## 数学代写|微积分代写Calculus代写|The Gaussian Integral

In this section, we derive the Gaussian integral
$$\int_{-\infty}^{\infty} e^{-x^2 / 2} d x=\sqrt{2 \pi}$$
This formula is remarkable because the primitive of $e^{-x^2 / 2}$ cannot be expressed in terms of the elementary functions (i.e., the functions studied in Chapter 3). Nevertheless the area (Figure 5.4) of the (total) subgraph of $e^{-x^2 / 2}$ is explicitly computable. Because of $(5.4 .1)$, the Gaussian function $g(x)=e^{-x^2 / 2} / \sqrt{2 \pi}$ has total area under its graph equal to 1.

The usual derivation of (5.4.1) involves changing variables from cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)(\S 3.5)$ in a double integral. How to do this is a two-variable result. Here, we give an elementary derivation that uses only the one-variable material we have studied so far. To derive (5.4.1), we will, however, need to know how to “differentiate under an integral sign.”
To explain this, consider the integral
$$F(x)=\int_c^d f(x, t) d t, \quad a<x<b,$$
where $f(x, t)=3(2 x+t)^2$ and $a<b, c<d$ are reals. We wish to differentiate $F$. There are two ways we can do this. The first method is to evaluate the integral obtaining $F(x)=(2 x+d)^3-(2 x+c)^3$ and, then, to differentiate to get $F^{\prime}(x)=6(2 x+d)^2-6(2 x+c)^2$. The second method is to differentiate the integrand $f(x, t)=3(2 x+t)^2$ with respect to $x$, obtaining $12(2 x+t)$ and, then, to evaluate the integral $\int_c^d 12(2 x+t) d t$, obtaining $6(2 x+d)^2-6(2 x+c)^2$. Since both methods yield the same result, for $f(x, t)-3(2 x+t)^2$, we conclude that
$$F^{\prime}(x)=\int_c^d \frac{\partial f}{\partial x}(x, t) d t, \quad a<x<b,$$
where the partial derivative $\partial f / \partial x(x, t)$ is the derivative with respect to $x$,
$$\frac{\partial f}{\partial x}(x, t)=\lim _{x^{\prime} \rightarrow x} \frac{f\left(x^{\prime}, t\right)-f(x, t)}{x^{\prime}-x}, \quad a<x<b .$$
It turns out that (5.4.2) implies (5.4.3) in a wide variety of cases.

## 数学代写|微积分代写Calculus代写|Stirling’s Approximation of n

The main purpose of this section is to derive Stirling’s approximation to $n$ !. If $\left(a_n\right)$ and $\left(b_n\right)$ are positive sequences, we say that $\left(a_n\right)$ and $\left(b_n\right)$ are asymptotically equal, and we write $a_n \sim b_n$ as $n \nearrow \infty$, if $a_n / b_n \rightarrow 1$ as $n \nearrow \infty$. Note that $a_n \sim b_n$ as $n \nearrow \infty$ iff $\log a_n-\log b_n \rightarrow 0$ as $n \nearrow \infty$.
Theorem 5.5.1. If $x$ is any real, then,
$$\Gamma(x+n) \sim n^{x+n-1 / 2} e^{-n} \sqrt{2 \pi}, \quad n \nearrow \infty .$$
In particular, if $x=1$, we have Stirling’s approximation
$$n ! \sim n^{n+1 / 2} e^{-n} \sqrt{2 \pi}, \quad n \nearrow \infty .$$
Note that $\Gamma(x+n)$ is defined, as soon as $n>-x$. By taking the log of both sides, (5.5.1) is equivalent to
$$\lim _{n \nearrow \infty} \log \Gamma(x+n)-\left[\left(x+n-\frac{1}{2}\right) \log n-n\right]=\frac{1}{2} \log (2 \pi) .$$
To derive (5.5.1), recall that
$$\Gamma(x+n)=\int_0^{\infty} e^{-t} t^{x+n-1} d t, \quad x>0 .$$
Since this integral is the area of the subgraph of $e^{-t} t^{x+n-1}$ and all we want is an approximation, not an exact evaluation, of this integral, let us check where the integrand is maximized, as this will tell us where the greatest contribution to the area is located. A simple computation shows that the integrand is maximized at $t=x+n-1$, which goes to infinity with $n$. To get a handle on this region of maximum area, perform the change of variable $t=n s$. This leads to
$$\Gamma(x+n)=n^{x+n} \int_0^{\infty} e^{-n s} s^{x+n-1} d s=n^{x+n} \int_0^{\infty} e^{n f(s)} s^{x-1} d s,$$
where
$$f(s)=\log s-s, \quad s>0 .$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|高斯积分

$$\int_{-\infty}^{\infty} e^{-x^2 / 2} d x=\sqrt{2 \pi}$$

(5.4.1)的通常推导涉及在二重积分中将变量从笛卡尔坐标$(x, y)$变换到极坐标$(r, \theta)(\S 3.5)$。如何做到这一点是一个有两个变量的结果。在这里，我们只使用到目前为止我们研究过的单变量材料，给出一个基本的推导。为了推导(5.4.1)，我们需要知道如何“在积分号下求导”。为了解释这一点，考虑积分
$$F(x)=\int_c^d f(x, t) d t, \quad a<x<b,$$
，其中$f(x, t)=3(2 x+t)^2$和$a<b, c<d$是实数。我们希望区分$F$。有两种方法可以做到这一点。第一种方法是求积分值，得到$F(x)=(2 x+d)^3-(2 x+c)^3$，然后求导得到$F^{\prime}(x)=6(2 x+d)^2-6(2 x+c)^2$。第二种方法是对被积函数$f(x, t)=3(2 x+t)^2$对$x$求导，得到$12(2 x+t)$，然后求积分$\int_c^d 12(2 x+t) d t$，得到$6(2 x+d)^2-6(2 x+c)^2$。由于这两种方法得到相同的结果，对于$f(x, t)-3(2 x+t)^2$，我们得出
$$F^{\prime}(x)=\int_c^d \frac{\partial f}{\partial x}(x, t) d t, \quad a<x<b,$$
，其中偏导数$\partial f / \partial x(x, t)$是对$x$的导数，
$$\frac{\partial f}{\partial x}(x, t)=\lim _{x^{\prime} \rightarrow x} \frac{f\left(x^{\prime}, t\right)-f(x, t)}{x^{\prime}-x}, \quad a<x<b .$$

## 数学代写|微积分代写微积分代写| n

$$\Gamma(x+n) \sim n^{x+n-1 / 2} e^{-n} \sqrt{2 \pi}, \quad n \nearrow \infty .$$

$$n ! \sim n^{n+1 / 2} e^{-n} \sqrt{2 \pi}, \quad n \nearrow \infty .$$

$$\lim _{n \nearrow \infty} \log \Gamma(x+n)-\left[\left(x+n-\frac{1}{2}\right) \log n-n\right]=\frac{1}{2} \log (2 \pi) .$$

$$\Gamma(x+n)=\int_0^{\infty} e^{-t} t^{x+n-1} d t, \quad x>0 .$$

$$\Gamma(x+n)=n^{x+n} \int_0^{\infty} e^{-n s} s^{x+n-1} d s=n^{x+n} \int_0^{\infty} e^{n f(s)} s^{x-1} d s,$$

$$f(s)=\log s-s, \quad s>0 .$$

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