\[\int_{0}^{+\infty}e^{-\frac{x^2}2}\mathrm{d}x\]

其妙

没想到高等数学第一帖由我来发啊？求下面积分的值：\[\int_{0}^{+\infty}e^{-\frac{x^2}2}\mathrm{d}x\]

pxchg1200

我们知道有
\[ \int_{0}^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2} \]
以下略。。。

其妙

回复 2# pxchg1200

kuing

好一个水贴……

PS、代码可以写成 \int_{0}^{+\infty}e^{-\frac{x^2}2}\rmd{x}\[\int_{0}^{+\infty}e^{-\frac{x^2}2}\rmd{x}\]

其妙

回复 4# kuing
\int_{0}^{+\infty}e^{-\frac{x^2}2}\mathrm{d}x
这个代码和你的那个不同，显示效果却是一样的：
\[\int_{0}^{+\infty}e^{-\frac{x^2}2}\mathrm{d}x\]

kuing

回复 5# 其妙

是有区别的，不过这里表现得不明显
看这个例子：\int x\mathrm{d}x \ne \int x\rmd{x}
\[\int x\mathrm{d}x \ne \int x\rmd{x}\]

其妙

回复 6# kuing
嗯，还是\rmd{x}好，有点空隙。

hbghlyj

Asymptote
import graph;
size(8cm);
real f(real x) {return exp(-x^2);}
real xmin = -4, xmax = 4;
real ymin = -0.1, ymax = 2;
real x1 = 3.5, x2 = -x1;
fill((x1,0)--graph(f,x1,x2)--(x2,1.6)--(x1,1.6)--cycle, lightgray);
draw(graph(f, xmin, xmax), black+linewidth(1));
draw((xmin,0)--(xmax,0), black+linewidth(0.8), EndArrow(5));
draw((0,ymin)--(0,ymax), black+linewidth(0.8), EndArrow(5));
label("$x$", (xmax,0), E);
label("$y$", (0,ymax), N);
label("$f(x) = e^{-x^2}$", (1,1), E);
en.wikipedia.org/wiki/Gaussian_integral
generalisation of Gaussian integral
multidimensional Gaussian integral
下面是Feynman integration的方法 (相关帖子)
As before, we simplify matters by computing the right half of the integral, $\int_0^{\infty} e^{-x^2} d x$. Set
$$
I(t):=\int_0^{\infty} \frac{e^{-x^2}}{1+(x / t)^2} d x .
$$
This is defined for all $t>0$, and we're hoping to determine $I(\infty)$. We simplify this integral by making the natural substitution:
$$
I(t)=t \int_0^{\infty} \frac{e^{-t^2 x^2}}{1+x^2} d x
$$
Right away this tells us new information about $I(t)$, for example, that
$$\tag2\label2
\lim _{t \rightarrow 0} \frac{I(t)}{t}=\frac{\pi}{2} .
$$
(In fact, the left hand side of this is $I^{\prime}(0)$, but we won't make use of this below.) To apply Feynman's trick we want to find $I^{\prime}(t)$, but if we do this naively the outcome will be ugly. To simplify matters, we recall our warm-up example $G(t)$, in which case differentiating simplified the integral because the denominator cancelled! Inspired by this, we instead work with the function
$$
e^{-t^2} I(t)=t \int_0^{\infty} \frac{e^{-t^2\left(1+x^2\right)}}{1+x^2} d x
$$
Finally, to avoid the product rule we divide both sides by $t$. Having thus renormalized $I$, we differentiate:
$$
\frac{d}{d t}\left(t^{-1} e^{-t^2} I(t)\right)=\int_0^{\infty}-2 t e^{-t^2\left(1+x^2\right)} d x=-2 e^{-t^2} \int_0^{\infty} e^{-u^2} d u=-2 e^{-t^2} I(\infty)
$$
Now, somewhat ridiculously, we undo what we just did by integrating both sides:
$$
\underbrace{\int_0^{\infty} \frac{d}{d t}\left(t^{-1} e^{-t^2} I(t)\right) d t}_{=-\lim _{t \rightarrow 0} \frac{I(t)}{t}}=\underbrace{\int_0^{\infty}-2 e^{-t^2} I(\infty) d t}_{=-2 I(\infty)^2}
$$
Since we computed the left hand side in \eqref{2}, we deduce $I(\infty)^2=\frac{\pi}{4}$. It follows that $I(\infty)=\frac{\sqrt{\pi}}{2}$, whence
$$
\int_{-\infty}^{\infty} e^{-x^2} d x=\sqrt{\pi}
$$

hbghlyj

hbghlyj 发表于 2023-2-19 13:52
Inspired by this, we instead work with the function
$$
e^{-t^2} I(t)=t \int_0^{\infty} \frac{e^{-t^2\left(1+x^2\right)}}{1+x^2} d x
$$

这里没有证$\int_0^{\infty} \frac{e^{-t^2\left(1+x^2\right)}}{1+x^2} d x$为什么收敛.
在Proof2有证明:
Then, we have: \(\displaystyle \frac {e^{-\lambda^2 (1 + x^2) } } {1 + x^2} \le \frac 1 {1 + x^2}\) for each $x∈\Bbb R$.
Note that \(\displaystyle \int_0^\infty \frac 1 {x^2 + 1} \rmd x = \frac \pi 2\)
So by the Comparison Test for Improper Integral:  $\displaystyle \int_0^\infty \frac {e^{-\lambda^2 (1 + x^2) } } {1 + x^2} \rmd x$ converges.