how Laplace solved the Gaussian integral
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- เผยแพร่เมื่อ 24 มิ.ย. 2024
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This is actually Laplace's method to evaluate the Gaussian integral, namely the integral of e^(-x^2) from -inf to inf. en.wikipedia.org/wiki/Gaussia... I believe this is a great method because calculus 2 students, with the knowledge of improper integrals and the willingness to see a double integral for the first time, should be able to understand. I will do it the classic way in my new 100 integrals video! Subscribe to @blackpenredpen so you don't miss it!
Here's how to "integrate" e^(-x^2) with the error function: • the impossible integra...
0:00 100 integrals part 2 coming soon
1:10 integral of e^(-x^2) from -inf to inf without using polar coordinates.
14:06 check out Brilliant!
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Teacher, I follow the lessons you post on TH-cam
I am studying in the college of maths and physics number whats app please tell me you can help me with maths
sir from where I will get more videos of definite integration
th-cam.com/channels/oLMpMr0JTdLZz4LPdvOf3A.html
Lol I don't have much time to go to brilliant when we have enough brilliant content on this channel.
Teacher may I know your telegram? I want to ask you something.
this truly is one of the most integral of all time
Truly. This integral is, in fact, an integral.
an integral integral
Indeed
The most integral?
This integral is indeed an integral
I like this more than the polar coordinates method because it is far easier to understand. No Jacobian, no different kind of coordinate system, just one substitution (really the second one you could do without the substitution, just by inspection).
There is no Jacobian probably because substitution is done in iterated integral only (he changed only one variable at the time)
Another approach is Gamma function with reflection formula
This is a computationally brilliant method for sure. However, compared to the polar approach the appearance of pi as a result comes across as the outcome of a somewhat artificial-looking substitution process. In contrast, the polar approach relies on the circular symmetry of the "Bell surface" about the z axis, which makes a pi-related value of the integral fairly obvious. Besides, one can relate the square root in the result to the Gaussian curve being a cross-section of that surface, whilst the squaring of the integral in Laplace's approach looks like a mere trick based on nothing but the factorizability of the exponential. For those who are interested, Dr Peyam in his channel does around 20 different derivations of this result!
@@FleuveAlphee what's an "artificial" looking substitution process?
@@holyshit922 That might actually be a line of circular reasoning, since many of the proofs of Gamma function integral rely on the Gaussian integral result!
@@Xoque551 not necessarily circuloar reasoning Reflection formula can be derived from product representation of Gamma function and Euler's product for sine
I love this integral! Funnily enough in all the physics exams it is always just given 😅
😆
keep up great work sir
Because in physic we just use... In math it depends of the subject
In physics the solution to this integral is an intuitive truth.
You're luckyy in engineering my profs made us do it
Never knew you could solve this without using polar coordinates... excellent video!
I didn't know you could solve it with polar coordinates so I guess we balance the universe out somehow. 😆
@@abebuckingham8198 It's very quick to solve with polar coordinates, but there is also a 3rd geometric way to solve this integral.
They're the two proofs outlined in Wikipedia.
there are other several ways to prove this remarkable fact.
@@jacoboribilik3253 yes, Dr Peyam presents 12 of them in his video collection here th-cam.com/play/PLJb1qAQIrmmCgLyHWMXGZnioRHLqOk2bW.html
It is great seeing this integral done in cartesian coordinates. All the textbooks I used so far either used the approach over polar coordinates or just used the result. Thank you for this video!
Interestingly, this quite related to polar coordinates but avoids using them :). From the 2D point of view in the first quadrant, we used the coordinates (x,t) where (x,y) = (x,tx). t = y/x is the slope of the line starting from the origin passing through (x,y), in other words t = tan(theta) where theta is polar angle. Cool video :)
Thanks!!
Wow never thought about this, very interesting thanks!
I remember doing this integral shortly after learning about the jacobian.
There is so much joy in doing this integral for the first time, thank you Prof. Steve!
As a stats guy, this is a beautiful detour from the more popular solution. I enjoyed the detailed steps and exploits of the symmetry inherent in the integral. Whilst the polar coordinates approach is easier to explain to anyone who has done trig, this solution is elagant af too.
