This integral is speaking VOLUMES! Doing the triple Gaussian using spherical and polar coordinates
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Gaussian Integral: • Destroying the Gaussia...
Leibniz rule: • The Leibniz rule for i...
Got it? :DDDDDDD It's a Volume integral that's why it is speaking volumes! HAH!
Pls end my miserable existence :(
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Oh baby, a triple!
Thomas Torrone ah, yes. The T R I P L E one
Good. Now do it for arbitrary dimensions.
The result would just be sqrt(pi)^n
Answer probably equals C*pi^(dimension/2)
The thing is... this integral is actually separable, because e^-(x^2+y^2+z^2) = e^-x^2 * e^-y^2 * e^-z^2.
So really this is just the product of three identical Gaussian integrals. The answer is going to be the value of the 1-D Gaussian integral (which is sqrt(pi)), cubed.
It's now incredibly straightforward to generalise to any number of dimensions.
en.wikipedia.org/wiki/Gaussian_integral#n-dimensional_and_functional_generalization
There's a neat formula for the (n-1)-surface of a n-ball involving the gamma function. You can do it. The function e^r^2 is isotropic, so you'll just have (n-1)-surface x int( r^(n-1) e^-r^2 ).
So the problem is equivalent to computing that last integral for any power of n. (Hint: use integration by parts).
Correct me if I'm wrong but isn't this just the I^3 for the normal gaussian integral I? so therefore it's sqrt(Pi)^3
That is how I went about solving this and it yielded the same answer
It blew my mind
hello!I am a student of department of chemistry from Taiwan. I am very glad that there is a channel talking about Calculus on TH-cam.
the best part isn't how the 4's cancel out! It's deffenietely your methos for the Gaussian integral, finding the double derivative was awesome, gj
Learn English!
In 1 minute : we can distribute the power to get e^-x²-y²-z² then distribute the integral to get 3 Gaussian integrals sometimes with respect to x sometimes with respect to y sometimes with respect to z and we know that the Gaussian integral is evaluated to √π then we multiple the product of this 3 integrals and finally we get √π×√π×√π equal to π√π = √π³=π^3/2 and we're done 🔥
I personally enjoyed this video a lot. Please do more multivariable calc, it brings back good memories
I reported this for pornographic content😍
Lel
_w h a t t h e _*_f u c k_*
A R E T H O S E ? ! ? !
For the integral x^2e^(-x^2), it would be simpler using the part integration technique.
Keep this tip. Your videos are fantastic. Congratulations.
I think it's very important to mention how beautifully drawn your integral signs are
never thought id see someone incorporate memes into math and i love it
We need more people like you
Others teach very slowly
good one! the r²e^(-r²) integral could be transformed more quickly by using feynman's trick. Just expending the exponent by a parameter e.g. "a" , differentiate for "a" and after drawing the derivative out of the integral set the parameter a=1. However, I liked your ansatz because you figured it out on your own. Well done.
I have three alternative thoughts for fun.
A1. I'd just write that I = J^3 and then obtain J as you did it.
A2. After spheric part intergation and switch from r^2exp(-r^2) to 1/2*exp(-r^2) observe equation: I = J^3 = pi*J. From which follows J = sqrt(pi).
A3. One can apply parameter method for int(r^2*exp(-a*r^2)) with parameter a =1 for this case.
We know that int(exp(-a*r^2),0,inf) = 1/2*sqrt(pi/a). Applying partial differential over "a" we obtain
int(-r^2*exp(-a*r^2),0, inf)= -1/4*sqrt(pi)/(a)^(3/2). Thus, at a=1: int(r^2*exp(-r^2),0,inf)=1/4*sqrt(pi).
What does three dot (vertices of triangle) symbol mean?
Therefore
Nice anime stuffs my boi, glad to know I'm not the only one xd
For the integarahl int_0^infty r^2 *e^(-r^2)dr, it is straightforward to solve it introducing a parameter t in the gaussian integral: int_0^infty e^(-t*r^2)dr which by doing substitution (r->sqrt(t)r) we get 1/sqrt(t) int_0^infty e^(-r^2)dr=(1/2)*sqrt(pi/t). Now, differentiating the parametric integral and (1/2)*sqrt(pi/t) wrt t we get that int_0^infty r^2 *e^(-t*r^2)dr=sqrt(pi)/(4*t^(3/2)) and evaluating at t=1 we get the desired result: sqrt(pi)/4.
