The Gaussian Integral

Summary

本视频讲解了如何计算从负无穷到正无穷的e的负x平方次方的积分。通常，求解定积分需要先找到原函数，但本题的原函数较为复杂，涉及误差函数。视频中通过绘制函数图像，发现其与正态分布曲线相似。通过将积分问题转化为求体积问题，并利用具有径向对称性的空心圆柱体来近似计算体积，最终得出积分结果为根号下π。这个过程展示了高斯积分的概念，它在数学和物理学中非常重要，特别是在处理正态分布时。

Takeaways

• 🤔 计算 e^(-x²) 在从负无穷到正无穷的积分是一个复杂的问题，使用常规的不定积分方法难以解决。
• 📉 e^(-x²) 的图像与正态分布曲线相似，实际上它就是正态分布的曲线。
• 🔢 直接求不定积分会得到一个非初等函数——误差函数（error function），不易处理。
• 🔄 可以通过转换为极坐标，简化积分问题，将 e^(-x²-y²) 转换为一个关于 r 的函数。
• 📏 通过几何方法，积分相当于计算一个 3D 钟形曲线下的体积，这个体积具有辐射对称性。
• 🛠️ 通过分解该体积为无数个半径从零到无穷的同心圆柱体，将体积转化为对每个小圆柱体的积分。
• ✂️ 这些圆柱体的体积可以通过高度（函数值）、周长（2πr）和厚度（dr）的乘积来计算。
• ♾️ 最终积分从 0 到无穷，通过简单的代换（u = r²）可以将积分简化并求出答案。
• ✔️ 最后得到的积分结果是 π，因此原始的积分结果为 √π。
• 📚 这个积分被称为高斯积分（Gaussian integral），在数学和物理中广泛出现，尤其是在正态分布的归一化过程中。

Q & A

视频中提到的积分是什么类型的积分？

视频中提到的积分是定积分，具体是计算函数e^(-x^2)从负无穷到正无穷的积分。

为什么直接计算不定积分不是解决这个问题的好方法？

直接计算不定积分不是好方法，因为对于这个函数，不定积分相当复杂，涉及到非初等函数——误差函数，这使得计算过程变得困难。

视频中提到了误差函数，它是什么？

误差函数是一个非初等函数，它不能简单地用基本的数学运算来描述，通常在数学中用来表示某些积分的解。

视频中提到了正态分布曲线，它与我们要积分的函数有什么关系？

我们要积分的函数e^(-x^2)的图形实际上就是正态分布曲线，这是高斯分布的一个特例。

视频中提到了将积分问题转化为体积问题，这是如何实现的？

通过将函数e^(-x^2 - y^2)转化为依赖于R（点到z轴的距离）的形式，并利用具有径向对称性的空心圆柱体来近似计算体积，从而将积分问题转化为体积问题。

视频中提到的'I'代表什么？

'I'代表的是积分∫e^(-x^2)dx从0到∞的结果，它是我们要求解的原始积分问题的答案。

视频中提到的'I^2'是什么意思？

'I^2'指的是'I'的平方，也就是积分∫e^(-x^2)dx从0到∞的结果的平方，这与我们要求解的原始积分问题相关。

视频中是如何通过积分来计算体积的？

通过将体积分解为无限多个无限细的空心圆柱体，然后对每个圆柱体的体积进行积分，最后求和得到总体积。

视频中提到的高斯积分有什么特殊的意义？

高斯积分在数学和物理学中有广泛的应用，特别是在处理正态分布和其归一化方程时非常重要。

视频中最后得出的积分结果是什么？

视频中最后得出的积分结果是√π，即∫e^(-x^2)dx从-∞到∞的积分等于√π。

视频中提到的积分方法为什么有效？

视频中提到的积分方法有效，因为它利用了函数的径向对称性和体积分解的思想，通过将复杂的积分问题转化为更简单的体积计算问题。

Outlines

00:00

🧮 高斯积分的介绍与挑战

第一段介绍了从负无穷到正无穷的 e 的负 x 平方次幂的积分。通常情况下，计算定积分的方法是找到其不定积分，但对于这个特定问题，不定积分非常复杂，并且涉及到误差函数。误差函数是一种非初等函数，无法通过基本的数学操作来表示。因此，视频提出了一种更有效的方法来解决该问题。接着，视频讨论了如何利用类似的积分来帮助求解原问题，并解释了 e 的负 x 平方次幂的函数图像与正态分布曲线的相似性。最后，通过分析类似积分的形式，提出了一个新的思路，即将积分拆分为两个相同的部分，从而简化问题。