In 1995 the internet arrived. And this channel is the only good thing worth watching (just a long time waiting :)). Love your style and you teach like someone who understands rather than repeats. Thanks for your hard work!
As a student who hasn't taken Calc 3 but still uses parts of it, I really appreciate this video and seeing this explanation!
This is a great video Steve; I've seen this integral solved before, but only with the usual polar coordinates method. Thanks!
The beauty of the mathematics lies not in the destination but in the elegance of the paths. Thanks for illustrating it.
Franchement très intéressant. Je ne connaissait que la méthode avec passage en polaire. Et j'étais persuadé que c'était la seule méthode possible ! Merci pour cette brillante présentation.
Nice route to solving a tricky integral. Great videos ... keep it up!
I have always loved your enthusiasm !! Also, nice way to solve the integral
I didnt know this approach. Thank you for the very clear and instructive presentation.
I had this on my calc 2 exam!! I approximated it using the mclaurin series for e to the x and then integrating that summation!! Cool video
I am getting 2 Σ(-x)^(2n+1)/((2n+1)(n!)) [0, inf]
isn't this diverging? I know I have done a mistake somewhere but I am not able to spot it
@@MohammadIbrahim-sq1xn i shouldve clarified but my integral was from 0 to x and the intial variable was t. Once you integrate the series you get Σ(-1)^n * x^(2n+1)/((2n+1)(n!)) which is the expansion of sin(x), cheers.
@@fooddrive8181 I have a doubt where did you get the (-1)^n from (this might be very silly but I can't seem to figure it out)
@@MohammadIbrahim-sq1xn the (-x^2) is (-1) (x^2) so in my expansion i applied the n to both from (a^m)^n = a^(m*n)
I watched 10 minutes of this stuff and we went from a simple formula to a bunch of mumbo jumbo. Can one of you calculus students explain to me what in the world this stuff is useful for? And be specific and have you found a job to where you are actually crunching out formulas to prove they are right or wrong for a living. Can't use teacher or professor either. Some other profession please.
Thanks for your videos! It's fun to watch your process!
Beautiful :) I did'nt know this trick. BTW when you switch to double integral and polar coordinates, you don't need Jacobian. You just can calculate the volume under the function as sum of volumes of cylinder shells. (Height of shell is e^(-x^2-y^2) = e^-r^2; length is 2 pi r and thickness is dr, so you integrate e^-r^2 . 2 pi r . dr from 0 to infinity.)
aninob
Yes. That is more elegant.
Plus I don't you understand his y = xt where his translation derivatives doesn't equal in dy where dy = x dt + a missing in his formulas t dx term:
dy = x dt + t dx I thought? 🤔
@@lawrencejelsma8118 In the substitution y = xt in the inner intergral is x in the role of constant. (Like "for given fixed x from the outer integral we shall calculate this inner integral..."). So the substitution y = xt is just recepee for transition from y to t. This is the reason why dy = x.dt.
@@aninob ... Yeah your correct. He should have written it out dy = xdt + tdx but the tdx term equals zero just like t can translate through the integral during summation. I was just use to "two space" partial differentiation mathematics of ut (ut)' = (u')t + u(t'). The recommended solution of real only or even when considering imaginary numbers are always polar solutions of rcosx + jrsinx solutions because trigonometric integral tables solutions are easier to understand.
@@MrPoornakumar 3B1B just made a beautiful video about it (plus some mathematical sugar on the top I never knew about). th-cam.com/video/cy8r7WSuT1I/w-d-xo.html
Love the aspect of switching the order of integration, I was wondering how it could be done. But then realised that it is ofcourse summation of product. At the start was confused as to how you could make the function 2* (0 to Infinity) of the f(x), but i had missed that this is a converging function. made math interesting again.. Thanks!
13:31 "And there's no +c"
CRIES OUTTA HAPPINESS
3:15 I was like “yeah I get it” 💪
How beautiful the result and the way to solve it , thanks
Nice to see a different way, I really liked the polar method when it was explained in multivariable calc. I also liked the way the Fourier transformation is used to find the improper integral of sin(x)/x.
