Destroying 3 Integrals in one go! Using Papa's way [ Integral of ln(cos(x)) from 0 to pi/2 ]

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F u c c off symmetry arguments, we are going to do sht Papa's way today! Let us use complex analysis to rip this classic boi in half!
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[@michaelz2270]

I'm 106 years old and I have been doing integrals since 1929, and I still learn from all of your videos. Keep up the good work!

[@allaincumming6313]

Wow sir, blessings to you!

[@verraoz2080]

Are you alive ?

[@michaelz2270]

@@verraoz2080 No, I died several years ago. But that hasn't stopped me from doing integrals in the Great Beyond.

[@liam5010]

Love how he uses fancy variables in the thumbnail lol

[@maxhenderson1890]

They’re just Greek letters lol

[@anegativecoconut4940]

Flawless Triple Integration. We want more greek letter substitutions! The third definite integral is a classic Complex Analysis test exercise. αωεζσμε!

[@JanKowalski-zz8ef]

Weird letters for variables

[@NotEvenOptimusPrime]

That ξ in the thumbnail fooled me in to thinking this was a video using Fourier transforms. Still a good video tho :)

[@quahntasy]

that's a lot of damage, sawed three integrals in half and Just ξ

[@tytuer]

This one got more likes then I mine :l i'm disappointed.

[@PackSciences]

Set of answers to comments for youtube search engine: {"Yes", ":)", "=)", "lol", "xd", "lel", "lol", "lmao"}

[@46pi26]

:)

[@RobsMiscellania]

3:41 my jaw opened so far that it spasmed and couldn't close, any my eyes rolled back into my head. For the next minute all I could do it swivel slowly back and forth in my chair with my jaw hanging open, trying not to do one of my Asperger's-inspired shrieking laughing fits

[@tytuer]

Wow! That's a lot of damage! I sawed three integrals in a half

[@chrisatwood0918]

Great job! That was awesome!

[@angelmendez-rivera351]

When you are doing integration, but you still cite the Peano axioms

[@ZonkoKongo]

*N W O R D*

[@tobiasgorgen7592]

G U A C A M O L E

[@user-wu8yq1rb9t]

*Dear Papaaaaaa* .... I missed you 🤗 Papa Flammy ... This integral, specially with my Papa's method, is *Great, amazing* ... I just enjoyed and definitely I'll watch it again. Thank you ❣️

[@Fightclub1995]

Never use ln in complex analysis, use Log and specify the branch

[@cristianmeraz4181]

Very nicely done papa my morning is made

[@ziolkovy]

12:29 lim instead li_ :o

[@aadits5624]

Spicy AF *fire_emoji*

[@roddeguzman9958]

The INTEGRALs have left the chat.

[@davidbrisbane7206]

We don't want to look at this today, we want to look at something completely different. Papa. You sound like a politician now!

[@PapaFlammy69]

Hehe :D

[@lecinquiemeroimage]

I = ∫₀ₐ ln(cosx) dx is not difficult to calculate (between 0 and a, if a = π/2): Michel Penn did it, but he forgot to prove that I exists! (because lim ln (cosx) = - ∞ when x → π/2 (x < π/2)). I have 2 better exercises : 1 - Calculate J = ∫₀ₐ ln( cosx) dx with a = π/4 (using an elementary method). 2 - Calculate K = ∫₀ₐ √sinx dx/(√sinx + √cos x) with also a = π/4 (here it is not possible to use an elementary method)

[@mark_tilltill6664]

Fellow mathematicians, such an egalitarian greeting. Thank you.

[@lionelinx7]

that was sick

[@funkycude57]

Flamily 💓

[@dagatho]

This integral got the big oof

[@ionutradulazar8984]

papa is on fire right now

[@MrRyanroberson1]

For the first integral, as the pro I am, I'll cheat. Derivative of ln(cos)/x is tangent/x - ln(cos)/x^2. Int(x) is x^2/2. We get x ln(cos(x)) + int(ln(cos)/2 - x tan(x))dx, which I will set equal to the original, since it already is this means I can take the ln(cos) term and subtract it from both integrals, and then subtract what is left of the newer integral to get: x ln(cos(x)) = int(ln(cos)/2 + x tan(x))dx, so that's *that* integral solved. We could also find that int(ln(cos)/2)dx = x ln(cos) + int(x tan), and multiply both sides by two, if int(x tan) is solvable. Perhaps let the integral, having bounds of 0 to pi, get a sunstitution. U = pi-x, du = -dx, we get int(ln(sin)), with the same bounds. With that, we know that int(ln(tan))dx must equal zero since it's the difference of the lncos with the lnsin. None of this helps... Classic u=cos? Du=-sin, so int(-sqrt(1-u^2) ln(u))du why, flammy, must you do this?

[@crosisbh1451]

I thought this was very cleverly done. Nice!

[@SiiKiiN]

You should try to solve for the distance down a ramp of a particle which I sliding down it, where the angle changes with respect to time.

