Improper Integrals: Evaluate & Determine Convergence

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In summary, the conversation is discussing how to determine if an improper integral converges and, if so, how to evaluate it. The specific integral being discussed has limits of integration from 1 to 2 and an integrand of dx/(xlnx). The conversation includes various attempts at solving the integral, including a substitution of u=lnx. Ultimately, it is determined that the integral diverges.
  • #1
physstudent1
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Homework Statement


Determine whether the improper integral converges and, if so, evaluate it.

the limits of integration are 1 to 2

the integrand is
(dx/(xlnx))




Homework Equations





The Attempt at a Solution



so first I found the indefinite integral which was ln(ln(x)) + c

now usually these problems have infinite for one of the limits of integration in which case that limit becomes a variable such as a or b, with this one I did notice that the the ln(ln(1)) is going to be taking the ln(0) which you cannot do I'm not sure what to do from here though or what limit I am supposed to be evaluating to see if it converges or not.
 
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  • #2
Have you dealt with things like:
[tex]\int_{0}^{1} \frac{1}{x^2} dx[/tex]
 
  • #3
oh if the exponent is >= 1 then the integral is infinite right?
 
Last edited:
  • #4
so if f(x)>=g(x)>=0

if the integral of a to infinite of f(x) converges then the integral of a to infinite of g(x) conveges and
if the integral of a to infinite of g(x) diverges then the integral of a to infinite of f(x) diverges

how would I go about selecting a function that is less then the one I was given though.
 
  • #5
You should be able to work it out with:
[tex]\lim_{a \rightarrow 1^+} \int_a^2 \frac{1}{x \ln x} dx [/tex]
 
  • #6
ok I figured that limit is positive infinite correct? which would mean it diverges ?
 
  • #7
physstudent1 said:
ok I figured that limit is positive infinite correct? which would mean it diverges ?

I get that it diverges also.
 
  • #8
physstudent1 said:
ok I figured that limit is positive infinite correct? which would mean it diverges ?

what is the integral of 1/xlnx dx
 
  • #9
sutupidmath said:
what is the integral of 1/xlnx dx

[tex]\ln(\ln x)+C[/tex]
(Just like you have.)
 
  • #10
integ dx/xlnx, take the sub lnx=t, dx/x=dt, from here we get

integ dt/t=lnt
but since he is taking a definite integral he need not go back to the original variable, so that means
lim (e-->0) integ(from 1+e to 2) of dx/xlnx=lim(e-->0) ln(t) on the interval 1+e to 2, hence

lim(e-->0) [ln2-ln(1+e)]=ln2
so it converges!
 
  • #11
sutupidmath said:
so it converges!
I see what you did with the substitution, but you got the limits wrong.

[tex]\int^2_1\frac{1}{xlnx}dx[/tex] [tex]u=lnx => du = \frac{dx}_{x}[/tex]
Now you have to plug in you original limits to find out what the limits in u are.
[tex]ln(1) = 0 , ln(2) = ln(2)[/tex]
in the end you end up with;
[tex]ln(ln2) - ln(0) [/tex]
[tex] ln(x) -> -\infty as x -> 0[/tex]
So it diverges.
 
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  • #12
where are you getting e from?
 
  • #13
<--- said:
I see what you did with the substitution, but you got the limits wrong.

[tex]\int^2_1\frac{1}{xlnx}dx[/tex] [tex]u=lnx => du = \frac{dx}_{x}[/tex]
Now you have to plug in you original limits to find out what the limits in u are.
[tex]ln(1) = 0 , ln(2) = ln(2)[/tex]
in the end you end up with;
[tex]ln(ln2) - ln(0) [/tex]
[tex] ln(x) -> -\infty as x -> 0[/tex]
So it diverges.

Oh yeah, sorry my bad. I totally missed this part. I can't believe i forgot to change the limits of integration after i made that substitution!
 
  • #14
physstudent1 said:
where are you getting e from?

I just let e =epsylon, sorry i should have defined that!
 

Related to Improper Integrals: Evaluate & Determine Convergence

1. What is an improper integral?

An improper integral is an integral where either the upper or lower limit of integration is infinite or the function being integrated has an infinite value at a point within the limits of integration.

2. How do you evaluate an improper integral?

To evaluate an improper integral, you must first determine if it converges or diverges. Then, you can use various techniques such as the limit comparison test, comparison test, or the integral test to evaluate the integral.

3. What is the difference between a convergent and divergent improper integral?

A convergent improper integral is one where the limit of integration exists and has a finite value. A divergent improper integral is one where the limit of integration does not exist or has an infinite value.

4. Can improper integrals have both upper and lower limits of infinity?

Yes, it is possible for an improper integral to have both upper and lower limits of infinity. In this case, both limits must be evaluated separately and then the two results can be added together.

5. Why is it important to determine the convergence of an improper integral?

Determining the convergence of an improper integral is important because it tells us whether the integral has a finite value or not. If it converges, we can use the integral to calculate important quantities such as areas, volumes, and probabilities. If it diverges, the integral does not have a meaningful value.

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