, where $\alpha > 0$.
Each expression is an abbreviation of

. This ia an improper integral.
In the following, we will prove
.

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).

Let $I_{1}$, $I$, and $I_{2}$ be an integral of $e^{-\alpha (x^{2}+ y^{2})}$ over $A_{1}$, $A$, and $A_{2}$, respectively, i.e.,

Their integrand is positive, and $A_{1}$ is included in $A$, and $A$ is included in $A _ {2}$.
Therefore the inequality

$$
\tag{1}
$$
holds.
By using polar coordinates,

,
the integral $I_{1}$ can be written as

. Here, let $t = R^{2}$. The integral
can be calculated as

. Hence we obtain

.
In the same way as deriving $I_{1}$, we obtain

.
The integral $I$ can be written as

.
From $(1)$, we see that

Since the integrand $e^{-\alpha x^{2}}$ is positive,
the integral $\int_{0}^{r} e^{-\alpha x^{2}}\mathrm{d}x$ is positive.
We have

$$
\tag{2}
$$
The limit of the left-hand side as $r$ approaches $+\infty$ is

, and the limit of the right-hand side as $r$ approaches $+\infty$ is

, which is equal to the limit of the left-hand side.
Therefore, by the squeeze theorem and $(2)$, we obtain

.
This expression is normally expressed as

$$
\tag{3}
$$
by omitting the symbol of limit.
The Gaussian integral of the range from $-\infty$ to $+\infty$ can be divided as

Substituing $(3)$ into this gives

Let $t = -x$, we can calculate the first term by $(2)$ as

Therefore we obtain

$\int x \exp [- \alpha x^2] \mathrm{d} x$

The following equations hold.

Proof
First, we will prove

Let $x=\sqrt{t} $, we have

. Using these, we obtain

$$
\tag{1}
$$
Next, we will prove

The integral of the range from $-\infty$ to $+\infty$ can be divided as

Substituing $(1)$ into this gives

Let $t = -x$, we can calculate the first term by $(1)$ as

From thise equations, we obtain

Recurrence formula

Let $I_{n} (\alpha)$ and $J_{n} (\alpha)$ and be integrals defined as

. Recurrence formula

hold.

proof
The derivative of
$I_{n}(\alpha)$ with respect to $\alpha$ is

Since the integrand $x^{n} e^{-\alpha x^2}$ and the partial derivative
$
\frac{\mathrm{\partial} }{\mathrm{\partial} \alpha } \left( x^{n} e^{-\alpha x^2} \right)
$
are continuous function, we can commute derivative and integral of the right-hand side.

By calculating the derivative, we obtain

Next, we will prove the recurrence formula for $J_{n}(\alpha)$.
The only difference between
$I_{n}(\alpha)$ and
$J_{n}(\alpha)$ is
the integral range.
We thus apply the same discussion for $I_{n}(\alpha)$, and obtain