05:01

🔍 利用柱形积分法求体积

第二段讲解了如何利用柱形积分法来求解复杂的三重积分问题。首先，通过将 e 的负 (x² + y²) 次幂转换为 e 的负 r² 次幂，利用极坐标形式简化了问题的表达式。接着，视频引入了“径向对称性”的概念，说明如何通过旋转图像来保持函数的形状不变。这种对称性使得我们可以使用“中空圆柱体”的方式来计算体积。通过将目标体积划分为无数个细小的圆柱体，并且利用极限的思想，将这些圆柱体的体积相加，最终得到一个积分表达式。

10:03

📏 高斯积分求解及其应用

第三段进一步详细解释了如何通过柱形积分法求出 e 的负 r² 次幂在整个三维空间内的体积。首先，将柱体切割成“纸条”以便更好地理解体积计算过程。然后，通过计算每个圆柱体的体积，构建一个积分表达式。最后，利用代换法（u = r²）简化积分并求解出最终结果，即 π。通过取平方根，得出初始积分的最终答案 √π。这个积分被称为高斯积分，并广泛应用于数学和物理中，特别是在正态分布的归一化过程中。

Keywords

💡积分

积分是数学中的一个重要概念，用于计算函数在某区间上的面积或体积。在视频中，积分被用来计算函数e的负x平方从负无穷到正无穷的定积分，这是高斯积分的一个例子。

💡不定积分

不定积分是求函数的原函数的过程，通常表示为∫f(x)dx。视频中提到，对于给定的函数e的负x平方，其不定积分相当复杂，因此不推荐使用这种方法来求解定积分。

💡误差函数

误差函数是一个非初等函数，不能简单用基本的数学运算来描述。在视频中，提到如果使用Wolfram Alpha求解e的负x平方的不定积分，会得到以误差函数表示的答案。

💡正态分布曲线

正态分布曲线是统计学中描述数据分布的一种曲线，其形状类似于钟形。视频中指出，所讨论的函数e的负x平方的图像实际上就是正态分布曲线。

💡高斯积分

高斯积分是数学中一个重要的积分，形式为∫e的负x平方dx，从负无穷到正无穷。视频中详细解释了如何计算这个积分，并指出其结果为根号π。

💡定积分

定积分是计算函数在某区间上的积分值，与不定积分不同，定积分有具体的积分上下限。视频中的积分就是定积分的一个例子，计算了函数e的负x平方在负无穷到正无穷区间的积分。

💡对称性

对称性是数学和物理中描述对象在变换下保持不变的性质。视频中提到，通过将函数e的负x平方加上y平方转换为以R为变量的形式，可以发现其具有径向对称性，这有助于简化积分的计算。

💡体积

体积是几何学中描述三维空间对象所占空间大小的量度。在视频中，通过将积分问题转化为求体积问题，使用无限多个薄圆柱体来近似计算函数下的体积。

💡微分

微分是数学中描述函数在某一点处变化快慢的量。在视频中，提到了使用微分来表示积分中变量的微小变化，例如dr表示半径的微小变化。

💡圆周率π

圆周率π是数学中的一个常数，表示圆的周长与直径的比值。视频中最终计算出的高斯积分结果为根号π，显示了π在数学分析中的普遍性和重要性。

💡极坐标

极坐标是一种二维坐标系统，使用距离和角度来表示点的位置。视频中提到将x平方加y平方替换为R平方，实际上是将问题从直角坐标系转换到了极坐标系，以利用其对称性简化积分计算。

