Improper Integrals

  • #1
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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.
 

Answers and Replies

  • #2
NateTG
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Have you dealt with things like:
[tex]\int_{0}^{1} \frac{1}{x^2} dx[/tex]
 
  • #3
270
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oh if the exponent is >= 1 then the integral is infinite right?
 
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  • #4
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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
NateTG
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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
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ok I figured that limit is positive infinite correct? which would mean it diverges ?
 
  • #7
NateTG
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ok I figured that limit is positive infinite correct? which would mean it diverges ?

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

what is the integral of 1/xlnx dx
 
  • #9
NateTG
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what is the integral of 1/xlnx dx

[tex]\ln(\ln x)+C[/tex]
(Just like you have.)
 
  • #10
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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
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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
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where are you getting e from?
 
  • #13
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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
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where are you getting e from?

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

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