Пікірлер

• 
  @cadmio941317 сағат бұрын

  Such a beautiful result, transforming such interesting integrals into things we've seen before will never stop amazing me.

• 
  @ethanbartiromo288819 сағат бұрын

  Wow, I never heard you say “fuck” before

• 
  @Loots120 сағат бұрын

  onlyfans?

• 
  @ruffifuffler871121 сағат бұрын

  Take it one step further by relating 'phi to kewness of fruit trees, thereby expanding the integral repetoir of Golden Ratios.

• 
  @orionspur21 сағат бұрын

  I'll be a lady every other Thursday if that helps.

• 
  @SuperSilver31621 сағат бұрын

  I believe it’s the Inversion formula for the Trilogarithm

• 
  @worldnotworld22 сағат бұрын

  "OK Cool" should be your nickname

• 
  @CM63_France22 сағат бұрын

  Hi, "ok, cool" : 4:29 , 6:30 , 7:30 , 10:25 , 11:51 , 13:56 , 17:44 , "terribly sorry about that" : 4:40 , 8:42 , 9:52 , 12:58 .

• 
  @maxochoa92522 сағат бұрын

  @Maths_505 - a suggestion for your consideration - I think it would be cool to show the Desmos graph of the integrand as part of the videos. Would give at least an approximate sense for the order of magnitude we should expect of the answer.

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  @neilgerace35522 сағат бұрын

  19:16 :)

• 
  @unturnedd22 сағат бұрын

  congrats on the ladies view increase 🤣

• 
  @giuseppemalaguti43523 сағат бұрын

  A me risulta I=-1/2+(3/2)ζ(3)+2Σ((-1)^k)coshk/k^3...ma la serie non converge,boh

• 
  @kingzenoiii23 сағат бұрын

  double trouble integral 😂😂

• 
  @vitorbordini524623 сағат бұрын

  Why 4 and a half ladies? What is a half lady?

• 
  @gold683623 сағат бұрын

  Half a lady is a lady cut in half?

• 
  @xleph252522 сағат бұрын

  That's the DiLady function evaluated at "Terribly sorry about that." Works like a charm out on dates

• 
  @Calcprof23 сағат бұрын

  Mathematica quickly gets -(1/2)+PolyLog[3,-(1/E)]+PolyLog[3,-E]+(3 Zeta[3])/2, where PolyLog[k,z] gives the PolyLog function Li_k (z)

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  @xleph252523 сағат бұрын

  Pollywog function? Does it display amphibious behavior and wiggle like a pollywog? Does it traverse the complex plane by hopping like a toad? You still spelled PolyLog wrong, bro.

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  @Calcprof22 сағат бұрын

  @@xleph2525 Autocorrect!!

• 
  @roshanmadhav887623 сағат бұрын

  Didn’t you do this one but It goes to infinity instead of 1 ?

• 
  @maths_50523 сағат бұрын

  Nope...not with infinity.

• 
  @notfancy200023 сағат бұрын

  The sign error after the first integral is resolved is cancelled by the sign error with the dilogarithm expansion.

• 
  @maths_50523 сағат бұрын

  Yeah I noticed that while editing but honestly I thought it was so damn funny I left it in as a sort of easter egg 😂😂

• 
  @bingchilling838420 сағат бұрын

  Thats actually hilarious and amazing

• 
  @noobymaster698023 сағат бұрын

  Interesting integral with solution that includes a coefficient of 4,5 when

• 
  @maths_50523 сағат бұрын

  Yo what are those strange symbols?? I see them on my keyboard as I'm typing and it feels like I remember them from a past life but....what do they mean???

• 
  @gaminfish23 сағат бұрын

  second

• 
  @maths_50523 сағат бұрын

  Third

• 
  @somerandomlettersКүн бұрын

  **slaps roof of video** this bad boy can fit so much nice cancellation taking place

• 
  @zunaidparkerКүн бұрын

  It's cheating to put phi in the intergrand I feel. Not surprising that phi pops out in the result.

• 
  @maths_505Күн бұрын

  It's cheating only if the solution did not make use of the properties of phi. Phi at the end is simply our reward😂

• 
  @77Chester77Күн бұрын

  11:13 shouldn't it be "phi -2" instead of "phi -1"? Cool result nevertheless.

• 
  @umairbutt1355Күн бұрын

  Wow. The fact that this even converges facinates me. sin(x^2) converges because it oscillates more wildly as x -> infty. But the factor of ln(x) is making it increase logarithmically - which is still "too slow" for the integral to diverge!

• 
  @DD-ce4ndКүн бұрын

  This integral possesses beautiful properties indeed. When replacing the golden ratio by a generic power z in C, we obtain the closed form: I(z) = (z^2 +1)/ (12*z) * Pi^2. And this yields the elegant reflexion formula I(z) = I(1/z). Its only zero seems to occur when z = +/- i 🙂

• 
  @MrGLA-zs8xtКүн бұрын

  This is nice

• 
  @CM63_FranceКүн бұрын

  Hi, The final result can be simplified into : sqrt(5) * pi^2/12 "ok, cool" : 2:31 , 2:58 , 11:26 , "terribly sorry about that" : 3:55 , 4:23 , 6:05 , 6:08 , 10:10 , 10:13 , 13:11 , 14:18 .