Highlights

求解负无穷到正无穷的e的负x平方次方的积分。

通常求解定积分需要先求不定积分，但此方法对这个问题可能不适用。

e的负x平方次方的不定积分相当复杂，涉及误差函数。

误差函数是数学中的非初等函数，无法用基本数学运算描述。

通过图形分析，该函数的图形类似于正态分布曲线。

利用误差函数的对称性，将问题转化为求解两个相似积分的乘积。

通过变量替换，将积分问题简化为求解e的负x平方次方的积分。

将积分问题转化为求解三维空间下曲线下的体积问题。

利用径向对称性，将体积问题分解为多个同心圆环的体积和。

通过无限细分圆环的方法，将体积问题转化为积分问题。

利用变量替换，将积分问题进一步简化为可解的形式。

最终积分结果为π，即e的负x平方次方积分的平方。

此积分问题被称为高斯积分，在数学和物理学中有广泛应用。

高斯积分在正态分布的归一化方程中非常重要。

通过本视频，观众可以了解高斯积分的求解方法及其在数学和物理学中的应用。

Transcripts

what's the integral of e to the minus x

squared when evaluated from minus

infinity to infinity if you want you can

pause this video here and get out a

piece of paper and try to actually

evaluate the integral but if you don't

want to then just carry on watching the

video now normally when you finding

definite integrals when you have an

integration which has got a range you

have to integrate over what you usually

do is you'll just find the indefinite

integral of that function and then plug

the numbers in as appropriate then

subtract them with each other however

for this particular problem trying to

find the indefinite integral of this

function and then hence the definite

integral is probably not the best way

for you to go about trying to solve the

problem because the indefinite integral

for this problem is actually rather

complicated now if you try asking

Wolfram Alpha what the indefinite

integral of e to the power of minus x

squared equals to it will give you some

answers in terms of what we call the

error function now this error function

as we call in mathematics is a non

elementary function meaning that it

can't be described simply in terms of

mathematical operations that you're

probably already familiar with now the

error function can be found however it's

just so complicated that we're probably

better off trying to solve this problem

using a different method so before we

even try to go in and evaluate the

interval let's actually look at what the

function looks like when it's plotted

out on a graph now the plotted graph

might ring a bell for some of you

because it does look like the normal

distribution curve in fact the graph

doesn't just look like the normal

distribution curve it is actually the

normal distribution curve but enough

about the normal distribution how do we

evaluate this integral well in order to

help us find what I is we're gonna be

looking at a very similar integral now

how are these two integrals actually

similar to each other

well let's carefully look at this

particular integral here now look at

this particular function e to the power

of minus x squared plus y squared you

can rewrite this top it as e to the

power of minus x squared minus y squared

and because of indices rule you can

actually split this into

e to power minus x squared times e to

the power of minus y squared now let's

look at the first integral which is with

respect to Y and because it's with

respect to Y this bit be e to the power

of minus x squared bin is not actually

going to affect the int go in any way

because it's with respect to X and so we

might as well just take this bit as a

constant and take it outside the

integral and then we're left with this

definite integral which can be evaluated

somehow and get out an answer as a

number as a constant and then if we put

the other integral back in we can also

try to evaluate this integral because

this bit here is actually just a

constant we can move this bit outside

and then all we're left with is this

integral and so what we have earlier now

just reduces to two integrals

multiplying with each other the first

integral will just equal to I as we

defined earlier

and the second integral in fact also

equals to I because it's actually just

the same integral but instead of writing

it in terms of X we're writing

everything in terms of Y instead and so

now what we have is that this integral

equals to I squared or whatever the

answer to the int were interested in

that first squared and so in order to

find the answer to the integral that

we're interested in all you have to do

is just try to find the answer to this

integral and then square root it but

what's the integral of this function it

looks much more complicated why am i

bringing this up well first let's

actually try to see how the function

looks like if I plot it out you'll get a

3d curve basically a three-d bell-shaped

curve and what we're doing in this

particular integral is we're trying to

find the volume underneath the curve

here similar how to when you have a

single integral you're trying to find

the area under the curve a double

integral means you're trying to find the

volume under the curve but the way that

we're gonna be finding the volume under

this particular curve is rather

interesting so stay tuned first let's

look back at the function again e to the

power of minus X square plus y squared

but what is x squared plus y squared if

draw up the equation x squared plus y

squared equals 2r squared on an XY plane

you'll see that what you get is you'll

