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  1. MHB Craftsman
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    #1
    $\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.
    Last edited by Random Variable; February 25th, 2012 at 21:20.

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    #2
    Quote Originally Posted by Random Variable View Post
    $\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

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