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  @maths_505Күн бұрын

  @@CM63_France damn I was terribly sorry a terribly lot this time 😂

• 
  @krishpandey854Күн бұрын

  another method- I= integral(xcotx) apply by parts to get: I=integral(-ln(sinx)) from x=0 to x=pi/2 Using symmetry: I=integral(-ln(cosx)) from x=0 to x=pi/2 Adding: 2I=Integral( -ln(sin2x) + ln2) from x=0 to x=pi/2 =integral(-ln(sin2x)) + pi/2 ln2 Interestingly: integral(-ln(sin2x)) from x=0 to x=pi/2 is equal to I (can be proved by subsitution u=2x, followed by use of symmetry of sinx about x=pi/2) 2I=I+pi/2 ln2 I=pi/2 ln2 Done

• 
  @MRGamesStreamerКүн бұрын

  How many years work in integral department (Years of experience)

• 
  @trelosyiaellinikaКүн бұрын

  Mashallah! I've said it once, whoever has given you the name Kamal (perfection) has depicted you exactly! Please thank him/her for me.😊

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  @maths_505Күн бұрын

  @@trelosyiaellinika you're message has been conveyed to my mother 😂

• 
  @guy_with_infinite_powerКүн бұрын

  Just out of curiosity, Where do you get these integrals? Like what book/s?

• 
  @maths_505Күн бұрын

  I mostly just make em up or find them on the internet. Math stackexchange is awesome 🔥🔥

• 
  @petterituovinem8412Күн бұрын

  17th

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  @giuseppemalaguti435Күн бұрын

  Si arriva facilmente a I=Σ((-1)^k/(k+1))π/sinπ(k+1)Φ..poi,boh..tu hai usato un metodo diverso .io ,invece, ho usato..la serie logaritmica,la funzione beta,e poi la gamma reflection.. poi mi sono bloccato..ah ah...forse ho trovato l'errore:non si può sviluppare in serie logaritmica perché ln(1+x)...x,tra 0 e 1, è maggiore di 1..

• 
  @Sugarman96Күн бұрын

  Very satisfying integral. The fact that η(2) popped in both sub integrals is nice and makes you think about whether or not it could have been arrived at from that integral before splitting it.

• 
  @AntAnkhКүн бұрын

  Where do you find such integrals? They're all really cool. Do you have any textbooks you can recommend which have integrals like this?

• 
  @Salmanul_Күн бұрын

  Cook up an integral that has Gamma (pi) as the result. Without any pi in the integrand

• 
  @MinecraftForever_lКүн бұрын

  Σ author💅

• 
  @leroyzack265Күн бұрын

  This was gorgeous 😍. Thanks for the amazing result.

• 
  @2kchallengewith4videoКүн бұрын

  Truly the golden moment of all time

• 
  @insouciantFoxКүн бұрын

  1/(φ-1)=φ (3-φ)φ= 3φ-φ-1=2φ-1

• 
  @maths_505Күн бұрын

  @@insouciantFox I know but I just loved that final form 😭

• 
  @waarschijnКүн бұрын

  φ + 1/φ is even nicer

• 
  @venkatamarutiramtarigoppul2078Күн бұрын

  Now i am starting a war 😅😅sqrt 5* pi^2 /12 is lot better. Kust kidding any form in maths is as beautiful &satisfactory as the other one

• 
  @slavinojunepri7648Күн бұрын

  This is a salivating beauty, and I cannot think of another way of describing it.

• 
  @Chris_387Күн бұрын

  π²√5/12

• 
  @RalfStephan21 сағат бұрын

  Claude 3.5 finds it too from the phi fraction

• 
  @user-gs5ff1cd2sКүн бұрын

  Nice!

• 
  @maxmoedough6401Күн бұрын

  Im so early there isn't even audio

• 
  @maxmoedough6401Күн бұрын

  Weird YT glitch, no worries ;)

• 
  @CM63_France3 күн бұрын

  Hi, "terribly sorry about that" : 1:32 , 6:10 , 11:45 , 12:40 , 14:12 , "ok, cool" : 12:50 .

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  @manstuckinabox36793 күн бұрын

  Ahh... the good ol massacring of integrals that make me wonder if they appear in any real life application. It's been a while since I came here, and so glad it still feels like home. always nice to see myself grow more mathematically mature along side your channel, keep it up big G!

• 
  @archinsoni12543 күн бұрын

  Is integral 0 to infinity x^(-x) possible ?

• 
  @davidblauyoutube3 күн бұрын

  Ah, but is the final result irrational?

• 
  @Tosi314153 күн бұрын

  solved last video's homework and got I=π/(2sqrt2*e^(sqrt2))

• 
  @theIndicLearner3 күн бұрын

  Sir, please can I solve the same equation with the same method on my channel.In my native 'Hindi Language ' Please let me do it for some videos.I will get a good starting.

• 
  @maths_5053 күн бұрын

  You have permission to take all integrals from my channel. Good luck my friend, I understand Hindi because I'm Pakistani and speak Urdu so I might watch some of your videos.

• 
  @theIndicLearner3 күн бұрын

  @maths_505 OH, is it, you are Pakistani ? I am really sorry for the embarrassment of us both, but I can't do it now. We both give taxes to our governments, and when they fight, it's the public who suffer. If you want to support us, bring a government in your country which is friendly with India. Ours, already is such. Thanks a lot for your support as a common man.

• 
  @maths_5053 күн бұрын

  @@theIndicLearner 🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️

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  @theIndicLearner3 күн бұрын

  @maths_505 does that mean you belong to Pakistan but you are currently a citizen of some other country. If this is so, then I am sorry again we suffered a blow very recently at 9th this month from your country's side as the circumstances are saying it loudly so even the name hurts, right now. Really sorry again.

• 
  @agrimmittal3 күн бұрын

  ​@@theIndicLearner you talk a lot

• 
  @Ben-wv7ht3 күн бұрын

  I beg of you , change the bounds to become 0 - pi/3 , this will be MUCH MORE challenge

• 
  @giuseppemalaguti4353 күн бұрын

  sinx>cosx....formule di bisezione...4 integrali risolvibili...in verità 3/4 sono semplici,ma int(lncos(x/2)*lnsin(x/2))????