evalution of a complex integral

Random Variable

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?

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

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 -> ∞.