# Mistake while evaluating the gaussian integral with imaginary term in exponent

I am trying to evaluate the integral $I=\int_0^\infty e^{-ix^2}\,dx$ as one component of evaluating a contour integral but I am dropping a factor of $1/2$ and after checking my work many times, I worry that I am making a conceptual mistake in moving to polar coordinates for the interval $(0,\infty)$ rather than $(-\infty,\infty)$. Below is my work:

\begin{align*} \text{Define } I&=\int_0^R e^{-ix^2} \, dx=\int_0^R e^{-iy^2} \, dy \\ \text{then } I^2&=\int_0^R e^{-ix^2} \, dx \int_0^R e^{-iy^2} \, dy \\ &\Rightarrow I^2=\int_0^R\int_0^R e^{-ix^2}e^{-iy^{2}} \, dx \, dy \\ &\Rightarrow I^2=\int_0^R\int_0^R e^{-i(x^2+y^2)} \, dx \, dy \end{align*}

And converting to polar coordinates, with jacobian $r$ for the integral, and taking $\theta\in[0,\pi]$ since we are integrating in the first quadrant, we have:

\begin{align*} I^2&=\int_0^\pi\int_0^R e^{-i(r^2\cos^2(\theta)+r^2 \sin^2(\theta))} r \, dr \, d\theta\\ I^2&=\int_0^\pi\int_0^R e^{-i(r^2)} r \, dr \, d\theta\\ \end{align*} Which we can now perform a u substitution on: $u=r^2\Rightarrow \frac{du}{2r}=dr$, yielding \begin{align*} I^2&=\frac{1}{2}\int_0^\pi\int_{0=r}^{R=r}e^{-iu} \, du \, d\theta\\ &\Rightarrow I^2=\frac{\pi}{2} \left[\frac{-e^{-iR^2}}{i}+\frac{1}{i}\right] \\ &\Rightarrow I^2=\frac{\pi}{2i}[1-e^{-iR^2}] \\ &\Rightarrow I=\sqrt{\frac{\pi}{2i}}\sqrt{1-e^{-iR^2}} \end{align*} Then in the limit $R\rightarrow \infty$ we have \begin{equation*} \lim_{R\rightarrow \infty}\sqrt{\frac{\pi}{2i}}\sqrt{1-e^{-iR^{2}}}= \sqrt{\frac{\pi}{2i}} \end{equation*}

The mistake is in my eyes that you allow the polar angle to be in $\phi\in[0,\pi]$ although just integrating over a quadrant integration domain in cartesian coordinates,$(x,y)\in[0,R]^2$. I'd suggest putting $\phi\in[0,\pi/2]$ in order to account for the integration in the first quadrant. This modification of your calculation results in an additional factor of $1/2$ when going to polars $(r,\phi)$.

$I = \sqrt{\frac{\pi}{4i}}\sqrt{1-e^{-iR^2}}$.

The limit $R\to\infty$ works for practical purposes.

David

• I thought that might be it, and revisiting it in context (i want to restrict myself to the positive reals in the complex plane) this makes sense. Thank you! I was pulling my hair out – qbert Apr 30 '16 at 6:31
• This answer misses something. See my posted answer. $\qquad$ – Michael Hardy Apr 30 '16 at 6:46

The integral $\displaystyle\int_0^R\int_0^R\cdots \,dx\,dy$ is over a square, $[0,R]^2$.

But the integral $\displaystyle \int_0^\text{something} \int_0^R \cdots \, r \, dr \, d\theta$ is over a sector of a circle, with an arc of a circle as a part of its boundary.

• so is it not possible to convert to polar coordinates and still be integrating over the rectangle? – qbert Apr 30 '16 at 6:58
• ah I see on the wikipedia page now, thank you for alerting me to this! – qbert Apr 30 '16 at 7:18
• @qbert : You can integrate over the rectangle in polar coordinates, but in that case the bounds of integration are more complicated. $\qquad$ – Michael Hardy Apr 30 '16 at 17:50
• Got it. Should I really have this $\sqrt{i}$ term in the denominator? – qbert May 1 '16 at 20:10
• I don't see that you need $\sqrt i$ in the denominator. $\qquad$ – Michael Hardy May 1 '16 at 20:31