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Is this non-integrable?

  1. Jan 12, 2007 #1
    I am interested in the integral

    [tex]\int e^{2 \pi i \left( x^3+ax^2+bx \right) } dx[/tex]

    Since [tex]\int e^{-x^2} dx[/tex] is non-integrable, I suspect this integral may be too, but I am not so sure because of the exponent being imaginary. Does anyone know?

    If it is integrable, does anyone have any idea how to go about solving it?

    If it is not integrable, does anyone have any idea how to most efficiently evaluate it numerically (from say 0 to an arbitrary [tex]x_0[/tex])?
     
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  3. Jan 12, 2007 #2

    HallsofIvy

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    Be careful about the term "non-integrable". Strictly speaking, any continuous function, such as [itex]e^{-x^2}[/itex] is "integrable". It's just that its integral is not any "elementary function". [itex]e^{-x^2}[/itex] certainly is integrable- its integral is the error function, Erf(x).

    Since I notice an "i" in your integral, it looks like it would be more appropriate to integrate it in the Complex plane.
     
  4. Jan 12, 2007 #3
    Yes, of course [tex]e^{-x^2}[/tex] is integrable. Silly me!

    Some sort of hypergeometric maybe?
     
    Last edited: Jan 12, 2007
  5. Jan 12, 2007 #4

    D H

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    I suggest integrating the real and complex parts separately. These are easily separable by Euler's formula.
     
  6. Jan 12, 2007 #5
    That doesn't really help much because I still don't know how to do [tex]\int \sin \left[ 2\pi (x^3+ax^2+bx ) \right] dx [/tex] or its cosine counterpart.
     
  7. Jan 12, 2007 #6

    NateTG

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    You could try applying integration by parts twice, and seeing if you get something that cancels, since you'll get a sign chance by pulling the i out of the exponent twice.
     
  8. Jan 13, 2007 #7

    Gib Z

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    Hey Severian, is this the same one from Scienceforums.net? Its Ragib from there.

    I believe the function you described is a constant since the function could be rewritten as [tex](e^{2\pi i})^{x^3+ax^2+bx}[/tex]. Since e^(2*pi*i) is equal to 1, and 1 to the power of anything is 1, I think we have a constant valued function, this should be easy from there, If i am correct.

    EDIT: btw, as you prob know, the integral is then x + C.
     
    Last edited: Jan 13, 2007
  9. Jan 13, 2007 #8
    ^ That isn't true in general. By that logic, Fourier transforms would be much simpler than they are. Infact, if it were true, anything which is of the form

    [tex]\int f(x)e^{ikx}dx[/tex]

    would be

    [tex]2\pi \int f(2\pi y)e^{i2\pi k y}dy = 2\pi \int f(2\pi y)(e^{2\pi i})^{ky}dy = 2\pi \int f(2\pi y)dy = \int f(x) dx[/tex]

    This would mean that Fourier transforming something is just integrating it. That's obviously not true. It's true if you were integrating the modulus of the integrand, but you're not. Instead you're considering it's complex value. So while you'd have

    [tex]\int |e^{2\pi i kx}| dx = \int dx = x+c[/tex]

    You get something else when you don't have the | | signs.
     
  10. Jan 13, 2007 #9

    Gib Z

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    Aww ok then, Im not going to be of any help here then.
     
  11. Jan 14, 2007 #10

    mathwonk

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    you might try the taylor series for your function, as they are easy to integrate term by term.
     
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