Could be nice to see an alternative way to do that one.
Hi,
I worked on this for years when I was young, until I found the polar solution in a book.
But I'm glad to see that there is a method that avoids polar coordinates. Thanks a lot for this 👍
This is a cool integral to know in that you can use integrals used by probability distributions to simply rewrite the integral in terms of it's probability distribution and then if they go from -inf to +inf they just become 1
Can you solve this problem?
Q. If f[{x + √(1 + x^2)}/x] = x^2. Then find f(x); domain & Range of f(x) =?
Video link:- th-cam.com/video/YVboBTlmUo8/w-d-xo.html
Simple and brilliant, never occurred to me!
Thank you so much. This is the best explanation of this ive ever seen
Quite impressive in terms of your presentation, well done
This has to be one of the most beautiful integrals out there
Thank you,I’ve been thinking about a method to do it without polar coordinates cuz I didn’t learn them,great job ❤
Beautiful problem, balckpen! Thank you for sharing :)
Well done. You made this very easy to follow. Thanks.
Hi bprp, thank you for the comprehensible and clear derivation, I am now practicing Laplace calculus.
No polar? Something against polar?
Glad I finally found someone doing this without going to polar coordinates.
Polar killed my father.
@@chitlitlah
Noooo!!!
change of variables would be calculus 3 ;)
Too COLD!
Lovely integral! Thank you.
Because of this video, now I understand how upper and lower bounds of integral change due to it's variable change. Thank you so much⭐
I have seen over 10 different proofs of this result, I don't think I have seen that one before. Great job!
Thank you! Cheers!
Hey, then you'll now have to make a video about the other 9 methods, to let the rest of us in on them! ;-)
Fred
It is so satisfying to watch you explain the math.
(The first thing that catch my eye is the 荼果 doll under the e)
I would love to see you use the Feynman trick with some rigorous explanations (uniform convergence) for the swap between the derivative and the integral!
Keep up the good work 😉
The Feynman trick has nothing to do with uniform convergence though, you prove it using the dominated convergence theorem - it essentially requires to dominate the integral of the partial derivative
Wow! Thank you so much for your videos!
Great job! Good pace and explanation, an alternative to polar conversion method. Thank You!
Very cool to see someone so passionate about a topic that so many people wrongly think of as boring
This integral is great. It's amazing how there are several connections between the Exponential function and pi, complex numbers and this at least.
I'm of the whacky opinion that complex numbers hide (or reveal) a door into other universe, Or universes. No evidence. Just a "gut feel".
@@starpawsy that doesn't mean anything
@@amineaboutalib Nope. Not a thing.
@@starpawsy Holy crankometer Batman, it's a kook!
@@holliswilliams8426 Oh look, I fully recognize the wackiness. That's ok. Im old enough not to care.
this is just realy original, congratulations!
I enjoyed watching this. I've taken a few courses in stats and probability, and none of the professors wanted to take the time to show this integral. We just accept it as fact.
The way he swicthes pens is no less than a magician.
Masterpiece . Thanks a lot for your great efforts ,Sir. 💖💖💖
Oh my god you are so amazing. I just loved it. The way you make it so easy for us is commendable. You are incomparable. Thank u so much. Because of you I am able to solve it without remembering polar coordinates typical method.
I feel great, even though I have studied calculus only on my own, not in school yet, I was able to follow along with what's happening. And oh boi, it was beautiful
OMGGGGGG Thanks you so much, i dont have words 😍😍😍
That was really nice!
Awesome video! Thank you!
beautiful job!!!
I dont understand calculus one bit, but something about your explanation style just drives me towards your videos
I love this integral and i never saw this approach
Nicely done! 🙂
Never mentioned Laplace. Would be interesting to hear just a little about when and how these guys came up with these things. We kinda owe them.
amazing sir👍today I had learnt little something,... and understood that, there is a lots of yet to learn ...
