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Gaussian Integral

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The area under ex2e^{-x^2}
The Gaussian integral, named after the German mathematician Carl Friedrich Gauss, the integral is
ex2dx=π.\begin{equation*}\int_{-\infty}^{\infty} e^{-x^2}\,\mathrm{d}x=\sqrt{\pi}.\end{equation*}
Abraham de Moivre originally discovered this type of integral in 1733, while Gauss [Gauss1809] published the precise integral in 1809, attributing its discovery to Laplace. De Moivre's work also led to the De Moivre–Laplace theorem, an early special case of the central limit theorem.
The standard normal density also appears as the limiting law in the central limit theorem. After a change of scale, it appears as the fundamental solution of the heat equation, where solutions are obtained by convolution with a Gaussian kernel.
Two-dimensional surface e(x2+y2)e^{-(x^2+y^2)}.
Let
ex2dx.\begin{equation*}\int_{-\infty}^{\infty} e^{-x^2}\,\mathrm{d}x.\end{equation*}
and taking the square gives
e(x2+y2)dxdy.\begin{equation*}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y.\end{equation*}
The radius rr satisfies r2=x2+y2r^2=x^2+y^2, so e(x2+y2)=er2e^{-(x^2+y^2)}=e^{-r^2}.
We use polar coordinates,
x=rcosθ,y=rsinθ.\begin{equation*}x=r\cos\theta,\qquad y=r\sin\theta.\end{equation*}
The corresponding Jacobian determinant is
(x,y)(r,θ)=(cosθrsinθsinθrcosθ)=r.\begin{equation*}\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| =\left|\begin{pmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{pmatrix}\right| = r.\end{equation*}
Hence
dxdy=rdrdθ.\begin{equation*}\mathrm{d}x\,\mathrm{d}y= r\,\mathrm{d}r\,\mathrm{d}\theta.\end{equation*}
Substituting these expressions, we obtain
02π0er2rdrdθ=2π[12er2]0=π.\begin{equation*}\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta = 2\pi\left[-\frac{1}{2}e^{-r^2}\right]_{0}^{\infty} = \pi.\end{equation*}
Therefore
ex2dx=π\begin{equation*}\int_{-\infty}^{\infty} e^{-x^2}\,\mathrm{d}x = \sqrt{\pi}\end{equation*}
The standard normal distribution distribution is defined by
Φ(t)=P(Xt)=12πtex2/2dx.\begin{equation*}\Phi(t) = \mathbb{P}(X\leq t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{t} e^{-x^2/2}\,\mathrm{d}x.\end{equation*}
The rate at which 1Φ(t)1-\Phi(t) decreases as tt\to\infty can be estimated explicitly.
The distribution function $\Phi(t)$
The distribution function Φ(t)\Phi(t)
LEMMA
For any t>0t>0,
12π(1t1t3)et2/21Φ(t)12π1tet2/2.\begin{equation*}\frac{1}{\sqrt{2\pi}} \left(\frac{1}{t}-\frac{1}{t^3}\right)e^{-t^2/2} \leq 1-\Phi(t) \leq \frac{1}{\sqrt{2\pi}}\frac{1}{t}e^{-t^2/2}.\end{equation*}
Write
1Φ(t)=12πtes2/2ds.\begin{equation*}1-\Phi(t) = \frac{1}{\sqrt{2\pi}}\int_t^\infty e^{-s^2/2}\,\mathrm{d}s.\end{equation*}
For the upper bound, since s/t1s/t\geq 1 for sts\geq t,
tes2/2ds1ttses2/2ds=1tet2/2.\begin{equation*}\int_t^\infty e^{-s^2/2}\,\mathrm{d}s \leq \frac{1}{t}\int_t^\infty s e^{-s^2/2}\,\mathrm{d}s = \frac{1}{t}e^{-t^2/2}.\end{equation*}
For the lower bound, integration by parts gives
tes2/2ds=tses2/21sds=et2/2tt1s2es2/2ds.\begin{equation*}\int_t^\infty e^{-s^2/2}\,\mathrm{d}s = \int_t^\infty s e^{-s^2/2}\frac{1}{s}\,\mathrm{d}s= \frac{e^{-t^2/2}}{t} - \int_t^\infty \frac{1}{s^2}e^{-s^2/2}\,\mathrm{d}s.\end{equation*}
Using the upper bound just proved,
t1s2es2/2ds1t2tes2/2ds1t3et2/2.\begin{equation*}\int_t^\infty \frac{1}{s^2}e^{-s^2/2}\,\mathrm{d}s \leq \frac{1}{t^2}\int_t^\infty e^{-s^2/2}\,\mathrm{d}s \leq \frac{1}{t^3}e^{-t^2/2}.\end{equation*}
Substituting this into the previous identity gives the claimed lower bound.

References

  • [Gauss1809]Gauss, Carl Friedrich. Theoria Motus Corporum Coelestium in Sectionibus Conicis Solem Ambientium. F. Perthes and I. H. Besser, 1809. PDF.
  • [Bogachev1998]Bogachev, Vladimir I. Gaussian Measures. First edition. American Mathematical Society, 1998.