i dont get how you see x^4+1 and think "hm, lets just add 0 which can also be written as 2x^2*cos(2a) with a=pi/4, this will really help me"
@mundocanibaloficil
5 жыл бұрын
experience...
@defunct1373
5 жыл бұрын
@@mundocanibaloficil well the first thing i thought after seeing x^4 + 1 was (x^2 + 1)^2, which has a 2x^2 term
@nastarkos5758
5 жыл бұрын
If you word it like that yeah, but that's not where the intuition comes from. It comes from something called the Poisson kernel.
@defunct1373
5 жыл бұрын
@@nastarkos5758 hmm ok
@nastarkos5758
5 жыл бұрын
@@defunct1373 ? I'm just saying. Wherever the video maker got this solution from probably failed to state this. It also shows up in basic generating functions.
@OleJoe
6 жыл бұрын
Wow! Great Job. You should title your video "Easy Integrals Made Difficult".
@ale4462
9 ай бұрын
How do you solve it easier?
@OleJoe
9 ай бұрын
@@ale4462 Contour integration using complex variables.
@taterpun6211
Ай бұрын
Or use beta function
@Seek64
6 жыл бұрын
"This is just as counter-intuitive as everything else" 😂
@blackpenredpen
6 жыл бұрын
The good old days! Btw, that was blackpenblackpen tho....
@rajibsarmah6744
4 жыл бұрын
Please make a video on integrating complex function
@rajibsarmah6744
4 жыл бұрын
Please
@JaskaranSingh-dz2wt
3 жыл бұрын
Hii BPRP an easy challenge for u Solve for x : cosine (2x)=i
@Euquila
6 жыл бұрын
That was a fun integral. It's amazing how all these steps seem both logical and magical at the same time!
@typo691
6 жыл бұрын
wow why didn't you become a doctor? you would've been a master at prescribing medicine with these skills 9:29
@davidberkowitz7697
6 жыл бұрын
Been a math teacher for twenty years. Love this proof, love the enthusiasm. You are a great lecturer.
@tensorproduct3666
6 жыл бұрын
you roasted that integral.
@anuvette
5 жыл бұрын
😂😂😂
@ianprado1488
6 жыл бұрын
0 to infinity real quick
@benthayermath
6 жыл бұрын
Papa flammy, I need some fatherly advice. Teach me how to win the ladies with sexy integrals
@mathematicalhuman6858
5 жыл бұрын
0:24 *_„And if you’re a smart ball...“_* this gave me tears of joy
@Paulovrish7334
4 жыл бұрын
I freaking enjoy watching you solve this type of problems
@PapaFlammy69
4 жыл бұрын
@debajyotisg
5 жыл бұрын
I am a simple man. I see 1/(x^4 + 1), I use complex analysis.
@gamma_dablam
5 жыл бұрын
I use partial fraction decomposition and solve simultaneous equations in 4 variables because why the hell not ! ?
@mustafaunal1834
4 жыл бұрын
Residue Theorem
@budtastic1224
3 жыл бұрын
I use: int from (0,∞) of 1/(x^n +1) = (π/n)/sin(π/n)
@ajfalo-fi3721
3 жыл бұрын
@@budtastic1224 Could this be solved using Beta and Gamma functions? I think I remember it could
@budtastic1224
3 жыл бұрын
@@ajfalo-fi3721 sure can! Very nice proof here.... kzitem.info/news/bejne/2ailnZdjpnacZJw
@RandomDays906
5 жыл бұрын
There's an amazing generalization of this integral that says Integral(0, inf)[x^(a-1)/(1+x^p) dx] = pi/(p*sin(a*pi/p)). For this case, a=1, p=4 gives pi/4sin(pi/4) = pi/2sqrt(2).
@debajyotisg
5 жыл бұрын
Can I just say, I never thought watching someone solve integrals would make such a satisfying video series? Mad respect!
@mlliarm
6 жыл бұрын
That's a lot of creativity for little over 15 mins of an explanation. I'm amazed :) Keep up the good work !
@kartikchoubisa
6 жыл бұрын
I just found this another nice way: put x=1/t in the original integral, rearrange the fractions so the the denominator of this new expression becomes same as that of orignial expression. Add them both, just like you did (but without introducing the cos(a) thingy). Divide the numerator and denominator by x^2. The denominator becomes (x^2+1/x^2) which is equal to [ (x- 1/x)^2 +2]. Substitute u in place of (x- 1/x) and the expression changes into simple [1/(a^2 + x^2)]form.