[@lecinquiemeroimage]

I = ∫₀ₐ ln(cosx) dx is not difficult to calculate (between 0 and a, if a = π/2): Michel Penn did it, but he forgot to prove that I exists! (because lim ln (cosx) = - ∞ when x → π/2 (x < π/2)). I have 2 better exercises : 1 - Calculate J = ∫₀ₐ ln( cosx) dx with a = π/4 (using an elementary method). 2 - Calculate K = ∫₀ₐ √sinx dx/(√sinx + √cos x) with also a = π/4 (here it is not possible to use an elementary method)

[@silasrodrigues1446]

Papa I wish someday I can understand everything you say! 💖

[@lukamitrovic7873]

PAPA IS BACC WITH MORE MATH!

[@antonzhuzhlev]

1:30 Zed'd died in one video before. He's immortal-ish, I think.

[@antonzhuzhlev]

Sort of

[@vietnambarca233gmail]

Such an amazing video, becoming a patreon right now

[@DemxnFPS]

what is the music that plays when the white backgroudn and the Integral itself is presented in the intro? there are two that play, I would like to find both of them.

[@theicemancometh5492]

Alright I got fully lost within 5 minutes. Never done complex analysis. Is there any books I can use?

[@bamdadshamaei1415]

Papa it's unironically harder to follow when you use the Greek alphabet, spare us North Americans a bit

[@PackSciences]

What does it have to do with North Americans?

[@bamdadshamaei1415]

@@PackSciences alot of us don't start using Greek letters well into grad school

[@luisramrod9121]

Nice!

[@mehtapuru]

#NoobAlert I don't know complex analysis can you tell me a source to learn it just enough to understand this video? I just know basic complex number and their curves. (I love curves) thanks.

[@nathanradford3416]

Super hot fiyaaa boiiii

[@eliasarguello9961]

Can someone explain to me how to prove Jordan's lemma for the gamma integral? I figured out that 1/(abs(R^2 * e^(2iθ) + 1))

[@eliasarguello9961]

Oh pls forgive me I just figured it out im stupid 🙄

[@charliesuarez1033]

Noize

[@DiegoTuzzolo]

ur awesome

[@keshavreddy8757]

As you are a genius xd solve the millennium problems please

[@soyoltoi]

Clickbaited by zeta.

[@-aaa-aaa]

Do you think in german and speak english or think in english and speak english?

[@Pinjesz]

Just ξ

[@Samuel-wd7sq]

I'd like to add you on Facebook. I have a First in Mathematics.

[@46pi26]

Is it actually Morning in Germany when you say "Good morning fellow mathematicians"? I wanna know you're legit.

[@luisramrod9121]

@@PapaFlammy69 😂😂😂

[@Eta_Carinae__]

Man, you need to do a complex analysis boyo

[@joshuatamer1933]

Why did he write ln(x^2 +1) as ln(x+i) and not ln((x+i)(x-i))?

[@timothywan2569]

What's your textbook for complex analysis? I wanna try and learn

[@aswinibanerjee6261]

Actually you are destroying ghosts of those integrals because I had already killed them

[@ADesignsOfficial]

10:05 You actually need to be really careful at this point if you want to pull the -1 out of the log. An unfortunate feature is that there the complex Logarithm is not unique. That is simply the consequence of the 2*pi*i-periodicity of the complex exponential function. At the beginning you are getting away with using the principal branch (which is not defined in -1!) of the complex logarithm but at this point you need to make your choice more wisely to guarantee that your argument is valid, i.e., you need to consider another branch that fits your integration path.

[@anaperea9251]

If I want to use this method with the integral of ln(sin x) between 0 and π... which f(z) I have to take for?

[@xCorvus7x]

17:39 That mathematical rigour XD

[@atrimandal4324]

It's raining integrals😂😂

[@tonk6812]

Sherlock will find the umbrella

Prior to watching the video: Got -(pi/2)ln2 for the first integral, hope it's correct. I've used phase shift and log sum to get I=1/2*intgr(ln(sen(2x)))-(pi/4)ln2 which is foreseeable to have a recurrence relation for the next integral, so in general terms we get that I=(1/n)*integr(ln(sen(nx)))-(pi/2)ln2*sum-of-gp [(1/2)/1-q, q=1/2], taking the limit as n tends to infinity we get I=-(pi/2)ln2 since the sum converges to 1.

[@mayankvats926]

Destroyed those Bois😂

[@masteryoda1748]

1sttttttt

[@weisengtan2534]

"here's the real deal" lmao

[@plemli]

1 ≥ cos(t) ≥ 0 over [0,π/2] so why would one be surprised that integrating ln(cos(t)) there yields a negative number ?

[@domenicosauta3116]

Can you recommend any complex analysis textbooks? I’m graduating high school this year and looking to do some reading over the break!

[@V-for-Vendetta01]

Two memes in 10 seconds. Way to go papa 😂😂

[@xmatterx9246]

What an exciting day ❤️❤️😫

[@ilovecovid]

OMG, i love you (no homo)