Solve Improper Integration: Integrand Convergence

  • Thread starter Redoctober
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In summary, the conversation is about using the limit comparison test to determine whether the integral \int^{\infty}_{1}\frac{e^x}{x}.dx is convergent or divergent. The initial attempt was to compare it with g(x) = \frac{e^x}{x +1}, but this was difficult to integrate. The conversation then discusses using the fact that e^x is greater than x for all positive values above x=1. The solution is eventually found by choosing a smaller function g(x) = \frac{e^x}{e^x + 1} and showing that its integral is also divergent, thus proving that the original integral is also divergent.
  • #1
Redoctober
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Help :@ !

the question states that

[tex]\int^{\infty}_{1}\frac{e^x}{x}.dx[/tex]

Determine whether the integrand is convergent or divergent ??

It tried using the limit comparison test but i fail to select a g(x) to compare it with this

i chose
[tex]g(x) = \frac{e^x}{x +1}[/tex]

but this hard to integrate too :S !
I can't find any other function to compare with cause i need to cancel out the e^x

this seek for functions to compare the integral seem more like luck factor dependent lol !
 
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  • #2
Try to use that e^x >= x when x >= 1.
 
  • #3
Well, isn't e^x greater than 1 for ALL positive values for x above x=1?
 
  • #4
I do know that e^x > than x when x>1 thus i know that the integral is divergent . but the question is asking me to prove it via the comparison test

[tex]\lim_{x \to \infty } \frac{f(x)}{g(x)}[/tex]
 
  • #5
Compare it to g(x)=1, then.
 
  • #6
Its alright i found out the solution

It turns out, because f(x) is divergent , i shall choose a smaller function g(x) but diverge too

i chose

[tex]g(x) = \frac{e^x}{e^x + 1}[/tex]

[tex]\int^{\infty}_{1} \frac{e^x}{e^x + 1}.dx [/tex]

the integral is easy to calculate . It will give infinity .

Because the g(x) < f(x) and g(x) integral is divergent , thus f(x) integral is divergant too :D !
 

1. What is improper integration?

Improper integration is the process of finding the area under a curve for functions that do not have a finite limit at one or both of its integration bounds. This can occur when the function is not defined at one or both of the bounds, or when the function approaches infinity at one or both of the bounds.

2. How do you determine if an improper integral converges or diverges?

To determine if an improper integral converges or diverges, you must evaluate the limit of the integral as the bounds approach infinity. If the limit exists and is finite, then the integral converges. If the limit does not exist or is infinite, then the integral diverges.

3. What is the difference between a type 1 and type 2 improper integral?

A type 1 improper integral has one or both of its bounds at infinity, while a type 2 improper integral has one or more discontinuities within its bounds. Type 1 integrals are typically evaluated using a limit, while type 2 integrals can be evaluated using a combination of limits and regular integration techniques.

4. Can improper integrals be solved using regular integration techniques?

Yes, improper integrals can sometimes be solved using regular integration techniques if they can be rewritten as a limit of a regular integral. However, this is not always possible and other methods such as the comparison test or the Cauchy principal value may need to be used.

5. How do you determine the convergence or divergence of an integrand?

The convergence or divergence of an integrand can be determined by evaluating the behavior of the function near the integration bounds. If the function approaches a finite value or zero, then the integral will converge. If the function approaches infinity or is undefined, then the integral will diverge. Additionally, the comparison test can be used to compare the integrand to a known convergent or divergent function.

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