@kemokidding
6 жыл бұрын
Except that u = (x - 1/x) leads to a really gnarly expression for dx in terms of u and du.
@allaincumming6313
5 жыл бұрын
Is that the algebraic twins method?
@You12783
5 жыл бұрын
for integrals of the form 1/(1+x^n) with limits going from 0 to infinity the value is given as [pi/n]/sin(pi/n) .. just substitute value of n to get answer ...
@W.M.-
Жыл бұрын
@@You12783 Be careful, your formula only works for even n
@sethuramanjambunathan
4 жыл бұрын
Fantasic! I knew to solve this on complex plane using residue theorem. This is fantastic. My appreciation.
@chessandmathguy
6 жыл бұрын
Amazing. The whole thing. Every step made sense, and the final answer is just beautiful. Thanks for posting this!
@OhDannyBoy512
6 жыл бұрын
I would never have thought to do most of these steps, but you make it seem so logical and straightforward :D
@Henrikko123
6 жыл бұрын
Have you considered making a ‘German plus Math’ tutorial series? Something like teaching both German and math at the same time. I would watch the shit out of that, probably even supporting you on Patreon.
@Henrikko123
6 жыл бұрын
Flammable Maths I know ;) I’ve watched a lot of them. Great work!
@dragonite7780
6 жыл бұрын
Jesus, what a madman, subbed.
@ozzyfromspace
4 жыл бұрын
I expected you to just contour integrate that thing, then I saw cos(2a) and realized, "oh shiiii, we're about to have a good time!" This was a super fun video, thanks Papa Flammy! 🙌🏽🙌🏽🙌🏽🔥😁🌠
@PapaFlammy69
4 жыл бұрын
@olimatthews5636
6 жыл бұрын
New to your channel but loving it so far. How old are you? Where did you go to uni? Are you gonna do a Q&A? What is your favourite piece of maths?
@Applefarmery
6 жыл бұрын
„Woop woop“ thats the sound that the police makes 😂
@46pi26
6 жыл бұрын
The intro=the birth of a goddamn legend
@u.v.s.5583
6 жыл бұрын
To "Good morning, fellow mathematicians!" you should record a chorus singing "Good morning Papa!"
@46pi26
6 жыл бұрын
U.V. S. Like the blackpenredpen "yay!"
@carlosgiovanardi8197
5 жыл бұрын
EXCELLENT - NO WORDS - GREAT TEACHER
@priyanshusingh3451
5 жыл бұрын
This is really a crazy approach
@michaelgeinopolos6911
6 жыл бұрын
This is absolutely ridiculous.
@anegativecoconut4940
6 жыл бұрын
FH: How many techniques to compute the 1/(x^4+1) integral you have? AM: Don't know, 5 or 6 I guess. FH: You are like a first grader, watch this.
@michelkhoury1470
6 жыл бұрын
Yes there are many ways to solve this
@michelkhoury1470
6 жыл бұрын
But not for integral of 1/(1+x^50) but I solved it 😀
@anegativecoconut4940
6 жыл бұрын
Michel Khoury It's easy, the final answer is pi/(50sin(pi/50)) but can you give me an analytical form of the indefinite integral?
@michelkhoury1470
6 жыл бұрын
Thomas Torrone yes of course it's very easy
@michelkhoury1470
6 жыл бұрын
Thomas Torrone okay to evaluate the indefinite integral you can transform the integral into an incomplete beta function by the substitution t=x^n
@krithikaa4188
4 жыл бұрын
Wow mind blown!!! I never thought of solving like this. Thank you 🙏🏼
@ronbannon
2 жыл бұрын
GREAT JOB. I will share this with my students and publish a different path to this solution. I will try to post a video later this week.
@ronbannon
2 жыл бұрын
I added this question to my course notes and will encourage my students to pay attention to your insightful technique. You can see my video solution, using a different method, on KZitem: kzitem.info/news/bejne/0Gl6tYiJbKidf3Y
@SiwakSerg
6 жыл бұрын
The very first thing I thought about was the residuals (the complex function approach and complex analytic integration), I think it's a standard thing to do, but your method is quite peculiar and clever. Well done)
@CornishMiner
6 жыл бұрын
Whoa! Retro Flammy.
