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!

Funky integral

  1. Sep 26, 2005 #1
    Funky integral!!

    This integral is driving me nuts :cry: , anyone got a clue?
    Given two real variables X and Y, one defines the function:
    where a, b and c are reals and a>0, b>0.
    Then the function g is defined as:
    I am looking for:
    1- a primitive for g
    2- and/or the value of the following integrals I=integral(-infty,+infty;g(X,Y),dX) and J=integral(-infty,+infty;g(X,Y),dY).
  2. jcsd
  3. Sep 26, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Do you have any reason to believe it has an "elementary" primitive?

    If you take a=-1, b= c= 0, you have
    [tex]f(x,y)= e^{-x^2}[/tex]
    which certainly does NOT have a primitive that can be written in terms of elementary functions. It can of course be written as [itex]2\pi Erf(x)[/tex] where Erf(x) is the error function- but it's not "elementary", it is defined as
    [tex]\frac{1}{2\pi}\int e^{-x^2}dx[/tex]
  4. Sep 26, 2005 #3
    You are right HallsofIvy and the answer to your question is: "No I do not.", this is why my second point starts with: "and/or [...]".
    I have made some (but little) progress on this and I'll let you know in a comming post where I stand now.
  5. Sep 26, 2005 #4
    HallsofIvy, you might have missed the square root in the definition of f:

    [tex]I(X)=\int_{-\infty}^{\infty} e^{-f(X,Y)}dY[/tex]
    Changes of variable:
    first [tex]u=cY+X[/tex]
    then [tex]v=u\sqrt{a+b}[/tex]
    and [tex]w=v-\frac{2bX}{\sqrt{a+b}}[/tex]
    leading to:
    [tex]I(X)=\frac{1}{c\sqrt{a+b}}\int_{-\infty}^{\infty} e^{-\sqrt{w^2+4X^2\frac{ab}{a+b}}}dw[/tex]

    I am now considering a trig transformation:
    to get rid of the square root but I am then stuck again :grumpy:
  6. Sep 28, 2005 #5
    I just found the answer: this integral is well known as the "modified Bessel function of second kind", period.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Funky integral
  1. The Integral (Replies: 17)

  2. An integral (Replies: 1)

  3. Integral of (Replies: 3)

  4. On Integration (Replies: 4)

  5. An integral (Replies: 2)