An incredible integral solved using Feynman's trick

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[@daddy_myers]

Maths 505 always providing us with what the rest of YT couldn't.

[@manstuckinabox3679]

we can also evaluate these integrals (after applying Feynman technique)using contour integration, which is a problem from Gamelin's Complex Analysis.
Exotic integral indeed.

[@maths_505]

Bro I don't think even gamelin knew his text this good 😂

[@pnintetr]

Absolute beauty.
I thought 1/log(x) in the integrand could be dreadful, but it was nothing in front of Feynman.

[@krisbrandenberger544]

@ 7:36 The first term of I'(alpha) should have a minus sign.

[@TMH2007]

yeah

[@jieyuenlee1758]

8:04 first term should have a negative sign in front

[@vit1leman14]

It’s so nice to catch up all your latest video! Quite awesome !

[@antoinehedin6608]

Hi!
Since Euler's reflection formula can only be applied with 0

[@ericthegreat7805]

What is great is that this means there is a connection and equivalence between this number and the binary integral
Int(oo,0) (exp(-x)/(1 + exp(-x)) dx = ln2
Which I mentioned was used in logistic regression for binary outcomes. Perhaps this can be related to the Lhopitals rule and considered an integral equivalent (due to the lnx in the denominator) i.e. a "derivative binary integral" with information/entropy equal to ln2, equivalent to an ordinary binary integral representing the rate of change of information or score function.

[@ericthegreat7805]

Also, in the limit the sequence at the bottom becomes a geometric series. So the relation is dependent on the fraction of the geometric series to the geometric series represented by (x^alpha - 1).

[@jannesfilgerdamm1419]

Are there some table sharts, that shows all, of the many possible transformation, into f.e. the gamma gunction etc ?

[@MrWael1970]

Very smart way. Thank you

[@nathanmenezes7914]

That PFD blew my mind. That was so fast

[@jhacklack]

wonderful

[@giuseppemalaguti435]

Applico feyman, semplicemente...I(a)=[...x^a-1.…],con I(0)=0,I(1)=I...I'(a)=Σ(-1)^k*Β(a+2k+1,-a-2k)...ma poi non riesco ad integrare la beta...come si fa?

[@TMH2007]

Very cool!

[@wagsman9999]

Whoa. Monster. Cool result tho.

[@arkadelik]

KNG 👑

[@kristim1]

🫡

[@Noam_.Menashe]

I think I once saw an easier way to solve integrals of this form, but I don't remember it.

[@daddy_myers]

Are you Indian by any chance?

[@Noam_.Menashe]

no@@daddy_myers