@henrykwieniawski7233
6 жыл бұрын
Have my babies Papa Flammy
@danielcardenas2607
5 жыл бұрын
I really enjoy your videos, you're so good doing this tricks and solve that crazy problems
@chaos4785
6 жыл бұрын
Math is fun!!! You make things look a lot easier than they really are😂 I always enjoy watching your videos👍
@TheSam1902
5 жыл бұрын
This is exactly why I watch this channel !!
@joshuawood1082
6 жыл бұрын
Thank you Papa, very cool!
@ernestschoenmakers8181
5 жыл бұрын
You can factor out 1+x^4 into (x^2-sqrt(2)x+1)(x^2+sqrt(2)x+1) and solve the integral by the method of partial fractions.
@Hexanitrobenzene
5 жыл бұрын
Of course you can, but that's boring... That's why there are nerds like Papa Flammy who invent some ingenuous methods and, lucky us, mere mortals, share them with us :)
@ernestschoenmakers8181
5 жыл бұрын
@@Hexanitrobenzene I don't see what's boring about it, this solution just came up to me.
@Hexanitrobenzene
5 жыл бұрын
@@ernestschoenmakers8181 Well, you can factor x^4 +1 by taking fourth complex root of -1 and then multiplying binomials corresponding to conjugate roots. Completely standard. Partial fractions are also completely standard. By boring I mean the path which is obvious, yet tedious. Partial fractions are more suited to computers, not humans...
@ethanjensen661
6 жыл бұрын
This integral is easier than people think. It can be "factored" into two quadratic polynomials, do partial fractions, and so on
@anderrafaellinaresrojas3772
4 жыл бұрын
Nobody: Flammable: :D
@brayanrojas7754
4 жыл бұрын
:v?
@nikhilnagaria2672
3 жыл бұрын
0:01 the best intro of Papa flammy ever
@AlEx-ro7kd
6 жыл бұрын
i dont understand much of this but your videos are entertaining
@charliepilgrim1065
6 жыл бұрын
Woop Woop! That's the sound the police makes
@atrimandal4324
6 жыл бұрын
It was kinda doable till around 10:15,after that it became typical Flammable 🔥🔥😂 Btw, that A4 paper stuff in your website was great ❤️
@zeroleader
6 жыл бұрын
Whoo! Excellent work, dude
@karthiksatyanarayana3559
4 жыл бұрын
we can complete the question by substituting x^2=tanx,and solving we get integral of cot^1/2(x) which we can solve
@karthiksatyanarayana3559
4 жыл бұрын
this method is faster than what he said
@VivekYadav-ds8oz
4 жыл бұрын
I like the more well-known method more. => integral dx / (x^4 + 1) = (1/2) integral 2 * dx / (x^4 + 1) => 2 can be written as 1 + 1. Add 0 in numerator which can be written as x^2 - x^2. => (1/2) integral ( (1 + x^2) + (1 - x^2) ) dx / (x^4 + 1) => Separate integral with one having (1 + x^2) in numerator and other having (1 - x^2) in numerator. => Now it's easy. Let's say I1 = (1+x^2) dx / (x^4+1) Divide by x^2 in numerator and denominator, I1 = (1 + 1/(x^2)) dx / (x^2 + 1/x^2) // Rearranged the numerator in same step. => (1 + 1/x^2) dx / (x^2 + 1/x^2 - 2 + 2) Now, x^2 + 1/x^2 - 2 can be written as x^2 + 1/x^2 - 2*x*(1/x), which is nothing but (x - 1/x)^2. So, => (1 + 1/x^2) dx / ( (x - 1/x)^2 + 2) We can now clearly see that the derivative of (x - 1/x) is present in numerator, so substituting (x-1/x) = t would be the right thing to do. x-1/x = t => 1 + 1/x^2 dx = dt => I1 = dt / (t^2 + sqrt(2) ^ 2) => I1 = 1/sqrt(2) arctan(t/sqrt(2)) + c, where t = x - 1/x. Done! Now that one knows this method, one can easily find I2 i.e (1 - x^2) dx / (x^4 + 1).
