Editing Gaussian function (section)

==Integral of a Gaussian function==
The integral of an arbitrary Gaussian function is<math display="block">\int_{-\infty}^\infty a\,e^{-(x - b)^2/2c^2}\,dx = \ a \, |c| \, \sqrt{2\pi}.</math>

An alternative form is<math display="block">\int_{-\infty}^\infty k\,e^{-f x^2 + g x + h}\,dx = \int_{-\infty}^\infty k\,e^{-f \big(x - g/(2f)\big)^2 + g^2/(4f) + h}\,dx = k\,\sqrt{\frac{\pi}{f}}\,\exp\left(\frac{g^2}{4f} + h\right),</math>
where ''f'' must be strictly positive for the integral to converge.

===Relation to standard Gaussian integral===

The integral
<math display="block">\int_{-\infty}^\infty ae^{-(x - b)^2/2c^2}\,dx</math>
for some [[real number|real]] constants ''a'', ''b'' and ''c'' > 0 can be calculated by putting it into the form of a [[Gaussian integral]]. First, the constant ''a'' can simply be factored out of the integral. Next, the variable of integration is changed from ''x'' to {{math|1=<var>y</var> = <var>x</var> − ''b''}}:
<math display="block">a\int_{-\infty}^\infty e^{-y^2/2c^2}\,dy,</math>
and then to <math>z = y/\sqrt{2 c^2}</math>:
<math display="block">a\sqrt{2 c^2} \int_{-\infty}^\infty e^{-z^2}\,dz.</math>

Then, using the [[Gaussian integral|Gaussian integral identity]]
<math display="block">\int_{-\infty}^\infty e^{-z^2}\,dz = \sqrt{\pi},</math>

we have
<math display="block">\int_{-\infty}^\infty ae^{-(x-b)^2/2c^2}\,dx = a\sqrt{2\pi c^2}.</math>