
#1
Mar112, 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? 



#2
Mar112, 04:15 PM

Sci Advisor
P: 5,939

http://en.wikipedia.org/wiki/Fresnel_integral
exp(ix^{2}) = cos(x^{2})  isin(x^{2}). Above reference discusses the integrals as well as the integral from 0 to infinity. 



#3
Mar112, 05:11 PM

P: 116

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



#4
Mar212, 03:56 PM

Sci Advisor
P: 5,939

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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