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Confused with complex limits

  1. Dec 18, 2006 #1
    [tex]\lim_{x\to\infty}(x+i)=?[/tex]

    What does this diverge to? Intuitively i would have thought (realinfinity + i).

    EDIT:

    Which brings the other question:

    [tex]\int^\infty_0}(x+i)dx=?[/tex]

    How should i interpret this integral considering x is a real number? And what does it diverge to?
     
    Last edited: Dec 18, 2006
  2. jcsd
  3. Dec 18, 2006 #2

    Hurkyl

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    It depends on what points you have placed at infinity.

    For example, if you were working in [itex]\bar{\mathbb{R}}^2[/itex], where [itex]\bar{\mathbb{R}}[/itex] is the extended real numbers, then every number is of the form a + bi, where a and b are any extended real number. (even an infinite one!) And then, you could show

    [tex]
    \lim_{x \rightarrow +\infty} x + i = +\infty + i
    [/tex]

    (even if x is not restricted to the reals!)


    Actually, if you are using any extension of the complexes such that + is continuous at [itex](\infty, i)[/itex] (whatever [itex]\infty[/itex] happens to mean in this context), then it would be true that your limit converges to [itex]\infty + i[/itex]. But it might happen (as it does on the complex projective line and the real projective plane) that [itex]\infty + i = \infty[/itex]. (for the appropriate meaning of [itex]\infty[/itex])
     
  4. Dec 18, 2006 #3
    Great answer :). That did actually clarify some things.

    Now here's the real context of the problem. I'm trying to simplify the following integral. I'm still confused as to which of those systems i should use for this one:

    [tex]\int^\infty_{-\infty}\left(e^{-\alpha(x-\frac{i}{2\alpha}(k_0-k))^2\right)dx}[/tex]

    I can simplify it by substituting [tex]y=x-\frac{i}{2\alpha}(k_0-k)[/tex]. However in the model solutions for this problem, the integral limits do not change after the substitution, even though i'm subtracting a complex number from the real number x. Any reason why the limits should still be INFINITY and -INFINITY after the substititution? (and not [tex]y=\infty-\frac{i}{2\alpha}(k_0-k)[/tex] and [tex]y=-\infty-\frac{i}{2\alpha}(k_0-k)[/tex]
     
    Last edited: Dec 18, 2006
  5. Dec 19, 2006 #4

    Hurkyl

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    Integrals1 don't happen between limits: they happen along paths.

    In the simple world of the reals, any path is (essentially) determined by its endpoints. So, we can get away with defining a definite integral in such a fashion.

    But for what you're doing, you really ought to be specifying the path. Your original integral really ought to be specified as being integrated over the path

    x = t ([itex]t \in \mathbb{R}[/itex])​

    and your second integral as being over the path

    y = t - (i / 2a)(k0 - k) ([itex]t \in \mathbb{R}[/itex]).​

    However, I have seen people use notation similar to what you want to use as shorthand for denoting such vertical or horizontal lines.


    The other thing to note is that (I think), this is an improper integral, which is defined in terms of a limit of things with bounded domain. On each of those bounded things, you don't have any of the problems you've been having!



    1: Well, this type of integral, anyways.
     
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