Proof of Gaussian Intergral

Jan. 2nd, 2019
  We will prove
Gaussian integral
, where $\alpha > 0$. This integral whose integral is the Gaussian function is called Gaussian integral .

  Proof

  We start with defining three areas in the two dimensional space.
  The first area is a closed area $A_{1}$ whose boundary is the x-axis, the y-axis and a circle with radius $r$ centered at the origin.
  The second area is a closed area $A$ whose boundary is the x-axis, the y-axis, the line $x=r$, and the line $y=r$. $A$ is a square with an edge length $r$.
  The third area is a closed area $A_{2}$ whose boundary is the x-axis, the y-axis and a circle with radius $\sqrt{2}r$ centered at the origin.
  In this way, $A_{1}$ is included in $A$ and $A$ is included in $A _ {2}$ (See figure below).
Figure of Gaussian integral
Let $I_{1}$ be an integral of $e^{-\alpha (x^{2}+ y^{2})}$ over $A_{1}$, and $I$ be an integral of the same function over $A$, and $I_{2}$ be an integral of the same function over $A_{2}$, i.e.,
The integrand is positive over $A_{1}$, $A$, and $A_{2}$, and $A_{1}$ is included in $A$ and $A$ is included in $A _ {2}$. Therefore the inequality
holds. By the substitution,
, the integral $I_{1}$ can be written as
. Here, by the substitution, $t = R^{2}$, the integral $ \int_{0}^{r} e^{-\alpha R^2} R \mathrm{d}R$ can be calculated as
. Hence we obtain
.
  In the same way as deriving $I_{1}$, we obtain
. And the integral $I$ can be written as
.
  Therefore the inequality
holds.
  Since the integrand $e^{-\alpha x^{2}}$ is positive, the integral $\int_{0}^{r} e^{-\alpha x^{2}}\mathrm{d}x$ is positive. Hence the inequality
holds.
  The limit of the left-hand side of $r$ as $r$ approaches $+\infty$ is
, and the limit of the right-hand side of $r$ as $r$ approaches $+\infty$ is
, which is equal to the limit of the left-hand side.
  Therefore, by the squeeze theorem, we obtain
. This expression is normally expressed as
by omitting the symbol of limit, and called Gaussian integral.