Can you generalize it as the following:
integral of e ^ -(x1^2 + x2^2 + x3^2 + ... + xn^2) from each xi = -infinity to infinity is pi^(n/2).
Alternatively, if you know some stuff about your gamma function you can do it a little bit faster. After you integrated over the angular components you are left with the radial integral:
G = \int_0^inf r^2 e^(-r^2) dr
u := r^2
du = 2r dr
G = \int_0^inf u e^(-u) 1/(2 u^(1/2)) du
G = 1/2 * Gamma(3/2) = 1/2 * (2!)/(4^1 1!) * pi^(1/2) = 1/4 pi^(1/2)
so you conclude: 4 pi 1/4 pi^(1/2) = pi^(3/2)
5:20 can you just drag the diferential operator outside of the integral? I'm wondering because if you justify that you can make this equal to zero because you are diferentiating a constant twice, then diferentiating a constant just once would work too. But you would end up with the integral from 0 to infinity of -2r*e^(-r^2) being equal to zero, wich is false.
Jens i did the r^2*e^(-r^2) integral slightly different:
I looked at the functional J(t) = \int_0^\infty e^{-tr^2} dr, and noted that J’(t) = -\int_0^\infty r^2 e^{-tr^2} dr, so -J’(1) is the original integral. With a simple u sub, you can turn J into the Gaussian integral and get that it’s equal to sqrt(π/4t). Therefore J’(t) = -sqrt(pi)/4 * t^(-3/2). Just plugging in 1 and negating gives us sqrt(pi)/4, and if you multiply that by the 4pi you already had you get an answer of pi^3/2 QED
Thank you so very much sir.
Ved Prakash heyyy
Hey Sir...
Sir, I really miss you to.
It seems Smarty Boi hasn’t slept very well lately, you need some rest boi c:
The n-dimentional analogue gives an integral whose value is π^(n/2). Easily solved with gamma function integral and the (known, also by gamma function) area of the n-1 dimensional Sphere.
4:05 sneaky boi, instead of using integration by parts, he uses clever tricks and Papa Lebnitz
Your handwriting has become much neater.
instructions unclear lost track at the first step
Loved this one! Calc III was sometime ago for me now; you made some of that come back though such as that Jacobian.
Integral r^2 e^-r^2 can be done as fellow also by using feynman technique
Integral e^-ar^2 from 0 to inf=1/2 *sqrt(pi/a)
differentiate
integral -r^2 *e^-ar^2 = 1/2 *sqrt(pi) *(-1/2)*(a^-3/2)
now let a=1
Integral r^2 *e^-r^2 =1/4 sqrt(pi)
and we are done
This seems like something we'd do in my calc III class
Another approach for the radial part of the integral: let u = r, and dv = re^(-r^2), and integrate by parts (I break it up like this because dv can be integrated easily with another substitution, like t = -r^2). The boundary term vanishes and the other part, quoting the Gaussian integral, gives sqrt(pi)/4; tack on the 4pi from the angular dependence and you're done.
...Of course, we might as well just quote the Gaussian integral from the very beginning and cube it, but you knew that :)
The integral of exp(-x^2) from -inf to +inf is equal to sqrt(pi) because exp(-x^2)/sqrt(pi) is the density fct of N(0,1/2).
And since exp(-(x^2+y^2+z^2))=exp(-x^2)*exp(-y^2)*exp(-z^2), we easily get the result sqrt(pi)^3.
Nice as allways! Thanks!
I believe that interchange in 5:10 is an improper use of Leibnitz rule. Partial integration would be one way to prove the integral is zero. Another option is to start with exp(-σr^2), then d/dσ could be interchanged with this novel integral.
Flammable Maths That was quick. BTW, love your videos.
