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Gaussian Integral Evaluation

The evaluation of the Gaussian integral \( \displaystyle I = \int_{-\infty}^{+\infty} e^{-x^2} dx \) using the double integrals and the polar coordinates is presented.

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From Double Integrals in Rectangular Coordinates to Polar Coordinates

Changing a double integrals from rectangualr to polar coordinates is done as follows [1] \[ \iint_R f(x,y) \;dy \;dx = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} f(r,\theta) r \;dr \;d\theta \qquad (I) \] with the relationships between the rectangular coordinates \( x \) and \(y \); and the polar coordinates \( r \) and \( \theta \) given by [3]
\( x = r \cos \theta \) , \( y = r \sin \theta \) , \( r^2 = x^2 + y^2 \)


Evaluate the Gaussian Integral

The Gaussian integral is defined as follows \[ I = \int_{-\infty}^{+\infty} e^{-x^2} dx \] and we need to evaluate \( I \).
We first note that the integrals \( \displaystyle \int_{-\infty}^{+\infty} e^{-x^2} dx \) and \( \displaystyle \int_{-\infty}^{+\infty} e^{-y^2} dy \) have equal values. We therefore may write that

\( I^2 = \displaystyle \left(\int_{-\infty}^{+\infty} e^{-x^2} dx \right) \left(\int_{-\infty}^{+\infty} e^{-y^2} dy \right) \)

The above may be written as a double integral as follows
\( I^2 = \displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-(x^2+y^2)} \; dx \; dy \)

Let \( x = r \cos \theta \) , \( y = r \sin \theta \) , \( r^2 = x^2 + y^2 \) be the relationships between the rectangular and polar coordinates and use the change of integral from rectangular to polar coordinates (I) given above.
\( I^2 = \displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-(x^2+y^2)} \; dx \; dy = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \;dr \;d\theta \)

Note that \( \displaystyle \int_{0}^{\infty} e^{-r^2} r \;dr = -(1/2) e^{-r^2} \), which gives

\( I^2 = \displaystyle \int_{0}^{2\pi} \left[-(1/2) e^{-r^2}\right]_0^{\infty} \;d\theta \)
which may be written as
\( I^2 = \displaystyle \int_{0}^{2\pi} (-1/2) \left[e^{-\infty}-e^0 \right] \;d\theta \)

Note that using limits, \( e^{-\infty} = \lim_{a\to\infty} e^{-a} = 0 \), we now evaluate
\( I^2 = \displaystyle \int_{0}^{2\pi} \dfrac{1}{2} \;d\theta \)

Evaluate the above integral
\( I^2 = \dfrac{1}{2} \left[\theta\right]_0^{2\pi} \)
\( = \dfrac{1}{2} \left[2\pi - 0 \right] \)
\( = \pi \)

So far we have calculated \( I^2 = \pi \) and therefore taking the square root, we have \[ I = \int_{-\infty}^{+\infty} e^{-x^2} dx = \sqrt {\pi}\]

More References and Links

  1. Joel Hass, University of California, Davis; Maurice D. Weir Naval Postgraduate School; George B. Thomas, Jr.Massachusetts Institute of Technology ; University Calculus , Early Transcendentals, Third Edition , Boston Columbus , 2016, Pearson.
  2. Double Integrals Calculations
  3. polar coordinates
  4. Convert Polar to Rectangular Coordinates and Vice Versa
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