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I Non-Integrable Function

  1. Mar 21, 2017 #1
    Hi All,

    I am wondering if the function below is Integrable:


    When I work it out on computer, the integral is finite from -Inf to Inf. But clearly it has a pole at u=2. Is this pole integrable? If yes, what kind of coordinate transform is required?

    Any help is appreciated!
    Last edited: Mar 21, 2017
  2. jcsd
  3. Mar 22, 2017 #2


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    in the vicinity of u=2, the numerator is bounded away from zero. The integral from 2-ε to 2 is -∞, while that from 2 to 2+ε is +∞. Trying to integrate through u=2 might effectively cancel these, leading to an apparently valid result, but it is not valid. -∞+∞ is undefined.
  4. Mar 22, 2017 #3
    It is integrable in the Cauchy Principal-Value sense: We take limits:

    ##\displaystyle\lim_{\epsilon\to 0} \big\{\int_{-\infty}^{2-\epsilon} f(u)du+\int_{2+\epsilon}^{\infty}f(u)du\big\}##

    We can numerically compute it in Mathematica:
    Code (Text):
    NIntegrate[Exp[(-2^(-1))*(u - 2)^2 - 2*u^2]/(u - 2), {u, -Infinity, 2, Infinity}, Method -> PrincipalValue]

  5. Mar 22, 2017 #4

    That looks totally good. But when I modify my Integrand with different parameters, the same Integral doesn't seem to be Integrable under Cauchy Principal-Value. For instance:


    This integral doesn't converge in mathematica. Why is it so?

    Code (Text):
    NIntegrate[2 Exp[-1/4*(u - 2)^2 - u^2]/Abs[(u - 2)], {u, -Infinity, 2, Infinity}, Method -> PrincipalValue]
    Says failed to converge. Isn't it the same type of integral as my original post. I just changed some parameters.

    One more thing, I originally had forgotten the absolute value on the denominator. So now, it looks like even my original integrand doesn't converge when I force my denominator to be positive.
    Last edited: Mar 22, 2017
  6. Mar 22, 2017 #5


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    Sure, but I find that very unconvincing. If you change one of the limits to 2-2ε it doesn't converge (does it?)
  7. Mar 22, 2017 #6


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    It's related to the contour integral of the analytic function and its residue, which is much less arbitrary.
  8. Mar 23, 2017 #7
    First, realize we are basically looking at a variant of ##\frac{1}{u}##. Basically. And we can integrate this in the principal value sense:

    ##\text{P.V.} \int_{-\infty}^{\infty} 1/u du=0##

    That's because the infinities cancel when taking "symmetric limits" ( as ##\epsilon\to 0##). However, when we take the absolute values, the infinities DO NOT cancel.

    Edit: Maybe not exactly a variant of 1/u but I think the two are similar qualitatively and looking at 1/u is easier to visualize and easier to understand.
    Last edited: Mar 23, 2017
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