## 17Calculus Improper Integrals - Limit Comparison Test

##### 17Calculus

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

Topics You Need To Understand For This Page

[domain] - [basic integration] - [limits at infinity] - [improper integrals]

Limit Comparison Test For Improper Integrals - Theorem

For positive, continuous and real functions, $$f(x)$$ and $$g(x)$$ on the interval $$[a,\infty]$$ and $\lim_{x \to \infty}{ \frac{f(x)}{g(x)} } = L, ~ \text{ with } ~ 0 \lt L \lt \infty$ then the integrals
$\int_a^{\infty}{ f(x) ~ dx }, ~~~ \int_a^{\infty}{ g(x) ~ dx }$ either both converge or both diverge.

Limit Comparison Test Development

Here is a video explaining the logic behind the Limit Comparison Test for improper integrals.

### JCCCmath - Limit Comparison Test Integrals - Development [7min-10secs]

video by JCCCmath

Practice

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

$$\displaystyle{ \int_3^{\infty}{ \frac{x^3-x+2}{x^5+x^4-x^3} ~dx } }$$

Problem Statement

Determine if $$\displaystyle{ \int_3^{\infty}{ \frac{x^3-x+2}{x^5+x^4-x^3} ~dx } }$$ converges or diverges using the Limit Comparison Test.

Solution

### JCCCmath - 4300 video solution

video by JCCCmath

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$$\displaystyle{ \int_1^{\infty}{ \frac{1}{(x-\ln x)^2} ~dx } }$$

Problem Statement

Determine if $$\displaystyle{ \int_1^{\infty}{ \frac{1}{(x-\ln x)^2} ~dx } }$$ converges or diverges using the Limit Comparison Test.

Solution

### Michael Penn - 4301 video solution

video by Michael Penn

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$$\displaystyle{ \int_3^{\infty}{ \frac{1}{2+\cos x + \ln x} ~dx } }$$

Problem Statement

Determine if $$\displaystyle{ \int_3^{\infty}{ \frac{1}{2+\cos x + \ln x} ~dx } }$$ converges or diverges using the Limit Comparison Test.

Hint

Use $$g(x) = 1/\ln x$$ as the comparison function.

Problem Statement

Determine if $$\displaystyle{ \int_3^{\infty}{ \frac{1}{2+\cos x + \ln x} ~dx } }$$ converges or diverges using the Limit Comparison Test.

The improper integral $$\displaystyle{ \int_3^{\infty}{ \frac{1}{2+\cos x + \ln x} ~dx } }$$ diverges by the Limit Comparison Test.

Problem Statement

Determine if $$\displaystyle{ \int_3^{\infty}{ \frac{1}{2+\cos x + \ln x} ~dx } }$$ converges or diverges using the Limit Comparison Test.

Hint

Use $$g(x) = 1/\ln x$$ as the comparison function.

Solution

 Set up the limit $$\displaystyle{ \lim_{x \to \infty} { \frac{f(x)}{g(x)} } }$$. Use the hint for $$g(x) = 1/\ln x$$. $$\displaystyle{ L = \lim_{x \to \infty}{ \frac{1/(2+\cos x + \ln x)}{1/\ln x} } }$$ $$\displaystyle{ L = \lim_{x \to \infty}{ \frac{1}{(2+\cos x + \ln x)/(\ln x)} } }$$ $$\displaystyle{ L = \lim_{x \to \infty}{ \frac{1}{(2/\ln x)+(\cos x/\ln x) + 1} } }$$ As $$x$$ gets very large, $$\displaystyle{ \lim_{x \to \infty}{ 2/\ln x } = 0 }$$ And, as $$x$$ gets very large, $$\displaystyle{ \lim_{x \to \infty}{ \cos x/\ln x } = 0 }$$ This leaves $$L = 1$$. Since $$L$$ is finite and non-zero, then the original integral will converge or diverge the same as $$\displaystyle{ \int_3^{\infty}{g(x)~dx} }$$. So now we need to determine if this integral converges or diverges. We can directly compare $$y=\ln x$$ with $$y=x$$. Since $$\ln x \lt x$$, $$1/x \lt 1/\ln x$$. Also, $$\displaystyle{ \int_3^{\infty}{ 1/x ~dx } \lt \int_3^{\infty}{ 1/\ln x ~ dx } }$$ Since $$\displaystyle{ \int_3^{\infty}{ 1/x ~dx } }$$ diverges, then $$\displaystyle{\int_3^{\infty}{ 1/\ln x ~ dx } }$$ also diverges and so does the original integral.

