1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration with upper limit infinity

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\stackrel{\infty}{0}\int2e^{ky}dy=3/2[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I got up to:
    [tex]\stackrel{lim}{x\rightarrow\infty}\[[e^{ky}]^{x}_{0}=3k/4[/tex]
    [tex]\stackrel{lim}{x\rightarrow\infty}\[[e^{kx}]=\frac{3k+4}{4}
    [/tex]

    I have no idea how to work that out. Any help will be much appreciated.
     
  2. jcsd
  3. Oct 13, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Can you show us how you got those crazy formulas?
     
  4. Oct 13, 2009 #3
    I went

    [tex]\frac{2}{k}\stackrel{\infty}{0}\int(k)e^{ky}dy=3/2[/tex]
    [tex]\stackrel{\infty}{0}\int(k)e^{ky}dy=3k/4[/tex]

    ^ That's the integral from 0 to infinity, I couldn't figure out how to make it look nice.
     
  5. Oct 13, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, the integral from 0 to x of 2*e^(ky)dy is (2/k)(exp(ky)) from y=0 to y=x. Which is limit (2/k)*(exp(k*x)-1) as x->infinity. Better assume k<0 or it won't converge. What's that limit? Equate that to 3/2. What's k?
     
  6. Oct 14, 2009 #5

    Mark44

    Staff: Mentor

    [tex]\frac{2}{k}\int_0^{\infty}(k)e^{ky}dy[/tex]
    [tex]=\frac{2}{k}\lim_{b \rightarrow \infty}\int_0^{b}e^{ky}kdy[/tex]
    [tex]=\frac{2}{k}\left[\lim_{b \rightarrow \infty}e^{ky} \right]_{y = 0}^{b}[/tex]
    In the 2nd line above, you used the substitution u = ky, du = kdy. Can you take it from here?

    Note: I made it "look nice." You can see my LaTeX script by clicking on any of the lines.
     
  7. Oct 14, 2009 #6
    [tex]
    \left[\lim_{x \rightarrow \infty}e^{kx} \right]=\frac{3k}{4}+1
    [/tex]

    This is what I'm up to (as in the OP), and I get everything up to here, but it's the limit I'm having trouble with.
     
  8. Oct 14, 2009 #7

    Mark44

    Staff: Mentor

    How on earth did you get that? I'm especially curious to learn where the 3 came from.

    Do you know what the graph of y = 2ekx looks like. Your original integral can be interpreted as the area between the graph of this function and the x-axis, for x >= 0. The value you get depends on k, which you didn't specify in the OP.
     
  9. Oct 14, 2009 #8
    The 3 comes from the original problem, where the RHS was 3/2.

    But I've now realised I wrote the problem down wrong, the RHS is 1/2 not 3/2.

    I'm meant to be solving for k, which is why I didn't specify the value for k.

    So it's really meant to be
    [tex]

    \left[\lim_{x \rightarrow \infty}e^{kx} \right]=\frac{k}{4}+1

    [/tex]

    I'm not trying to find what the integral equals, I'm solving for k but I'm still confused about how to get that limit out.
     
  10. Oct 14, 2009 #9

    Mark44

    Staff: Mentor

    So is this the problem?
    Find k so that
    [tex]\int_0^{\infty} 2 e^{ky} dy~=~ \frac{1}{2}[/tex]
    This is equivalent to
    [tex]\int_0^{\infty} e^{ky} dy~=~ \frac{1}{4}[/tex]

    To solve for k, you need to carry out the integration, and since the integral is improper (one limit is infinity), you need to evaluate this integral.
    [tex]\lim_{b \rightarrow \infty} \int_0^{b} e^{ky} dy~=~ \frac{1}{4}[/tex]

    For some values of k, the limit will not exist, because e^(ky) grows too large. You need to determine conditions on k so that the limit exists, and therefore that the integral is a finite number. See my work in post 5 of this thread.
     
  11. Oct 14, 2009 #10
    Ok, I think I got it.

    If k<0, then that limit is zero, which means k=-4?
     
  12. Oct 14, 2009 #11

    Mark44

    Staff: Mentor

    Yes.
     
  13. Oct 14, 2009 #12
    Sweet, thanks. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration with upper limit infinity
Loading...