# 物理代写|统计力学代写Statistical mechanics代考|GAUSSIAN INTEGRALS

## 物理代写|统计力学代写Statistical mechanics代考|GAUSSIAN INTEGRALS

Gaussian integrals are a class of integrals involving $\mathrm{e}^{-x^2}$. We start with the basic integral
$$I(\alpha) \equiv \int_{-\infty}^{\infty} \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=\sqrt{\frac{\pi}{\alpha}} . \quad(\alpha>0)$$
There’s a trick to evaluating this integral that every student should know. Let $I \equiv \int_{-\infty}^{\infty} \mathrm{e}^{-x^2} \mathrm{~d} x$. Then $I^2=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathrm{e}^{-\left(x^2+y^2\right)} \mathrm{d} x \mathrm{~d} y$. Change variables to polar coordinates with $r^2=x^2+y^2$ and change the area element $\mathrm{d} x \mathrm{~d} y \rightarrow r \mathrm{~d} r \mathrm{~d} \theta$, so that $I^2=\int_0^{2 \pi} \mathrm{d} \theta \int_0^{\infty} \mathrm{e}^{-r^2} r \mathrm{~d} r$. Change variables again with $u=r^2$, and you should find that $I^2=\pi$.

Forms of this integral containing powers of $x$ are straightforward to determine. The integrand in $I(\alpha)$ is an even function, and thus integrals involving odd powers of $x$ vanish:
$$\int_{-\infty}^{\infty} x^{2 n+1} \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=0 . \quad(n \geq 0)$$

Integrals involving even powers of $x$ can be generated by differentiating $I(\alpha)$ in Eq. (B.7) with respect to $\alpha$ :
$$\int_{-\infty}^{\infty} x^{2 n} \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=(-1)^n \frac{\partial^n}{\partial \alpha^n} I(\alpha)=\sqrt{\frac{\pi}{\alpha}} \frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{(2 \alpha)^n} \quad(n \geq 1)$$
Gaussian integrals are often specified over the range $[0, \infty)$. We state the following result and then discuss it,
$$\int_0^{\infty} x^n \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=\frac{1}{2} \Gamma\left(\frac{n+1}{2}\right) \frac{1}{\alpha^{(n+1) / 2}} . \quad(n \geq 0)$$
Equation (B.10) follows under the substitution $y=\alpha x^2$ and then noting that the integral so obtained is none other than $\Gamma(x)$, Eq. (B.1).

## 物理代写|统计力学代写Statistical mechanics代考|VOLUME OF A HYPERSPHERE

A hypersphere is a set of points at a constant distance from a given point (its center) in any number of dimensions. As we now show, the volume $V_n(R)$ enclosed by a hypersphere of radius $R$ in $n$-dimensional Euclidean space $\left(\sum_{k=1}^n x_k^2=R^2\right)$ is given by
$$V_n(R)=\frac{\pi^{n / 2}}{\Gamma\left(\frac{n}{2}+1\right)} R^n \text {. }$$
As an example of Eq. (B.14), $V_3(R)=\left(\pi^{3 / 2} / \Gamma\left(1+\frac{3}{2}\right)\right) R^3$. Using the recursion relation, $\Gamma\left(1+\frac{3}{2}\right)=$ $\frac{3}{2} \Gamma\left(\frac{3}{2}\right)=\frac{3}{2} \Gamma\left(1+\frac{1}{2}\right)=\frac{3}{4} \Gamma\left(\frac{1}{2}\right)=3 \sqrt{\pi} / 4$. Thus, $V_3(R)=4 \pi R^3 / 3$. For $n=2$ and $n=1$, Eq. (B.14) gives $V_2(R)=\pi R^2$ and $V_1(R)=2 R$.

