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Evalution of a complex integral

  1. Mar 1, 2012 #1
    Is there a problem with the following evaluation?


    [itex]\displaystyle \int e^{-ix^{2}} \ dx = \frac{1}{\sqrt{i}} \int e^{-u^{2}} \ du = \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \text{erf}(u) + C = \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \text{erf} (\sqrt{i}x) + C[/itex]


    So [itex] \displaystyle \int_{0}^{\infty} e^{-ix^{2}} \ dx = \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \text{erf} (\sqrt{i}x) \Big|^{\infty}_{0} = \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \text{erf} (\sqrt{i} \infty) [/itex]

    or more precisely [itex] \displaystyle \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \lim_{R \to \infty} \text{erf} (\sqrt{i} R) [/itex]


    The error function has an essential singularity at [itex] \infty [/itex] , so the limit as you approach [itex] \infty [/itex] is path dependent. But aren't we looking specifically for the limit as we approach [itex] \infty [/itex] on the line that originates at the origin and makes a 45 degree angle with the positive real axis?

    So my idea was to use asymptotic expansion of the error function ([itex] \displaystyle 1 - e^{-x^{2}} O \left( \frac{1}{x} \right) [/itex]), replace [itex] x [/itex] with [itex] \sqrt{i} R[/itex], and take the limit as [itex]R[/itex] goes to [itex] \infty [/itex]. Is that valid?
     
  2. jcsd
  3. Mar 1, 2012 #2

    mathman

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  4. Mar 1, 2012 #3

    I want to evaluate the integral without using a closed contour and the residue theorem.
     
  5. Mar 2, 2012 #4

    mathman

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    You can carry out the integrals for the cos and sin from 0 to T and let T -> ∞.
     
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