Yayy, part 2. Let's goooo
Very nice , I knew of the Feynman technique , but this is very nice
thank you that was very amazing
This was quite easy to follow - which is weird, I'm very bad at integrals. Honestly the only point I tripped up was in the end where 2*int[0,inf] (1/1+t^2) * dt became 2*tan^-1 (t) [0,inf]. But that is probably because I don't remember formulas for anti-derivative of 1/(1+x^2) and didn't know derivative of tan^-1 (x). My big fail in trigonometry is remembering all the formulas that can be used. I only remember that sin^2+cos^2 = 1.
Very entertaining and informative video - thanks!
Wow! This is so good!!
Excellent presentation 👌
Thank you very much... Love your videos...
I am so happy to live in a world where bprp exists! greetings from Brazil!!
Thank you
Nice explanation!
Thank you so much You are the greatest teacher in the world🤩🤩🤩
how good are your tutorials? I passed a calculus 2 course with 70% with little to no help from my professor. keep up the great work man!
I don’t even know basic functions, except for linear functions, yet I still watch his videos as If I understand something.
I really doubt polar coordinates since I don't know how it works. With this method, I believe now that the integral of e^-x^2 from -inf to inf is equal to sqrt of pi. Amazing.
You can imagine the coordinate and variable switch in the polar coordinate trick as allowing one to sweep from 0 to infinity in a circular way all at once (radially with r from 0 to infinity and circularly with theta from 0 to 2pi). Since the function is actually radially symmetrical from the origin and there is only an infinity (imagine it in 3D), with polar coordinates you do not need to split it up in two (-+inf,0] halves.
Start with the original integral:
Integral e^(-x^2) dx
Square it:
(Integral e^(-x^2) dx)^2 = double integral e^(-x^2) * e^(-x^2) dx dx
Change one of our variables of integration to y:
double integral e^(-x^2) * e^(-y^2) dx dy
Using properties of exponents, the product of the two exponential functions becomes e^(-x^2 - y^2):
double integral e^(-x^2 - y^2) dx dy
In polar coordinates, r^2 = x^2 + y^2.. The differential area dx*dy is equivalent to r dr dtheta. Thus:
double integral e^(-r^2) r dr dtheta
The limits on this integral are the full domain of x and y. In polar coordinates, it is 0 to infinity for r, and 0 to 2*pi for theta.
This generates the derivative of the inside function, so we can use u-substitution. Work with the inner integral first, and theta isn't involved.
integral r*e^(-r^2) dr
Let u = -r^2. Thus du = -2*r dr, and dr = du/(-2), and the integral becomes:
-1/2*integral e^u du, which evaluates to -1/2*e^u + C. In the r-world, it becomes: -1/2*e^(-r^2) + C
We'd like to evaluate this from r=0 to r=infinity:
(-1/2*e^(-inf^2)) - (-1/2*e^0) = 1/2
The outer integral is trivial, it's just integral 1 dtheta, which is theta + C. Evaluate from 0 to 2*pi, which is 2*pi.
Multiply with the r-integral result, which gives us the result:
[integral e^(-x^2) dx from 0 to infinity]^2 = pi
Since we originally squared the integral, take the square root to get the original integral we want:
integral e^(-x^2) dx from 0 to infinity= sqrt(pi)
Apart from this awesome video supplied by ur easy explanation . Actually, Like the Gamma function, since there is no explicit integral to f(X) = e^-x² , another special function can easily be defined to be the distribution function F(X) = the integral of f(X) between -∞ and x. Obviously it is positive , strictly increasing and limited. and actually , since ex is equal to the Taylor series that has the terms 1/n! .X^n . Then e^-x² is equal has the Taylor series with terms : 1/n! . (-x²)^n . Thus F(X) can easily be obtained by integrating of that series. I think This can be helpful in numerical analysis as the you can have an approximation to F(X) by studying a polynomial of sufficiently large degree and dropping the rest of the power series.
i don’t have the slightest idea about a single thing he said.
xD
How? It’s simple. You just might have to learn the basics of this topic first to understand deeper things
What part didn’t you understand? It’s okay to admit it. We can help you.
the integral is of an even function.
