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Gaussian integral

Under a normal distribution, μ = 0 and σ = 1, but when then integrating this equation, I get an error function.

Without using Riemann sums, how can I prove that this equation = 1?

I have only had a year of calc, no multivar.

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  • $\begingroup$ Have you studied multivariable calculus? $\endgroup$ Jan 29, 2015 at 17:16
  • $\begingroup$ @MhenniBenghorbal I have not studied multivariable calculus. I updated the function to give a more specific definition of the Gaussian function, according to wikipedia. $\endgroup$
    – aplet
    Jan 29, 2015 at 17:23
  • $\begingroup$ Do you know gamma function? See my answer. $\endgroup$ Jan 29, 2015 at 17:25
  • $\begingroup$ I think it would be best to learn double integrals and multivariate integration by substitution. This lets you identify the square of the integral fairly easily, and then you can take the square root to conclude. $\endgroup$
    – Ian
    Jan 29, 2015 at 18:19
  • $\begingroup$ The function $\exp(-(x-\mu)^2/(2\sigma^2))$ is integrable on $-\infty$ to $\infty$. The factor $1/(\sigma\sqrt{2\pi})$ is chosen exactly to make the integral equal to $1$. When you study multivariable calculus, you will learn an easy method to evaluate this constant. The answer by slo is merely repeating what you will see in your calculus book at that time. $\endgroup$
    – GEdgar
    Feb 19, 2020 at 16:16

1 Answer 1

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This proof uses as little multivariable calculus as I can manage. Let $f$ be the standard normal distribution $$f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},$$ $$\int_{-\infty}^\infty f(x)\,dx=A$$ and $$g(x,y)=f(x)f(y)=\frac{1}{2\pi}e^{-(x^2+y^2)/2}.$$

By rotational symmetry, the area under $g$ can be evaluated using shell integration in $r=\sqrt{x^2+y^2}$ and then a change of variables to $t = r^2$: $$V=\int_0^\infty 2\pi r \frac{1}{2\pi }e^{-r^2/2}dr=\int_0^\infty \tfrac{1}{2} e^{-t/2} dt = 1.$$ However it can also be evaluated using a double integral: $$\begin{align}V&=\int_{-\infty}^\infty\int_{-\infty}^\infty f(x)f(y)\,dxdy\\ &=\int_{-\infty}^\infty f(x)\,dx\cdot\int_{-\infty}^\infty f(y)\,dy\\ &=\left[\int_{-\infty}^\infty f(x)\,dx\right]^2\\ &=A^2 \end{align}$$ Since $1=V=A^2$, $A=1$. Now if $f$ is not the standard normal distribution, but another normal distribution, do the variable change $z=(x-\mu)/\sigma$. $$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/(2\sigma)^2}\,dx=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz=1$$

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