A masterful theorem for integration.

[@insainsin]

It's also true for a=0.

[@lossen1984]

No this is wrong. See the solution for finding dx again in the u substitution. You will find that the sqrt when a=0 become zero and therefore the denominator becomes u/0 which is undefined.

[@megauser8512]

@@lossen1984 Only when u = 0 though.

[@VICTOR-vf8yx]

​@@lossen1984 Set a = 0 Then f(x-a/x) = f(x-0/x) = f(x)

[@MarcoMate87]

@@lossen1984 WTF?

[@Memzys]

this is trivially true, so it is not necessary to include it

[@trevistics]

And now I can derive the inverse Gaussian density from the plain ole' Gaussian density! Thanks , Prof. Penn!

[@timelsen2236]

❤ I'm always amazed. This TEACHER'S TEACHER takes unique or modern results rarely seen and presents in a clear way that's very understandable.

[@liyuanzhang8647]

It feels so good. 2019 I was an undergraduate in U.S, I learn the math from you. Now, I’m an IB mathematics teacher, I still learn things from you. It’s my horn to subscribe you❤❤❤

[@landsgevaer]

Cute. Are there more function like u = x+a/x that leave the integral unaffected? I can think of ±x+a, obviously. So any ±x + a/(x+b) + c would work, in other words any u = (±x²+ax+b)/(x+c). Any u(x) should work for which the sum of the derivatives of the branches of the inverse of u(x) equals 1 in absolute value. I can come up with u = -ln(e^x -1) for x > 0 u = ln(e^-x -1) for x < 0 Not half as pretty though...

[@nyghts7]

Bit of an old comment, but I'm replying anyway. Glasser's Master Theorem actually generalizes what u can be equal to quite a great deal. In general, this substitution works out for any u = x - sum {j = 1 to n - 1} (a_j)/(x - C_j), where a_j is a sequence of positive real constants, and C_j is a sequence of real constants. The interesting thing is that n is allowed to approach infinity, so there are some beautiful substitutions involving Fourier Transforms, notably the substitution u = x + tanx. The only added restriction to the special case in this video is that for the integral of f(u), you must take the Cauchy Principal Value, although this is mostly a formality to rigorously ensure convergence. The proof for the generalized form of u can be found in Glasser's original paper, "A Remarkable Property of Definite Integrals" that this video mostly follows.

[@srijanbhowmick9570]

@@nyghts7 Thanks for the info !

[@ELBARTO2023-bg1rw]

Yeah, generalizations of this theorem can be made.

[@jasoncetron233]

You've heard of elf on a shelf. Now get ready for pi on root phi! (I know I know. It doesn't really rhyme. Artistic license 😂)

[@allanjmcpherson]

It depends on your pronunciation. It rhymes for me!

[@gerardomalazdrewicz7514]

Should work on many, many languages.

[@saulmendoza1652]

very neat!! i usually do contour... but this is more straightforward!

[@mattcarnevali]

3:35 a spirit ball appears, call Ghost Adventures

[@jimschneider799]

@0:40 - if it's true for all positive a, then it's true for all nonnegative a, because if a = 0, the integrands are identical.

[@gerardomalazdrewicz7514]

Same split, but now use the du in the other section, to keep the radical real?

[@jimschneider799]

@@gerardomalazdrewicz7514 , the only difference between the positive reals and the nonnegative reals is that the latter case contains a = 0, and for a = 0, f(x + a/x) = f(x + 0/x) = f(x), so you are integrating the same thing on both sides of the equation. I apologize for not making it clear that I was only expanding the domain by a single point.

[@goodplacetostop2973]

13:03

[@johnshortt3006]

at 3:10 shouldn't there be a 2u not u in the numerator? doesn't matter since it cancels

[@neilgerace355]

As soon as I saw the 5, I thought, the Golden Ratio is going to turn up somewhere.

[@ArthurvanH0udt]

At 7m40 u is replaced with x. Whilst u was x-a/x so I do not get this step!!

[@ddognine]

7:40 I don't follow how f(u)du maps back to simply f(x)dx. Since u = x - a/x, shouldn't it map back to something like f(x - a/x)du where du is (1 + a/x^2)dx?

[@ddognine]

Doh! I see it now!

[@yellowrose0910]

@@ddognineDoh! I still don't! Help pls thx!

[@Grecks75]

@@yellowrose0910 It's a simple renaming. It doesn't matter what name you give to the integration variable. For the same reason as it doesn't matter what you name the index variable of a finite or infinite sum. The integration variable is bound to the process of integration and has no meaning outside of the integral. The definite integral itself is a single real number, not a function, and as such does not depend on any variable.

[@Necrozene]

OMG, I had too much to drink. I love this stuff!

