A masterful theorem for integration. 

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.

@@lossen1984 Only when u = 0 though.

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

@@lossen1984 WTF?

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

It depends on your pronunciation. It rhymes for me!

Should work on many, many languages.

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.

@@nyghts7 Thanks for the info !

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

@@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.

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'

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

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

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.

Doh! I see it now!

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

The a=0 case is trivial.

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)

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)

@@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.

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

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

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

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

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?

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?