For every 4 min I'm in awe.. very Elegant methods
Ich habe mal eben dein Video angesehen und den Beweis selbst durchgeführt. Bis zu ca. 3:40 habe ich das Gleiche wie Du. Aber das was du dann machst, erschien mir zu kompliziert und ich habe mich selbst an die Rechnung gemacht und hatte das Ergebnis VIEL schneller, in dem ich partielle Integration genommen habe.
Hier ist es:
Ansatz wie du auch: I=4pi*int(r^2*exp(-r^2,dx,0,inf)
Zum Integral: Ich benutze partielle Integration und definiere:
u'=r*exp(-r^2), v=r -> u=-1/2*exp(-r^2), v'=1
Nach der Regel für partielle Integration int(u'v)=uv-int(uv') folgt durch Einsetzen
int(r^2*exp(-r^2,dx,0,inf) = -1/2*r*exp(-r^2) (von 0 bis inf) + 1/2*int(exp(-r^2,dx,0,inf)
Nun zu den beiden Summanden:
Der erste Summand geht für unendlich gegen 0, denn exp(-r^2) geht schneller gegen 0 als r gegen unendlich geht. Und für r=0 ist das Ergebnis 0.
-1/2*r*exp(-r^2) (von 0 bis inf) = 0
Den zweiten Summanden kennen wir schon.. Herr Gauss lässt grüßen: int(exp(-r^2,dx,0,inf)=sqrt(pi)/2. Daraus folgt sofort, dass unser gesuchtes Integral den Wert sqrt(pi)/4 hat. int(r^2*exp(-r^2,dx,0,inf)=sqrt(pi)/4
Setzen wir nun in die unterbrochene Hauptrechnung ein und wir erhalten:
I=4pi*sqrt(pi)/4=pi*sqrt(pi)=pi^(3/2).
Fazit: Ich spare mir die Beobachtung der zweiten Ableitung von exp(-r^2), was ein nicht-intuitiver Trick ist (ich hätte das nicht gesehen, grins), sondern ich gehe "straight forward" ohne Mathe-Trickkiste.
Ansonsten: Toll gemacht mit einem erstaunlichen Ergebnis :)
EDIT: Und noch schneller geht es mit Fubini's Regel und dem bekannten Wert des Gauss'schen Integrals int(exp(-x^2) über R)=sqrt(pi)
int(exp(-(x^2+y^2+z^2),R^3)=int(exp(-x^2),R)*int(exp(-y^2),R),int(exp(-z^2),R)=sqrt(pi)^3
You are incredible!! Hope you get more subscribers! I'll stay tuned to new videos ^^
Wonderful. . . Very clear.
You can also use gamma function applying formula gamma(1-x)gamma(x)=pi/sin(pi*x) and replacing x with 1/2
I love this mathematics Channel💝
I enjoy your videos you really know how to do them :)
Oh my, that was a pretty original way to solve it, I would have used integration by parts
By seeing the title, I see someone has watched Matt Parker's comedy with toroidal vortices :P
Thats it im getting the gaussian integral tat’d on my arm
Brilliant video, very entertaining!
amazing effort and talent
@Flammable Maths I have a question for you my boi, where'd you, bprp, and dr peyam gained your experience? Ur very inspiring to me and I want to know what have you done to acquire this knowledge in math, are you students or teachers? And what university have you done? (I don't know if in your place are called universities, i'm still in an Italian's high school). Thank you!! ❤🐤
I'm pretty sure d/dx of an integral of f(x) with constant bounds is not equal to the integral of d/dx of the integrand with the same bounds. Otherwise you could say d/dx int (e^x) dx with constant bounds is 0, but the int (d/dx(e^x)) dx with the same bounds is definitely not zero, it would be e^b-e^a if a and b are the bounds.
That self lifting blackboard ahahah
Would you be able to do videos on quarternion analysis?
Can somebody make/send a link to the sound at 0:04 (no pranks, I don't wanna get sent to 0:04 of this video)
It was impressive and interesting to apply polar/spherical coordinates, but I wanted just to separate x, y and z and obtain a product of three equal integrals: \int_{-\infty}^\infty e^{-t^2}dt that are \sqrt{\pi}\erf(t)|_{-\infty}^\infty=\sqrt{\pi}(1-(-1))=2\sqrt{\pi}, so the product is the 3-rd power of it: 8\pi^{3/2}. It seems shorter :).