@duckymomo7935
6 жыл бұрын
That was quite the adventure for that integral The indefinite integral is scarier
@duckymomo7935
6 жыл бұрын
Flammable Maths Complex analysis would’ve worked as well The general solution to Integral from 0 to infinity of 1/(x^(2n) + 1) = pi/(2n*sin(pi/2n))
@richardheiville937
5 жыл бұрын
Perform the change of variable y=1/x, I=integrate(0,infinity, x^2/(1+x^4)) therefore, 2I=integrate(0,infinity, (1+1/x^2)/((x-1/x)^2+2)) perform the change of variable y=x-1/x, 2I=integrate(x=-infinity,+infinity, 1/(x^2+2)) therefore I=integrate(x=0,+infinity, 1/(x^2+2))=Pi/(2*sqrt(2))
@Mr_Academic98
4 жыл бұрын
Great inspiration
@PapaFlammy69
4 жыл бұрын
@seegeeaye
Жыл бұрын
I did it without using 2x^2cos2a, but still use x = 1/t. We then have 2I = int of (x^2 + 1) / x^4 + 1, then factorize the denominator followed by using partial fractions and finally end up with two arctangent functions before getting the answer of sqrt 2 pai /4
@neo_tsz
6 жыл бұрын
Huh...my mind is satisfied.
@lesnyk255
6 жыл бұрын
That was a wild ride! Loved it! Seems to come from the same mindset as Feynman's integration of sinc(x) - introducing a parameter whose special case is the integrand you actually want to plug in....... (see blackpenredpen)
@slavinojunepri7648
Жыл бұрын
Impressive solution from an alternate universe
@charllsquarra1677
5 жыл бұрын
badass way to integrate
@richardturietta9455
6 жыл бұрын
nice! I remember doing all sorts of math gymnastics back in college...
@paparapiropip87
6 жыл бұрын
Man, you’re the God of integrals
@jaredwhite4934
5 жыл бұрын
I think I like watching cute guys solve sexy integrals even more than those "hot guys and baby animals" videos.
@lorenzonioi7855
6 жыл бұрын
I'm in love
@si48690
6 жыл бұрын
excellent video
@dnranjit
5 жыл бұрын
making 1=2cos(2a) for a=pi/4 was magic...from a creating thinking perspective ... I would really like to understand how you came to the conclusion that this transposition is needed.Does it come by practice or a general rule that if you see a particular form then this particular modification can be applied. Or is it just from pure trial and error ?
@孙林可
6 жыл бұрын
Wow, what can I say? I guess if Feynman had watched this he would open IMU's locker and award you with Fields Medal.
@ИванРодькин-ю7ч
6 жыл бұрын
Now, try this: integral from -inf to +inf ln(x)/(x^2+1) Pretty nice answer
@duckymomo7935
6 жыл бұрын
From 0 to infinity is just 0 -inf to inf is “don’t worry about it” (i*pi^2)/2 Use complex analysis for the second one
@visheshmangla2650
6 жыл бұрын
this sucker is a must for board exams(class 12) in India.
@TheNachoesuncapo
6 жыл бұрын
Me:Hey Pappa flammy do you do any sports?you look great! Pappa:no I just solve crazy ass integrals from time to time :)
@restitutororbis964
6 жыл бұрын
Nacho He lifts heavy infiniti bois for about 3 hours a day.
@TheNachoesuncapo
6 жыл бұрын
No pi-in no gain
@Bjowolf2
5 жыл бұрын
Chalk gymnastics, blackboard lifting, extreme integration without limits, bending the rules... 😂
@moslemasultana9388
6 жыл бұрын
balance the denominator by 1/2[(x^2+1)-(x^2-1)], then separate and divide up and downside by x^2.
@MathIguess
5 жыл бұрын
Whoop whoop, that's the sound the police makes xD excellent I'm binging your videos, man, and all I can say is thank you xD
@thexabi123
5 жыл бұрын
What a lovely "2" at 13:27
@אדוארדששון-ע8ס
3 жыл бұрын
Your problem is only a special case of the integral of 1 divided by 1 + x^n, whose value is (pi/n)/sin(pi/n), which is solved using substitution, the gamma and beta functions, and the relation between the product of two gamma functions and sine! The general problem can also be solved using complex variable integration.
@im_Smitty
6 жыл бұрын
Sick approach, any particular course where you learn stuff like this? Or is it more general?
@ChiragBharadwajYT
6 жыл бұрын
@Flammable Maths is there a particular resource where one could learn this approach? I quite like this method of "parameterizing the integral" to solve more complex integrals algebraically with clever substitutions. Would love to get some practice with this!