[@johannmeier6707]

I din't really understand why you can use the one substition for the negative integration bounds and the other for the positive integration bounds. Why was this allowed?

[@Yougottacryforthis]

it's not can, it's must. When x is +ve you want to use +ve and vice versa. You just take the appropriate 'branch'

[@Yougottacryforthis]

If its not clear in the domain (-inf,0] x is negative you have to assure RHS is negative (since its an equality)

[@ha14mu]

He split the integral into two. He's using one substitution on each integral which is allowed

[@danielleza908]

Once you split the integral to two different integrals, you can solve each of them individually, and use any substitution you find useful in each of them. You're not required to use the same substitution for both. In this case, he correctly identified that one substitution was only defined for positive x's, and the other was only defined for negative x's, so he knew which to use for each integral.

[@TomFarrell-p9z]

Nice! Just wrote that in my Gradshteyn and Ryzhik, in case I can ever use it.

[@Flores31206]

why does this not work with the dirac delta function?

[@minwithoutintroduction]

جميل جدا.أتمنى ان لا أنسى استعمال هذه الخاصية قبل البدء بحساب أي تكامل

[@aksenchukaleksandr3273]

Thank you very much. very clear proof.

[@grayjphys]

Idk why this is reminding me of the legendre transform...

[@mihaichelariu9137]

What happens if everything returns to the point of origin?

[@vnknovn]

7:40 why can you do that?

[@ethancheung1676]

we cannot do this if it is an indefinite integral. but this is a definite integral so we can write all u to whatever we want including x (it is called dummy variable)

[@Grecks75]

It doesn't matter what name you give to the integration variable. For the same reason as it doesn't matter what you name the index variable of a finite or infinite sum. The variable is bound to the process of integration and has no meaning outside the integral.

[@jamesfortune243]

Nice idea!

[@wandrespupilo8046]

can you actually do that in 7:40? isn't u = x - a/x ???? what?

[@ethancheung1676]

we cannot do this if it is an indefinite integral. but this is a definite integral so we can write all u to whatever we want including x (it is called dummy variable)

[@yellowrose0910]

@@ethancheung1676 But u is a function of x. We can't just 'rename' u to x because u contains x's since it's a function of x.

[@ethancheung1676]

@@yellowrose0910 we can, if it is definite integral.

[@Tehom1]

Wondering why it's just positive a. Can't you just define g(x) = f(x + a/x) and get the negative case? And use x = x - 0/x for a=0 except for the one indeterminate point x=0?

[@gerardomalazdrewicz7514]

The a=0 case is trivial.

[@richardheiville937]

G. Boole and, Cauchy were surely aware of such theorem. Ramanujan was using a generalization of such theorem

[@zucazx]

Michael, can you develop the infinite product (1+x/2^n), n = 0, 1, 2... into a infinite sum?

[@roberttelarket4934]

S & M(ike)! Mike the Master!

[@moonwatcher2001]

And that's a good place to stop. I love this Channel ❤

[@saulmendoza1652]

almost the best 14 mins of my life... (RMT)

[@riadsouissi]

I would used u=-a/x as a substitution. Much easier.

[@shahidkupe]

Amazing🤩🤩

[@ojas3464]

👍

[@OlympiadProblemsolving]

kzbin.info/www/bejne/e3irn56KbbyiiaM

[@nicolascamargo8339]

Genial

[@barryzeeberg3672]

It seems incorrect to me to break up the integral into 2 parts, and then use different formulas for x in each part (ie, using + or - the square root). I think you need to choose either the + or the - and stick with that one consistently both parts. Sometimes one or the other can be eliminated, based on physical reality. Otherwise, you need to solve separately for both variants. What might work is to solve the integral(s) consistently first for the + and then for the - . Maybe if you take an average of the 2 solutions, there will be a cancellation that ineffect produces the same result as in the video? But that is still not the same as deriving the master formula.

[@amritlohia8240]

It's just making a separate substitution in each integral - this is perfectly valid.

[@G0r013]

​@@amritlohia8240you are right. Basically you are using two different injective substitutions.

[@amritlohia8240]

@@G0r013 Yes - as long as your substitution defines a continuously differentiable bijection, it will work.

[@Alan-zf2tt]

I think this may be one of those special methods that computers can do very well whereas mere mathematicians will work through from first principles. It also looks like a good justification for Category Theory and Abstract Algebra. I suppose that speeds up computer ways of doing things too. EDIT: hmmm I wonder if Glasser declared IP over it as a computer algorithm would he be on a nice little earner?

[@DR-tx3ix]

It doesn't make sense to me either. He used the negative x formula in the left integral (-infinity to zero): would that integral yield the same result if he had used the positive x formula?

[@terryendicott2939]

Cute.