Gaussian integral

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\frac{d}{dx} \sin x=?\, This article/section deals with mathematical concepts appropriate for a student in late high school or early university.

The Gaussian integral is the integral:

 \int_{-\infty}^{\infty} e^{-x^2}dx.

It has a value of \sqrt\pi. The value is needed to normalize the Normal distribution.

Derivation

First look at the double integral

\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.

Separating it,

\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2

So, the double integral is merely the square of the Gaussian integral.

Now, do the double integral in polar co-ordinates. x2y2 = − (x2 + y2) = − r2 and dxdy = rdrdθ, so:

\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta

 = (\int_{0}^{2\pi}d\theta)(\int_{0}^{\infty} e^{-r^2}rdr)

 = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.

Substituting z = r2 into the integral,

 = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})

 = \pi \int_{0}^{\infty} e^{-z}dz = \pi  \left[-e^{-z} \right]_0^{\infty} = \pi.

Therefore,

(\int_{-\infty}^{\infty} e^{-x^2}dx)^2 = \pi

\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi.

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