@im_Smitty
6 жыл бұрын
Flammable Maths That some sick dedication, do you just do math in your free time or some physics problems as well?
@EAtheatreguy
3 жыл бұрын
I just used substitution and assumed an integral from infinity to infinity was 0. I don't know if my methods were sound, but I got the same answer.
@Zzznmop
6 жыл бұрын
I came back to this video, I realized what zero is. I am smart boy like papa says
@dhoyt902
5 жыл бұрын
This video is a year old but I still clapped.
@JohnSmith-iu3fc
4 жыл бұрын
Thank you! n You also verify that integral of // 1/x^4 +1 dx and x^2/x^4+1 dx is equal . So, the value of integral of x^2 +1/x^4+1 dx is squrt 2 *pi , the same value as integral of 1/x^4 +1 dx plus x^2/x^4+1 dx.
@MrAcidonucleico
6 жыл бұрын
Papa spacy as always!
@undeadarmy3000
4 жыл бұрын
This is insane
@williamtachyon2630
6 жыл бұрын
Why could he wrote 1 + x^2 in the numerator (in about 4:55) ? Where this +1 came from?
@MarioFanGamer659
6 жыл бұрын
It's coming from adding integral before and after it was rearranged together (i.e. added the top and bottom integral together ‒ both, the integral boundaries and the denominator are the same).
@michaldvorak8586
6 жыл бұрын
Insane papa, well done!
@pituitlechat3807
6 жыл бұрын
I Flammy You dont need to use trig identity. Your trig functions are just a constants and you can replace it by cos²a = sin²a=(√2/2)²=1/2 then : your denominator x^4+1 become (x²)² + 2x² +1 -2x² = (x²+1)² - (√2)²x² = (x² + √2x + 1)(x² + √2x - 1) and in your numerator you can add the odd function -(√2)x, simplify with the denominator and obtain 1/(x² + √2x + 1) then you split 1 in 1/2 + 1/2 or (√2/2)² + (√2/2)² and you have 1/ [(x + √2/2)² + (√2/2)²] and it's the same answer that 1/ [(x + sin(a))² + cos²(a)]... Yay!
@housamkak646
6 жыл бұрын
Man u r a legendary
@arjunv7597
4 жыл бұрын
I did this in the back whiteboard of my Calc AB class, except I used PFD and some really sketch u-subs. Was way more approachable for a high schooler with no rigorous training lol.
@johnoleary3732
6 жыл бұрын
Does the odd function integral property still work for upper- and lower-limits of infinity?
@Thenewgreenmaju
5 жыл бұрын
Excellent! Nice explanation.
@pritishmoharir3830
5 жыл бұрын
Amazing stuff
@cliffordwilliams9597
4 жыл бұрын
God I love this!
@cytyy8531
3 жыл бұрын
we just had an exam today with this kind of integrand.
@timdedecker7894
6 жыл бұрын
Stunning
@shubhamaryan8202
5 жыл бұрын
Really studying with you maths has become easier
@JohnSmith-iu3fc
5 жыл бұрын
Too flammable! Also too pedantic!!! I need more explanations of your logical leaps in your video. How about the result " pi/2" for the arctangent (x^2) when the x goes unlimited?
@joshuamason2227
4 жыл бұрын
HOLY SHIT THAT'S THE SMARTEST THING IVE SEEN IN MY LIFE
@46pi26
6 жыл бұрын
Gotta chow down on some infinity bois before work for protein
@bboygmoney
5 жыл бұрын
That was intense.
@derpsalot6853
4 жыл бұрын
Could someone explain to me the part where he added I + I = 2I? Why is he allowed to change dt -> dx? doesn't he have to integrate the function back with respect to x? Kinda new to integration so could use some help on a few fundamental properties
@unnamed7401
4 жыл бұрын
I may realize later that I’m just being dumb but in performing the first substitution, didn’t you get the same exact integral (denominator and limits) but with a different numerator? So the 1 in the integral’s original numerator may as well have been x^2?
@gongasvf
6 жыл бұрын
How do you even come up with this?... Amazing! :D
@guillaumedeplus7727
6 жыл бұрын
What about some generalisation, the integral from 0 to t of 1/(x^n + 1) dx ? :)
@dugong369
3 жыл бұрын
Dr Payam has a good video that shows the integral from 0 to infinity is pi/n/sin(pi/n)
Пікірлер: 492