That is the way I would solve it but this makes for a better and more interesting video
Hello Dear 🔥 Flammable Maths 🔥
At first, thanks for this video.
And second; I searched but I didn't find anything about Mellin transformations. Do you any video about Mellin transformations or not? If no, please make a video about Mellin transformations (and also Fourier transform).
Thank you.
Big Fan.
But but but, you can just break up the inside into e^-x^2 * e^-y^2 * e^-z^2 which breaks up into 3 integral mutiplicants which becomes sqrt pi * sqrt pi * sqrt pi
Bro this was great
OMG !!!!! Amazing !
12:50 Are those I see some delicious anime figures?
Flammable Maths ah I see you're a man of culture as well
Flammable Maths do you like anime or do you just collect figures?
I love all your videos... could you please do some videos on arithmetics or geometry... that would be amazing ❤
You forgot the potatoes to go with the meat. However, sour kraut will do this time.
Is true that integral in R^n of e^-(x0²+x1²+...+xn²) = π^(n/2)?
Hi Papa Flammy,
I’m a high school junior who hasn’t done volume integrals before, but isn’t it possible to take e^(-(y^2+z^2)) out, as it is a constant with regards to dx, and then do the same for e^-(z^2)? Then you have three integrals of what looks like the same form to me. Is there a reason this wouldn’t work for volume integrals in particular? Thanks a bunch.
Can you do a video explaining what Jacobians are? Or otherwise link me to a good source? I understand that obviously the polar transformation requires this other factor to make it work, but I don't understand where you got the expressions from.
Cool video as per! Please let integration week never end :'(
+Rohan Skatedude
Jacobian is a modification you do when you change between different coordinate systems.
For example, an integral of the type int(f(x,y,z)dxdydz) can be transformed into different systems of coordinates
For example:
regular to cyllindrical:
x=p*cos(@)
y=p*sin(@)
z=z
J=
p є
@ є
z є
then you have to redifine boundaries in terms of @, p^2, and z stays the same.
Then it looks like this:
int(J*f(p,@,z)dpd@dz)
It's a fairly short explanation.
For spherical system of coordinates transformation and Jacobian look like this:
x=p*cos(@)*sin(&)
y=p*cos(@)*sin(&)
z=p*cos(&)
J=p^2*sin(&)
& є
@ є
p є
and then if we substitute it back it'll look like this:
int(J*f(p,@,&)dpd@d&)
I wish I could attach an image to illustrate why & goes from 0 to pi and @ from 0 to 2pi, but I'll try to explain to the best of my ability.
@ defines an angle that goes radially around the sphere.
& defines an angle which goes from the lowest part of the sphere to the uppermost part of it, hence it's only from 0 to pi.
p is radius in both cases.
I think the most intuitive understanding of the Jacobian results from linear approximations of a multivariable vector function. For example, if I want to integrate multiple integrals in terms of dudv instead of dxdy, where x and y are expressed both as functions of u and v (i.e. x(u,v) and y(u,v)),
then I can say that at (u0,v0), a good approximation of (x,y) in terms of u and v would be
x=x(u0,v0)+(u-u0)(dx/du) + (v-v0)(dx/dv)
y=x(u0,v0)+(u-u0)(dy/du) + (v-v0)(dy/dv)
Jacobi probably was like wow gee, I can write these two equations neatly with a matrix
[x]=[x(u0,v0)]+[(dx/du) (dx/dv)][u-u0]
[y]=[x(u0,v0)]+[(dx/du) (dy/dv)][v-v0]
Now just like when doing u-substitution for single derivatives, how we can say that x=u, dx=u' du, we can analogously say that
[x] [(dx/du) (dx/dv)] [u]
d[y] = [(dx/du) (dy/dv)] d[v]
This is not rigorously expressed by any means, but it provides some nice intuition. We can then find how much [x,y] changes in total magnitude (in both x and y direction) for a change in [u,v] by taking the determinant of that matrix full of partial derivatives (the jacobian). This makes sense because the determinant says how much a matrix scales area (the product of the eigenvalues). Hope this helps :)
Perfect👍👍👍
How you transformed the integral in spherical form
Well assuming the variables are independant of eachother wouldnt the integrals be seperable resulting in the gaussian integral cubed?
