in between these numbers.), so we need to find continuous functions that have the same values at the discrete values. We don't care what is going on between the discrete values as long as the function is continuous. So for $$n$$, we can use $$f(x)=x$$ and for $$\ln(n)$$ we can use $$g(x)=\ln(x)$$ where $$x$$ is a continuous variable in both functions. Now we have continuous functions so that we can take derivatives. I know this is a fine point, but we need to get it right.

Okay, we need to show that $$x \geq \ln(x)$$for all $$x$$ greater than some value. Let's call $$f(x) = x$$and $$g(x) = \ln(x)$$. If we can show that $$f(x) \geq g(x)$$ for some specific value of $$x$$ and the slope of $$f(x)$$ is greater than the slope of $$g(x)$$, then $$f(x)$$ will always be greater than $$g(x)$$. The graphs will never cross and the inequality $$x \geq \ln(x)$$ will hold. You can see this intuitively in the graph on the right.

Let's see if we can show this. First, we know that when $$x = 1$$, $$f(1) = 1$$and $$g(1) = 0$$. Since $$1 \geq 0$$, we have established a point ($$x=1$$) where $$f(x) \geq g(x)$$. Now we will use slope to establish that the functions stay that way for all $$x > 1$$.

Taking the derivatives, we have $$f'(x) = 1$$ and $$g'(x) = 1/x$$.
We need to show $$f'(x) \geq g'(x)$$

$$\displaystyle{ \begin{array}{rcl} f'(x) & \geq & g'(x) \\ 1 & \geq & 1/x \\ x & \geq & 1 \end{array} }$$

Now, since, $$x$$ is always greater than or equal to $$1$$, then the slope of $$f(x)$$ is always larger than the slope of $$g(x)$$. This says that $$f(x)$$ is increasing at a faster rate than $$g(x)$$ and therefore will always be larger.

To recap, what we have done here is that we have found a point, $$x=1$$ where the inequality holds. Then we have used slope to say that the inequality holds for all values greater than that value.

If you are unable to prove the inequality, then you need to choose a different test integral. Using the Direct Comparison Test takes practice and time to sink in before you can understand it and use it.

Practice

Unless otherwise instructed, determine if these improper integrals converge or diverge using the Direct Comparison Test.

Given that $$\displaystyle{ \int_{1}^{\infty}{ \frac{1}{e^x} ~dx } }$$ converges, which of the following also converge by the comparison test?
A. $$\displaystyle{ \int_{x}^{\infty}{ \frac{1}{e^x} ~dx } }$$
B. $$\displaystyle{ \int_{1}^{\infty}{ \frac{1}{e^x + x} ~dx } }$$
C. $$\displaystyle{ \int_{1}^{\infty}{ \frac{1}{e^x-1} ~dx } }$$

Problem Statement

Given that $$\displaystyle{ \int_{1}^{\infty}{ \frac{1}{e^x} ~dx } }$$ converges, which of the following also converge by the comparison test?
A. $$\displaystyle{ \int_{x}^{\infty}{ \frac{1}{e^x} ~dx } }$$
B. $$\displaystyle{ \int_{1}^{\infty}{ \frac{1}{e^x + x} ~dx } }$$
C. $$\displaystyle{ \int_{1}^{\infty}{ \frac{1}{e^x-1} ~dx } }$$

Solution

### blackpenredpen - 3556 video solution

video by blackpenredpen

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$$\displaystyle{ \int_{1}^{\infty}{ \frac{x}{\sqrt{1+x^6}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{1}^{\infty}{ \frac{x}{\sqrt{1+x^6}} ~dx } }$$

Solution

### PatrickJMT - 3550 video solution

video by PatrickJMT

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$$\displaystyle{ \int_{0}^{\pi}{ \frac{\sin^2x}{\sqrt{x}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{\pi}{ \frac{\sin^2x}{\sqrt{x}} ~dx } }$$

Solution

### Krista King Math - 3551 video solution

video by Krista King Math

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$$\displaystyle{ \int_{1}^{\infty}{ \frac{x^2-1}{2x^5+3x+17} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{1}^{\infty}{ \frac{x^2-1}{2x^5+3x+17} ~dx } }$$

Solution

### blackpenredpen - 3552 video solution

video by blackpenredpen

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$$\displaystyle{ \int_{2}^{\infty}{ \frac{x^2}{\sqrt{x^5-1}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{2}^{\infty}{ \frac{x^2}{\sqrt{x^5-1}} ~dx } }$$

Solution

### blackpenredpen - 3553 video solution

video by blackpenredpen

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$$\displaystyle{ \int_{e}^{\infty}{ \frac{1}{\ln x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{e}^{\infty}{ \frac{1}{\ln x} ~dx } }$$

Solution

### blackpenredpen - 3554 video solution

video by blackpenredpen

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$$\displaystyle{ \int_{2}^{\infty}{ \frac{x+1}{\sqrt{x^3-1}} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{2}^{\infty}{ \frac{x+1}{\sqrt{x^3-1}} ~dx } }$$

Solution

### blackpenredpen - 3555 video solution

video by blackpenredpen

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$$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt{x+3x^2}} } }$$

Problem Statement

$$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt{x+3x^2}} } }$$

Solution

### Dr Phil Clark - 3557 video solution

video by Dr Phil Clark

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$$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt{x^3+3x^2}} } }$$

Problem Statement

$$\displaystyle{ \int_{1}^{\infty}{ \frac{dx}{\sqrt{x^3+3x^2}} } }$$

Solution

### Dr Phil Clark - 3558 video solution

video by Dr Phil Clark

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$$\displaystyle{ \int_{0}^{\infty}{ e^{-x^2} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{0}^{\infty}{ e^{-x^2} ~dx } }$$

Solution

### Michel vanBiezen - 3559 video solution

video by Michel vanBiezen

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$$\displaystyle{ \int_{1}^{\infty}{ \frac{1+2e^{-x}}{x} ~dx } }$$

Problem Statement

$$\displaystyle{ \int_{1}^{\infty}{ \frac{1+2e^{-x}}{x} ~dx } }$$

Solution

### Michel vanBiezen - 3560 video solution

video by Michel vanBiezen

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