Ahmed's monster integral: solution using Feynman's technique

[@violintegral]

I remember watching papa flammy evaluate this integral a couple years ago but your solution is much more elegant!

[@maths_505]

Pinning this so that flammy knows he sucks🤣🤣🤣

[@Qrudi234]

@@maths_505real

[@machickenjoy3202]

@PapaFlammy69

[@manstuckinabox3679]

13:31 surely, you must be joking Mr.Feynman! Plot twist, he wasn't joking, and don't call him Sherley.

[@maths_505]

😂😂😂 Jokes aside t->1/t is one hell of a transformation

[@manstuckinabox3679]

@@maths_505 IKR! I always watch watching these kind of "monstrous" integrals due to learning techniques like this particular one, or weistrauss' sub and so on and so forth...

[@Rando2101]

Yeah, that'll Sherley work.

[@skyethebi]

5:45 couldn’t you just use t = 0 in which case the integral collapses to 0 rather than calculating the limit as t approaches infinity

[@megauser8512]

I don't know about that honestly, since you also end up with I(inf) within the I sub 1 that he had there, because of the inverse tangent at x = 1 making you have that inverse tangent of a function of t leftover within I sub 1.

[@trelosyiaellinika]

Indeed elegant! Not QUITE but VERY!

[@pacman7168]

Cheers from Caracas Venezuela how it call the application which you use for this video? Thanks in advance

[@zahari20]

By the way, Ahmed is pronounced with a hard "h", Akhmed.

[@volsyb]

.😊.

[@maalikserebryakov]

احمد Did i “pronounce” it correctly?

[@erivaldolopes632]

Why using the parameter trick setting t=infinity when you could set t=0 and avoid calculating that nasty integral?!

[@thomasblackwell9507]

Your handwriting is not bad; mine is so bad that my chickens cannot even read it!

[@nightmareintegral5593]

Now COXETER! 😂 Beautiful as always💚 But… What about Hankel Contour? You wanted to do that…

[@maths_505]

Hankel contour = next week Coexter = gonna knock me into next week

[@Stopinvadingmyhardware]

We solved that in Calc 2

[@ガアラ-h3h]

What Uni ?

[@Stopinvadingmyhardware]

@@ガアラ-h3h that was Community College

[@minhnguyen1338]

I'm really like you video when I use integration by parts I got this integral: int from 0 to 1/sqrt(2) of (x * arctan(x/sqrt(x^2+1)))/((2x^2+3)sqrt(x^2+1)) the answer is pi^2/(288*sqrt(2)) can you help me this Integral without using/knowing the orginal integral Thank you