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Gaussian Integral · cho.sh

Gaussian Integral

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The Gaussian integral is the integral of the function $e^{-x^2}$ over the entire real line, and it is given by:

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

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−∞

∞

e−x2

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π

First, we square the integral:

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

Next, we change to polar coordinates:

\int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} r d\theta dr = \pi

To evaluate this integral, we use the substitution $u = r^2$ and $du = 2r dr$ , which gives:

\frac{1}{2}\int_{0}^{\infty} e^{-u} du = \frac{1}{2}

Therefore, we have:

\left(\int_{-\infty}^{\infty} e^{-x^2} dx\right)^2 = \pi \quad \Rightarrow \quad \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}

And this completes the proof.

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e−y2

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x

d

y

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