One thing we left off at the first was to check that $$f(x) \gt 0$$. However, we think you can see that since the numerator is always positive, $$2+\cos x$$ is always positive and $$\ln x$$ is always positive, the function $$f(x) \gt 0$$ on the interval of integration. Although not required, we decided to plot the function just to see what it looks like.

The improper integral $$\displaystyle{ \int_3^{\infty}{ \frac{1}{2+\cos x + \ln x} ~dx } }$$ diverges by the Limit Comparison Test.

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$$\displaystyle{ \int_3^{\infty}{ \frac{x~dx}{\sqrt{x^5+x^3}} } }$$

Problem Statement

Determine if $$\displaystyle{ \int_3^{\infty}{ \frac{x~dx}{\sqrt{x^5+x^3}} } }$$ converges or diverges using the Limit Comparison Test.

The improper integral $$\displaystyle{ \int_3^{\infty}{ \frac{x~dx}{\sqrt{x^5+x^3}} } }$$ converges by the Limit Comparison Test.

Problem Statement

Determine if $$\displaystyle{ \int_3^{\infty}{ \frac{x~dx}{\sqrt{x^5+x^3}} } }$$ converges or diverges using the Limit Comparison Test.

Solution

 We can see that the integrand is positive on the integration interval. So now we need to find a comparison function $$g(x)$$. As we look at the integrand, the $$x^5$$ under the radical dominates the $$x^3$$ term as $$x$$ gets very large. So we can ignore the $$x^3$$ term. This leaves $$x/x^{5/2} = 1/x^{3/2}$$. So we will use $$g(x) = 1/x^{3/2}$$ as our comparison function. Now let's set up the limit $$\displaystyle{ L = \lim_{x\to\infty}{f(x)/g(x)} }$$ $$\displaystyle{ L = \lim_{x\to\infty}{ (x/\sqrt{x^5+x^3})/(1/x^{3/2}) } }$$ $$\displaystyle{ L = \lim_{x\to\infty}{ \frac{x^{5/2}}{\sqrt{x^5+x^3}} } }$$ $$\displaystyle{ L = \lim_{x\to\infty}{ \frac{1}{\sqrt{1+1/x}} } = 1 }$$ The limit is real, finite and positive. So now we need to determine if the integral of $$g(x)$$ converges or diverges. Fortunately, we can evaluate $$\displaystyle{ \int_3^{\infty}{ 1/x^{3/2} ~dx } }$$ $$\displaystyle{ \int_3^{\infty}{ x^{-3/2} ~dx } }$$ $$\displaystyle{ \lim_{b\to\infty}{\int_3^{b}{ x^{-3/2} ~dx }} }$$ $$\displaystyle{ \lim_{b\to\infty}\left[ \frac{x^{-1/2}}{-1/2} \right]_3^{b} }$$ $$\displaystyle{ \lim_{b\to\infty}\left[ \frac{-2}{x^{1/2}} \right]_3^{b} }$$ $$\displaystyle{ \lim_{b\to\infty}\frac{-2}{(b)^{1/2}} - \frac{-2}{3^{1/2}} }$$ $$0 + 2/\sqrt{3} = 2/\sqrt{3}$$ Since the comparison integral converged, so does our original integral.

Note that the improper integral of $$g(x)$$ evaluated to $$2/\sqrt{3}$$ but that does not mean that the original integral also would evaluate to that same number. For this problem, it's just the fact that the comparison integral converged that was significant.

The improper integral $$\displaystyle{ \int_3^{\infty}{ \frac{x~dx}{\sqrt{x^5+x^3}} } }$$ converges by the Limit Comparison Test.

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