To derive Eq. (B.14), we first define a function of $n$ variables, $f\left(x_1, \cdots, x_n\right) \equiv$ $\exp \left(-\frac{1}{2} \sum_{i=1}^n x_i^2\right)$. In Cartesian coordinates, the integral of $f$ over all of $\mathbb{R}^n$ is
$$\int_{\mathbb{R}^n} f\left(x_1, \cdots, x_n\right) \mathrm{d} V=\prod_{i=1}^n\left(\int_{-\infty}^{\infty} \mathrm{d} x_i \exp \left(-\frac{1}{2} x_i^2\right)\right)=(2 \pi)^{n / 2},$$
where $\mathrm{d} V$ is the $n$-dimensional volume element, and we’ve used $\int_{-\infty}^{\infty} \mathrm{e}^{-a x^2} \mathrm{~d} x=\sqrt{\pi / a}$. Now recognize that $\sum_{i=1}^n x_i^2 \equiv r^2$ defines the radial coordinate in $n$ dimensions. Let $\mathrm{d} V=A_n(r) \mathrm{d} r$, where $A_n(r)$ is the surface area of an $n$-dimensional sphere. The surface area of an $n$-sphere can be written $A_n(r)=A_n(1) r^{n-1}$. The same integral in Eq. (B.15) can then be expressed
\begin{aligned} \int_{\mathbb{R}^n} f \mathrm{~d} V & =\int_0^{\infty} A_n(r) \exp \left(-\frac{1}{2} r^2\right) \mathrm{d} r=A_n(1) \int_0^{\infty} r^{n-1} \exp \left(-\frac{1}{2} r^2\right) \mathrm{d} r \ & =2^{(n / 2)-1} A_n(1) \int_0^{\infty} t^{(n / 2)-1} \mathrm{e}^{-t} \mathrm{~d} t=2^{(n / 2)-1} A_n(1) \Gamma\left(\frac{n}{2}\right), \end{aligned}
where in the second line we have changed variables $t=\frac{1}{2} r^2$. Comparing Eqs. (B.16) and (B.15), we conclude that $A_n(1)=2 \pi^{n / 2} / \Gamma\left(\frac{n}{2}\right)$. This gives us the surface area of an $n$-sphere of radius $R$ :
$$A_n(R)=\frac{2 \pi^{n / 2}}{\Gamma\left(\frac{n}{2}\right)} R^{n-1}$$
From Eq. (B.17), $A_3(R)=4 \pi R^2, A_2(R)=2 \pi R$, and $A_1(R)=2$. By integrating $\mathrm{d} V=A_n(r) \mathrm{d} r$ from 0 to $R$, we obtain Eq. (B.14). Note that $A_n(R)=(n / R) V_n(R)$.

# 统计力学代考

## 物理代写|统计力学代写Statistical mechanics代考|GAUSSIAN INTEGRALS

$$I(\alpha) \equiv \int_{-\infty}^{\infty} \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=\sqrt{\frac{\pi}{\alpha}} . \quad(\alpha>0)$$

$$\int_{-\infty}^{\infty} x^{2 n+1} \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=0 . \quad(n \geq 0)$$

$$\int_{-\infty}^{\infty} x^{2 n} \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=(-1)^n \frac{\partial^n}{\partial \alpha^n} I(\alpha)=\sqrt{\frac{\pi}{\alpha}} \frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{(2 \alpha)^n} \quad(n \geq 1)$$

$$\int_0^{\infty} x^n \mathrm{e}^{-\alpha x^2} \mathrm{~d} x=\frac{1}{2} \Gamma\left(\frac{n+1}{2}\right) \frac{1}{\alpha^{(n+1) / 2}} . \quad(n \geq 0)$$

## 物理代写|统计力学代写Statistical mechanics代考|VOLUME OF A HYPERSPHERE

$$V_n(R)=\frac{\pi^{n / 2}}{\Gamma\left(\frac{n}{2}+1\right)} R^n$$

$$\int_{\mathbb{R}^n} f\left(x_1, \cdots, x_n\right) \mathrm{d} V=\prod_{i=1}^n\left(\int_{-\infty}^{\infty} \mathrm{d} x_i \exp \left(-\frac{1}{2} x_i^2\right)\right)=(2 \pi)^{n / 2},$$

$$\int_{\mathbb{R}^n} f \mathrm{~d} V=\int_0^{\infty} A_n(r) \exp \left(-\frac{1}{2} r^2\right) \mathrm{d} r=A_n(1) \int_0^{\infty} r^{n-1} \exp \left(-\frac{1}{2} r^2\right) \mathrm{d} r$$在第二行我们更改了变量 $t=\frac{1}{2} r^2$. 比较方程式。(B.16) 和 (B.15)，我们得出结论 $A_n(1)=2 \pi^{n / 2} / \Gamma\left(\frac{n}{2}\right)$. 这给了我们一个表面积 $n$-半径球体 $R$ :
$$A_n(R)=\frac{2 \pi^{n / 2}}{\Gamma\left(\frac{n}{2}\right)} R^{n-1}$$

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