one seriously crazy integral!

A cool generalized result derived using Feynman's trick.
My complex analysis lectures:
• Complex Analysis Lectures
Support my work and get early access to content via Patreon:
/ maths505
You can follow me on Instagram for write ups that come in handy for my videos and DM me in case you need math help:
...
My LinkedIn:
/ kamaal-mirza-86b380252
Advanced MathWear:
my-store-ef6c0...

[@CM63_France]

Hi, "ok, cool" : 1:14 , 2:16 , 3:25 , 4:25 , 6:17 , "terribly sorry about that" : 5:01 , 6:49 , 8:35 .

[@RohitPanda08]

woah! he's here

[@stefanalecu9532]

I think you meant to add 8:35 to the ok cool category

[@CM63_France]

@@stefanalecu9532 ok, I noted a "TSAT" at 8:35, maybe I confused, thanks a lot.

[@SamLindskougMBGY]

can already tell its gonna be a banger

[@stefanalecu9532]

I am not the French guy, so I'll take his place this time around "ok cool": 1:14, 2:16 (pen scratches ASMR), 3:24, 6:16, 8:34 "terribly sorry (about that)": 0:03, 6:47

[@jpf119]

I do need a KZbin plugin to un-OKcool the video, could someone help? 🤣

[@alphazero339]

Hello, thanks for your attempt. You did a good job, you can be him while he's gone. I also enjoyed that you included ASMR pen scratch in the list. Overall, the experience was positive as I can now click any of the timestamps you provided and hear the author speak the phrase. I do not know what happened to French guy, but we can hope he is okay and safe. For now, I'm voting you to take his place.

[@CM63_France]

@@alphazero339 wait a minute 🤣

[@stefanalecu9532]

​@@alphazero339 I can't guarantee I'll be on time, but I'll try if the guy above me can't do it

[@ByeUnderGround24]

Do a x=1/t sub on original integral that gives I = -I + (alpha-beta) int 0 to ∞ dx/1+x² so I = π(alpha-beta)/4 Thanks a lot for sharing new methods and manipulations . Each and every video of yours is a gem!

[@venkatamarutiramtarigoppul2078]

Ya i also did that i am way too lasy . What to do . But the solution develepment in the video is kinda cool

[@rishabhhappy]

awesome approach bhaiya

[@Grecks75]

Wow, can it be that simple? I just checked it. Incredible, awesome! What a symmetry! 🎉

[@ByeUnderGround24]

​@@rishabhhappy rishabh bro 🌚🗿

[@zachariastsampasidis8880]

Well it's an equivalent transformation since he did x to arctanx to π/2 - χ

[@Nottherealbegula4]

Waiting for the french guy to list the "ok cool" and "terribly sorry about that" (:

[@alphazero339]

He must have retired

[@jpf119]

I might unsubscribe because of that🤣

[@CM63_France]

Terribly sorry about that, I am back 🗼💈

[@alphazero339]

@CM63_France okaay, cool

[@Haxislive766]

😂

[@ddg-norysq1464]

I think this is the cleanest solution I've ever seen 😮

[@trelosyiaellinika]

Yep, you have a talent in choosing really elegant integrals.

[@maths_505]

@@trelosyiaellinika thanks mate

[@farfa2937]

Expectation: Some combination of gamma, beta and zeta. Reality: A linear function with delusions of grandeur.

[@MrWael1970]

Very nice idea. Thank you

[@michaelihill3745]

That was pretty awesome!

[@NeofeaxFeax]

sub x-> 1/x gives you the answer on a single line

[@atulyaroy8962]

Yea

[@anas-altaleb]

Tup

[@IsomorphicToIntingersAdjoin_i]

Great video i hope for more videos.

[@SabaSa6a]

The expansion of tan is not necessary here I remember that there is a similar question giving α=sqrt(2)

[@slavinojunepri7648]

Fantastic

[@AyushRajput-xw2ru]

Day by day kamaal questions are getting juicy . ❤

[@Aplicapitagoras]

@Math 505 any tips on how to identify which substituition I should make to solve more difficult integrals?

[@raghavendraPi]

Kamaal bhai, no video for 12 days 🥲, missing the awesome problems you solve

[@ademnasri6557]

Can I1 be solved by doing a simple fraction decomposition from the begining ? If yes please tell me how i've trying and i found no solution

[@Ownageffects]

please make a video of how to do the leibniz rule by details please and when can we switch the operators

[@شعرکوتاه-ع7ظ]

Was interesting

[@chancia8990]

okay this is iis awesome

[@giuseppemalaguti435]

X=tgθ..poi uso Feymann I(α)..I'(α)..I(β)..I'(β).2I'=π/2....I=(π/4)(α-β)

[@shadowreaper-gq6bz]

can you solve the integral from : 0 to infinity of (dx/((x²+1)(pi²+ln²(x))) , btw , i solved the integral the is result is "2/pi -1/2"

[@raylure991]

Let t=ln(x). The integral can be transformed to 1/2* ʃ_{-∞}^ ∞ \sech(t) / (\pi^2+t^2) dt. Use Mittag-Leffler expansion to expend \sech(t)

[@Haxislive766]

I(ß) = 0??

[@Lepl_Acie]

Hello , sorry for being annoying , may you complete your complex analysis explanation , it's very nice but I want more 🙂 . because I hope I can use counter intergals for solving so many things ...

[@charliecox13]

I know this will get drowned out. But on mathstackexchange I say an integral I haven’t been able to solve easily. The integral from 0 to 1 of f(x)=(x^3)tan(x^2). I tried Feynmans, exponentiating it, by parts, and writing it using summation. Can you give me a solution, or better yet a step by step?

[@maths_505]

@@charliecox13 I'll take a shot at it

[@tpgmflint]

when can theta be transformed to pi/2 - theta? (6:23)

[@lotaniq4449]

If the limits are from (0 to a), you can always do a substitution theta’ = a - theta. Its just a substitution, easy to verify.

[@absol4844]

​@@lotaniq4449 But doesn't leave that us with extra negative sign? Since a - θ evaluated at a is 0 and at 0 it's a, so the bounds are upside down now...

[@lotaniq4449]

@@absol4844 yes but dtheta’=-dtheta, so that - cancels the other -

[@atulyaroy8962]

​​@@absol4844 the lower and upper limits will be interchanged if you do the substitution so to make the limits same you use the negative sign. This called the King property check it out :}

[@atulyaroy8962]

It's called the King property, it can be easily proved by substitution (lower limit + upper limit -t)

[@tanusreekar2839]

okayyy cool

[@Ownageffects]

kamaal can i be the zeta to your (s) ? (no diddy)

[@davidblauyoutube]

Noice

[@alikazemi1329]

👏👏👏❤♥🙏👍

[@apnakaamkrelala]

Greetings and welcome back our kamaal😌✨🤌 my ears were eagerly waiting for your ohk cool and terribly sorry about that😭🤌✨💓