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Evalution of a complex integral

by Random Variable
Tags: complex, evalution, integral
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Random Variable
#1
Mar1-12, 02:39 PM
P: 116
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?
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mathman
#2
Mar1-12, 04:15 PM
Sci Advisor
P: 6,076
http://en.wikipedia.org/wiki/Fresnel_integral

exp(-ix2) = cos(x2) - isin(x2).

Above reference discusses the integrals as well as the integral from 0 to infinity.
Random Variable
#3
Mar1-12, 05:11 PM
P: 116
Quote Quote by mathman View Post
http://en.wikipedia.org/wiki/Fresnel_integral

exp(-ix2) = cos(x2) - isin(x2).

Above reference discusses the integrals as well as the integral from 0 to infinity.

I want to evaluate the integral without using a closed contour and the residue theorem.

mathman
#4
Mar2-12, 03:56 PM
Sci Advisor
P: 6,076
Evalution of a complex integral

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


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