get an equation for a circle with a

radius ah every point on this curve will

all be a distance R away from the origin

so now let's go back to the function

that we have earlier I'll replace x

squared plus y squared with R squared

and now we have a function that looks

much more interesting instead of having

a function which just depends on x and y

we now have a function which instead

depends on R which is just the distance

that the point is away from the z axis

and with the equation in this form we

can see that this graph actually has got

some sort of a radial symmetry you can

rotate this graph however about the z

axis and it will still look the same and

this particular property you will help

us a lot when trying to find the volume

under this graph so how can we actually

find the volume under this particular

graph well what we can do is we can try

to break this volume down into much

simpler shapes which we can actually

work with and we should try to pick

shapes which also have this sort of

radio symmetry similar to the graph that

we already have so let's pick cylinders

or more specifically hollow cylinders or

just tubes so all we can try to do is we

can try to fit these tubes underneath

the graph within the volume that we're

interested in and then we can try to

find the volume of these individual

tubes sum them all up and then be able

to get the volume that we're interested

in but using tubes which are too thick

we'll leave these particular volumes

uncovered so what we need to do in order

to get a proper accurate answer is to

use tubes with really really thin walls

and we need lots of these in fact for us

to be able to find this particular

volume accurately we need an infinite

number of choose whose thickness is

infinitesimally small so now let's say

underneath the graph we have an infinite

number

of concentric tubes each with radii

varying from zero all the way to

infinity

so now we can do is we can try to find

the volume of these individual tubes and

then hence be able to find the volume of

the entire thing so let's just take out

an individual tube and try to find its

volume but let's make things a bit

simpler rather than thinking of this as

a tube let's get a scissor and cut this

up like this so now rather than having a

tube we have to deal with all we have is

just some sort of a strip of paper which

we have to find the volume or the height

of these tubes is simple it's just

whatever the function equals do so

basically in this case it would just be

e to the power of minus R squared now

the length of this volume well because

this volume before used to be a cylinder

the length of this volume would just be

whatever the circumference of that

particular cylinder is so for this case

it'll just be 2 PI R now about the

thickness of this particular volume well

let's go back to the picture of the tube

in beginning the thickness of the two

were just equal to whatever this length

is minus whatever this length is this

length would just be R whereas this

length would just be R plus a little bit

more which I'll call D R and so the

thickness would just simply be d r a

really really small distance and so we

multiply all of these three things

together we'll get that the volume of

each of these tubes equals 2 e to the

power of minus R squared times 2 PI R

times d R and since we have an infinite

number of choose with varying radii from

0 all the way to infinity we need to

perform an infinite sum of these tubes

or perform an integral for this

particular volume going from 0 all the

way to infinity and so now trying to

find the volume under this graph is just

as simple as trying to evaluate this

integral which is very doable if you

know bit of substitution we can let u

equal to R squared so that D u equal

- to our dr stop - things as appropriate

cancel a few things out then get an

integral which can be evaluated and then

find the answer to this integral and if

we do all this process we'll get the

answer out of Pi and so now we have the

answer to the volume underneath this

graph which means we have the answer to

this integral and because this integral

equals to I squared we can finally find

what I equals U which is just the square

root of pi meaning that we have the

final answer to the very problem that I

proposed at the very beginning like

quite a bit ago the integral of e to the

minus x squared from minus infinity to

infinity will equal to the square root

of pi now the integral here that we've

just solved does have a special name and

it's called the Gaussian integral and

this Gaussian integral can be seen quite

a lot - all across mathematics and

physics as you've seen earlier we'll

come across this Gaussian integral and

we're working with normal distributions

and being able to evaluate this Gaussian

integral is important when we're trying

to normalize the equation for the normal

distribution but anyways is all I have

time for thank you very much watching

this video and I'll see you again very

soon

goodbye