Is There a Problem with This Complex Integral Evaluation?

[Random Variable]

Is there a problem with the following evaluation?\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 So \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)

or more precisely \displaystyle \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \lim_{R \to \infty} \text{erf} (\sqrt{i} R)The error function has an essential singularity at \infty , so the limit as you approach \infty is path dependent. But aren't we looking specifically for the limit as we approach \infty 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 (\displaystyle 1 - e^{-x^{2}} O \left( \frac{1}{x} \right)), replace x with \sqrt{i} R, and take the limit as R goes to \infty. Is that valid?

 

[mathman]

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]

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]

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