# evalution of a complex integral

by Random Variable
Tags: complex, evalution, integral
 P: 116 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?
 Sci Advisor P: 5,772 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.
P: 116
 Quote by 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.

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