Is this approach to evaluating a complex integral valid?

[polygamma]

$\displaystyle \int_{0}^{\infty} e^{-ix^{2}} \ dx = \lim_{R \to \infty} \int_{0}^{R} e^{-ix^{2}} \ dx $

I think I'm perfectly justified in treating $i$ like a constant since we're integrating with respect to a real variable.$ \displaystyle = \frac{1}{\sqrt{i}} \ \lim_{R \to \infty} \int_{0}^{i\sqrt{R}} e^{-u^{2}} \ du $ EDIT: Initially we were integrating along the positive real axis. By making that substitution, we are now integrating along the line in the complex plane that begins at the origin and makes a 45 degree angle with the positive real axis. But since the exponential function is analytic, the path itself doesn't matter. All that matters is the starting point and the ending point.$= \displaystyle \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \lim_{R \to \infty} \int_{0}^{i\sqrt{R}} e^{-u^{2}} \ du = \frac{\sqrt{\pi}}{2} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \lim_{R \to \infty} \text{erf} \left( i \sqrt{R} \right) $$ \displaystyle = \frac{\sqrt{\pi}}{2} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right)$ since the limit evaluates to 1 (which can be evaluated by using the asymptotic expansion of the error function)Depending on whom I ask, I get a different response. Some say it's totally valid; others say that it needs more justification to be valid; while a few contend it's totally invalid. Which is it?I've seen people say that $\displaystyle \int_{0}^{\infty} e^{-ix^{2}} \ dx = \frac{1}{\sqrt{i}} \int_{0}^{\infty} e^{-u^{2}} \ du$. Now that would require more justification.

 

[CaptainBlack]

$\displaystyle \int_{0}^{\infty} e^{-ix^{2}} \ dx = \lim_{R \to \infty} \int_{0}^{R} e^{-ix^{2}} \ dx $

I think I'm perfectly justified in treating $i$ like a constant since we're integrating with respect to a real variable.$ \displaystyle = \frac{1}{\sqrt{i}} \ \lim_{R \to \infty} \int_{0}^{i\sqrt{R}} e^{-u^{2}} \ du $ EDIT: Initially we were integrating along the positive real axis. By making that substitution, we are now integrating along the line in the complex plane that begins at the origin and makes a 45 degree angle with the positive real axis. But since the exponential function is analytic, the path itself doesn't matter. All that matters is the starting point and the ending point.$= \displaystyle \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \lim_{R \to \infty} \int_{0}^{i\sqrt{R}} e^{-u^{2}} \ du = \frac{\sqrt{\pi}}{2} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \lim_{R \to \infty} \text{erf} \left( i \sqrt{R} \right) $$ \displaystyle = \frac{\sqrt{\pi}}{2} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right)$ since the limit evaluates to 1 (which can be evaluated by using the asymptotic expansion of the error function)Depending on whom I ask, I get a different response. Some say it's totally valid; others say that it needs more justification to be valid; while a few contend it's totally invalid. Which is it?I've seen people say that $\displaystyle \int_{0}^{\infty} e^{-ix^{2}} \ dx = \frac{1}{\sqrt{i}} \int_{0}^{\infty} e^{-u^{2}} \ du$. Now that would require more justification.

For \(x \in \mathbb{R}\) we can write:

\[e^{-ix^2}=\cos(x^2)-i\sin(x^2)\]

So:

\[ \int_0^{\infty}e^{-ix^2}=\int_0^{\infty} \cos(x^2)\;dx - i \int_0^{\infty}\sin(x^2)\; dx = \sqrt{\frac{\pi}{8}}(1-i)\]

Which without checking in detail looks like your last line to me.

CB

 

[math771]

I think your approach is valid. You've just substituted a different variable, which is allowed as long as the limits of integration are still the same. And since the exponential function is analytic, the path of integration doesn't matter. So your substitution is perfectly fine. As for the justification, you could mention that you are just substituting a different variable and that the path of integration doesn't matter because the function is analytic. That should be enough to convince anyone who doubts your approach.