1. Limited time only! Sign up for a free 30min personal 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!

Finding Integrals of Non-Elementary Functions

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data
    We know that F(x) = [itex]\int^{x}_{0}e^{e^{t}} dt[/itex] is a continuous function by FTC1, though it is not an elementary function. The Functions [itex]\int\frac{e^{x}}{x}dx[/itex] and [itex]\int\frac{1}{lnx}dx[/itex] are not elementary funtions either but they can be expressed in terms of F.

    a) [itex]\int^{2}_{1}\frac{e^{x}}{x}dx[/itex]

    b)[itex]\int^{3}_{2}\frac{1}{lnx}dx[/itex]


    2. Relevant equations
    I only used and u-substitution in my attempt at a solution, but I suppose integration by parts and trigonometric substitution is fair game as well.

    Integration by parts: [itex]\int udv[/itex] = uv - [itex]\int vdu[/itex]

    3. The attempt at a solution

    a) [itex]\int^{2}_{1}\frac{e^{x}}{x}dx[/itex]

    Let x = et
    dx =etdx

    [itex]\int\frac{e^{e^{t}}e^{t}}{e^{t}}dt[/itex]

    [itex]\int^{2}_{1}e^{e^{t}} dt[/itex]

    Is this answer correct? If it isn't, where did I go wrong? If it is, how do I put it in terms of F(x)?

    b)[itex]\int^{3}_{2}\frac{1}{lnx}dx[/itex]

    let u = ln(x)
    du = 1/x dx
    xdu = dx
    eudu = dx

    [itex]\int\frac{e^{u}}{u}dx[/itex]

    Which this is just part a again.. but am I allowed to do u substitution that way by plugging in u into my du line since it ended up being in terms of an x I didn't have to substitute with?
     
  2. jcsd
  3. Sep 11, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your limits on the last integral are incorrect; those are the ##x## limits. You need the new ##t## limits that correspond to them.

    Once you have any integral of the form ##\int_a^b e^{e^t}~dt## you should be able to write it as ##F(b) - F(a)##. Do you see why?
     
  4. Sep 11, 2014 #3
    So to find the correct variables...

    2 = et
    ln(2) = t
    and
    1 = et
    ln(1) = t = 0

    [itex]\int^{ln(2)}_{0}e^{e^{t}} dt[/itex]

    and the final answer would be F(ln2) - F(0) because FTC1 says

    F(x) = [itex]\int^{b}_{a}f(x) dx[/itex] = F(b) - F(a)

    Right?
     
  5. Sep 11, 2014 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    and what is F(0)?

    Yes and no. How can you have x on the left and no x on the right?
     
  6. Sep 11, 2014 #5
    I'm not sure what F(0) is. I feel inclined to say that F(0) = e, but that's without evaluating [itex]\int e^{e^{t}}[/itex]. But I don't know how to evaluate that integral. Do you want me to try to evaluate it or are you looking for a different answer? I'm confused.

    For your second question... x isn't on the right side of that equation because a and b are being plugged in for x? I'm confused here too.
     
  7. Sep 11, 2014 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##F(x) = \int_0^x e^{e^t}~dt##. What do you get if you put ##x=0##?

    With that definition of ##F(x)## does it work out that ##F(b)-F(a) = \int_a^be^{e^t}~dt##?
     
  8. Sep 11, 2014 #7
    If x = 0, then we're taking the integral from 0 to 0, which would have to be 0. Right?

    Does F(b) - F(a) = [itex]\int^{b}_{a} e^{e^{t}} dx[/itex]? I don't see why not..

    Is it supposed to be F(b)F'(b) - F(a)F'(a)? I know that happens every now and then, but I'm pretty sure that's only for the case in which I'm looking for the derivative of F(x).
     
  9. Sep 12, 2014 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes.

    That's not much of an answer. Do the calculation. Use the formula for ##F(x)## to write down ##F(b)## and ##F(a)## and work it out.
     
  10. Sep 12, 2014 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that is correct. What LCKurtz was objecting to before, when you wrote
    [tex]F(x)= \int_a^b e^{e^t} dt= F(b)- F(a)[/tex]
    was the "F(x)= " on the left. There is NO "x" in [itex]\int_a^b e^{e^t} dt= F(b)- F(a)[/itex] so it cannot be any function of x.
     
  11. Sep 12, 2014 #10

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Watch out, on the right side of the last equation you should have [itex] e^t dt [/itex], not [itex] e^t dx [/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding Integrals of Non-Elementary Functions
  1. Non elementary integral (Replies: 12)

Loading...