Why do you call the r And r^2 the Jacobian? I tried looking it up and it seems to be matrix or tensor related. Does that have to do with the order of the Jacobian used since one was a volume integral and one was an area integral?
Lmao that intro
multidimensional standard normal distribution
Papa, can you explain why the angle "Feida" is Pi, and the third integrals is from 0 to 2Pi but why 2Pi?
This may have just been a preference for how you wanted to do it, but isn't there a much easier way to do this 3D gaussian integral? You can just "split up" the integral into a product of three independent gaussian integrals, each resulting in \sqrt{\pi}. Thus your final result is just \sqrt{\pi}^3=(\pi)^{3/2}.
You only really had to do integration by parts with r^2e^-r^2 and then solve for the gaussian integral
Epic bad boy 🔥
Orrr we can really make good use of gamma integral...😬😬
Nice one though.. love your videos..
i got an idea from you solviong the gaussian integral
i think i solved the int cos^2 by making it into a double integral I^2=int int cos^2 sin^2 since int sin^2=int cos^2 and sub. with u=sin^2 giving the integarl I^2=int int 1/4 du du=>I=sqrt((1/8)sin^4+Csin^2+D)
is this an alternative answer or have i missued the multiplication when i turned it into a double integral?
Try the integral between 1 and -1 of (1-x^4)^1/2It’s an elliptical function
it will be much easier using the normal distribution property to split the triple integral into individual integration.
But can you do integral of (e^-r)/r along only the x axis from 0 to infty?
Can u pls suggest some books on integrals!!!!
I didn't get at all. What you calculated second derivative for? Isn't it obvious that the first derivative taken out of integral would be zero, since it doesn't depend on r any more.
don't you need a Maßfactor when you integrate exp(-r²) in Cartesian coordinates?
Really awesome
sick!
I like this guy. I've watched some of his videos but I wish he could avoid the use of strong language if you know what I mean.
Awesome video i!, thank you very much :D
Even when we convert it into polar coordinates, we multiply it by extra r term... Can u plz explain that stuff... Your videos are amazing. Even though few concept are new to me but I really understand and admire ur teaching.... Keep up the good work
pure gold!
I had a question about Zeno's paradox, more specifically Reimman hypothesis. one of the equations involves an H with a little hat (this thing --> ^) I wanted to know If you knew the name and if this is used when talking about only complex or imaginary numbers or also real numbers
I just googled it, and yes it's hamiltonian. Thanks! :)
Bruh, do it the physics way and just use the gauss integral on the back of Griffiths QM.
R^n=(pi)^(n/2)
This is so good 😂😂😂
kann es sein, dass hier nicht die leibniz regel sondern der satz der dominierten konvergenz benötigt wird, da das integral und die ableitung bezüglich der gleichen variable bestimmt werden?
Couldn’t you have calculated the Gaussian integral and then shown that the one we were looking for is it’s cubic root?
Are you a physicist by any chance? You really understand the concepts really well. 😀😀😀😀
+Flammable Maths Is there any shortcut to becoming a math expert? I love maths but I really don't do well in exams. How many hours do I need to practise??
Hey can you do an integral on my request? Integral of {x}/{1+xy}dxdy from 0 to 1. Prove this to be equal to (1-¥) where ¥ is Euler constant and {} is the fractional part
So what was that in the beginning, middle and end?
Cool, though I got a bit hung up on the integral of the second derivative exp -r^2. Can you redo that bit please.
Once again, you dirty dirty boy!
What school are you at?
hell yeah beutiful
You could have used Fubini Tonelli at the very beginning to get back to 1-D gaussian integral (Which you still have to calculate), but I think it would have been faster. Is there a reason why you chose to do a shperical coordinates change ? Great video otherwise !