This means that it’s symmetrical about the y axis.
that’s why he rewrote it in the beginning. From there since it’s a definite integral we can swap out variables so long as the final definite value is the same. So that’s where the integral with respect to y of e^-y^2 came from. Then he just squared it which is the same as multiplying the integrals and got it to 4 times double integral from 0 to inf+ of e^-(x^2 + y^2)
Then he substituted for T and rewrite the bounds, then factored out the x^2 and flipped the bounds of integration to make it an integral with respect to x. Then he does a U substitution to compute the inner integral.
From there the final integral is a basic trig substitution which yields 2tan^-1(t) from 0 to inf = 2(pi/2) = pi
And since this result is the square of the original integral we need to take the square root to get our final answer of root(pi).
@@CalculusIsFun1 your efforts will be remembered bro
You could also try x=rcosa and y=rsina for solving double integral
I´m calculus 2 student from Argentina and I understand it so well, I´ll believe in Chen Lu
Waiting for 100 integral part 2 😌
Dhanyawad bhaiyaa 🙏🏻🙏🏻.
Love from BHARAT 🇮🇳
It's cute, but you cant hide from the trig on this one. You end up with the arctan anyway. Thats kinda obvious because the answer has pi in it, but still fun to watch you sweep it under the rug for as long as possible. I like different approaches, but I still like the move to polar better at the end of the day.
It would be nice to have a video where you solve this integral using complex analysis (residue theorem). It's a bit longer but it is a very fun calculation.
Very neat and direct. For those interested in a simple way, using polar coords, ( and simple to understand) I suggest you refer to Prime Newtons' video- it is nice for beginners.😊
If you substitute t= y/x then you have substituted the tan of the angle for polar coordinates. Also the substitution for u is minus the square of the radius in polar coordinates. So you have used polar coordinates, it is just disguised.
Excellent! It is fun even if (or maybe because) it is more complicated than going through polar coordinates. I watched it because I was curious to see how the π number would appear without that. I remember it was a wonder to me when I found out that trigonometric functions can be built by just integrating functions defined using only the 4 basic arithmetic operations plus the square root, the latter not even always necessary, like here. More generally, it is fascinating how π can materialize where it is not expected, like for instance in the sum of reciprocal squares.
No more words need to be said. Just... Elegant.
I remember learning to do this one!
Very nice solution dear teacher. 👍👍👍👍👍👍👍
#blackpenredpen I was wondering how you come up with the idea y=xt and why it works? thanks in advance.
Thank you so much sir
this method is so much nicer than using polar coordinates and the equation which has a difficult to understand derivative
y = xt , when y goes to 0, the value of t does not have to be always 0. because in y = xt , the value of y also depends on x. in the integral we can see that x varies from 0 to infiniti. For example if x = 0, then t can have many values while xt remains 0. Can you explain why we assume when y = 0 that t = 0.
The value at some isolated point never effects the integral as a whole. The substitution is okay if still have convergence approaching zero. Polar coordinates has the same problem as theta can be anything at the origin.
I remember solving the indefinite integral version in my calc 2 class by using the Taylor series expansion of e^x
Awesome resolution ❤
What a cute way to calculate the Gaussian integral!
I love calculus
Nice video:)
if you took the polar coordinates in r and theta term with limits r=0 to infinity and theta=0 to pi/2 and applying Jacobian elementary area would convert to rdrdtheta and integrand to e^(-r^2) then could have changed r^2 to t and would have changed the whole thing to e^tdt. this would have made it a little easier but all in all this is a great video! I appreciate that
This is one of the youtube videos of all time
Not apparent that use of iterated improper integrals as double improper integral and the interchange of integration in the iterated improper integrals behave this way. An explanation with improper integrals as limiting values would be lengthy. But with I(t) = (integral of e^(-x^2)) from 0 to t)^2 and g(t) =integral from 0 to 1 of e^(-t^2(x^2+1))/(1+x^2) with respect to x, we can show that I(t)+g(t)= is a constant = g(0) = integral of 1/(1+x^2) from 0 to 1 and is Pi/4. We can use differentiation under the integral sign here. Letting t tends to infinity you get your answer since g(t) tends to 0.
You are outstanding!
Somehow the polar coordinates method never really meshed with me so this one is quite nice to see. Props!