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The Gaussian Integral

The area under e−x2e^{-x^2} over the whole real line is π\sqrt{\pi}.
The Gaussian integral, also known as the Euler--Poisson integral, is the integral of the Gaussian function
f(x)=e−x2\begin{equation*}f(x)=e^{-x^2}\end{equation*}
over the entire real line. Named after the German mathematician Carl Friedrich Gauss, the integral is
∫−∞∞e−x2 dx=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 published the precise integral in 1809, attributing its discovery to Laplace. The integral has a wide range of applications.
Rotating the one-dimensional Gaussian suggests the two-dimensional surface e−(x2+y2)e^{-(x^2+y^2)}.
Let
I=∫−∞∞e−x2 dx.\begin{equation*}I=\int_{-\infty}^{\infty} e^{-x^2}\,\mathrm{d}x.\end{equation*}
Since the integrand is positive, it is enough to compute I2I^2 and take the positive square root. Multiplying two identical copies of the integral gives
I2=∫−∞∞∫−∞∞e−(x2+y2) dx dy.\begin{equation*}I^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y.\end{equation*}
The integrand depends only on the distance from the origin. Passing to polar coordinates,
The radius rr satisfies r2=x2+y2r^2=x^2+y^2, so e−(x2+y2)=e−r2e^{-(x^2+y^2)}=e^{-r^2}.
x=rcos⁡θ,y=rsin⁡θ,dx dy=r dr dθ,\begin{equation*}x=r\cos\theta,\qquad y=r\sin\theta,\qquad\mathrm{d}x\,\mathrm{d}y=r\,\mathrm{d}r\,\mathrm{d}\theta,\end{equation*}
we obtain
I2=∫02π∫0∞e−r2r dr dθ=2π[−12e−r2]0∞=π.\begin{equation*}I^2=\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 I=πI=\sqrt{\pi}.
The Gaussian integral gives the normalizing constant for the normal distribution. A change of scale turns into the integral
∫−∞∞e−x2/2 dx=2π.\begin{equation*}\int_{-\infty}^{\infty} e^{-x^2/2}\,\mathrm{d}x=\sqrt{2\pi}.\end{equation*}
Therefore the function
φ(x)=12πe−x2/2\begin{equation*}\varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\end{equation*}
has total integral equal to 11. It is called the density of the standard normal distribution.
The corresponding distribution function depends on a real parameter tt and is defined by
Φ(t)=P(X≤t)=∫−∞tφ(x) dx=12π∫−∞te−x2/2 dx.\begin{equation*}\Phi(t)=\mathbb{P}(X\leq t)=\int_{-\infty}^{t}\varphi(x)\,\mathrm{d}x=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{t} e^{-x^2/2}\,\mathrm{d}x.\end{equation*}
It measures the area under the normal density to the left of tt.
The rate at which 1−Φ(t)1-\Phi(t) decreases as t→∞t\to\infty can be estimated explicitly.
The distribution function $\Phi(t)$ is the area to the left of $t$, while $1-\Phi(t)$ is the right tail.
The distribution function Φ(t)\Phi(t) is the area to the left of tt, while 1−Φ(t)1-\Phi(t) is the right tail.
LEMMA
For any t>0t>0,
12π(1t−1t3)e−t2/2≤1−Φ(t)≤12π1te−t2/2.\begin{equation*}\frac{1}{\sqrt{2\pi}}\left(\frac{1}{t}-\frac{1}{t^3}\right)e^{-t^2/2}\leq1-\Phi(t)\leq\frac{1}{\sqrt{2\pi}}\frac{1}{t}e^{-t^2/2}.\end{equation*}
Write
1−Φ(t)=12π∫t∞e−s2/2 ds.\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/t≥1s/t\geq 1 for s≥ts\geq t,
∫t∞e−s2/2 ds≤1t∫t∞se−s2/2 ds=1te−t2/2.\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
∫t∞e−s2/2 ds=∫t∞se−s2/21s ds=e−t2/2t−∫t∞1s2e−s2/2 ds.\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,
∫t∞1s2e−s2/2 ds≤1t2∫t∞e−s2/2 ds≤1t3e−t2/2.\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