What a crazy approach! Integral 1/(x^4+1) from 0 to infinity [ Papa Flammy's V2 ]

[@Riiisuu]

25 different substitutions > complex analysis

[@MrZauberwuerfel]

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]

experience...

[@defunct1373]

@@mundocanibaloficil well the first thing i thought after seeing x^4 + 1 was (x^2 + 1)^2, which has a 2x^2 term

[@nastarkos5758]

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]

@@nastarkos5758 hmm ok

[@nastarkos5758]

​@@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]

Wow! Great Job. You should title your video "Easy Integrals Made Difficult".

[@ale4462]

How do you solve it easier?

[@OleJoe]

@@ale4462 Contour integration using complex variables.

[@taterpun6211]

Or use beta function

[@Seek64]

"This is just as counter-intuitive as everything else" 😂

[@blackpenredpen]

The good old days! Btw, that was blackpenblackpen tho....

[@rajibsarmah6744]

Please make a video on integrating complex function

[@rajibsarmah6744]

Please

[@JaskaranSingh-dz2wt]

Hii BPRP an easy challenge for u Solve for x : cosine (2x)=i

[@typo691]

wow why didn't you become a doctor? you would've been a master at prescribing medicine with these skills 9:29

[@Euquila]

That was a fun integral. It's amazing how all these steps seem both logical and magical at the same time!

[@benthayermath]

Papa flammy, I need some fatherly advice. Teach me how to win the ladies with sexy integrals

[@debajyotisg]

I am a simple man. I see 1/(x^4 + 1), I use complex analysis.

[@gamma_dablam]

I use partial fraction decomposition and solve simultaneous equations in 4 variables because why the hell not ! ?

[@mustafaunal1834]

Residue Theorem

[@budtastic1224]

I use: int from (0,∞) of 1/(x^n +1) = (π/n)/sin(π/n)

[@ajfalo-fi3721]

@@budtastic1224 Could this be solved using Beta and Gamma functions? I think I remember it could

[@budtastic1224]

@@ajfalo-fi3721 sure can! Very nice proof here.... kzbin.info/www/bejne/rqPSaJZjoqmeZ8k

[@ianprado1488]

0 to infinity real quick

[@tensorproduct3666]

you roasted that integral.

[@anuvette]

😂😂😂

[@davidberkowitz7697]

Been a math teacher for twenty years. Love this proof, love the enthusiasm. You are a great lecturer.

[@mathematicalhuman6858]

0:24 *_„And if you’re a smart ball...“_* this gave me tears of joy

[@RandomDays906]

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).

[@kartikchoubisa]

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]

Except that u = (x - 1/x) leads to a really gnarly expression for dx in terms of u and du.

[@allaincumming6313]

Is that the algebraic twins method?

[@You12783]

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.-]

@@You12783 Be careful, your formula only works for even n

[@Paulovrish7334]

I freaking enjoy watching you solve this type of problems

[@PapaFlammy69]

[@debajyotisg]

Can I just say, I never thought watching someone solve integrals would make such a satisfying video series? Mad respect!

[@sethuramanjambunathan]

Fantasic! I knew to solve this on complex plane using residue theorem. This is fantastic. My appreciation.

[@mlliarm]

That's a lot of creativity for little over 15 mins of an explanation. I'm amazed :) Keep up the good work !

[@chessandmathguy]

Amazing. The whole thing. Every step made sense, and the final answer is just beautiful. Thanks for posting this!

[@OhDannyBoy512]

I would never have thought to do most of these steps, but you make it seem so logical and straightforward :D

[@Henrikko123]

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]

Flammable Maths I know ;) I’ve watched a lot of them. Great work!

[@michaelgeinopolos6911]

This is absolutely ridiculous.

[@olimatthews5636]

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]

„Woop woop“ thats the sound that the police makes 😂

[@ozzyfromspace]

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]

[@dragonite7780]

Jesus, what a madman, subbed.

[@carlosgiovanardi8197]

EXCELLENT - NO WORDS - GREAT TEACHER

[@karthiksatyanarayana3559]

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]

this method is faster than what he said

[@anegativecoconut4940]

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]

Yes there are many ways to solve this

[@michelkhoury1470]

But not for integral of 1/(1+x^50) but I solved it 😀

[@anegativecoconut4940]

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]

Thomas Torrone yes of course it's very easy

[@michelkhoury1470]

Thomas Torrone okay to evaluate the indefinite integral you can transform the integral into an incomplete beta function by the substitution t=x^n

[@ronbannon]

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]

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 KZbin: kzbin.info/www/bejne/pWSngIeJaNufgqM

[@VivekYadav-ds8oz]

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).

[@ernestschoenmakers8181]

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]

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]

@@Hexanitrobenzene I don't see what's boring about it, this solution just came up to me.

[@Hexanitrobenzene]

@@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]

This integral is easier than people think. It can be "factored" into two quadratic polynomials, do partial fractions, and so on

[@SiwakSerg]

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)

[@46pi26]

The intro=the birth of a goddamn legend

[@u.v.s.5583]

To "Good morning, fellow mathematicians!" you should record a chorus singing "Good morning Papa!"

[@46pi26]

U.V. S. Like the blackpenredpen "yay!"

[@krithikaa4188]

Wow mind blown!!! I never thought of solving like this. Thank you 🙏🏼

[@priyanshusingh3451]

This is really a crazy approach

[@richardheiville937]

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))

[@jaredwhite4934]

I think I like watching cute guys solve sexy integrals even more than those "hot guys and baby animals" videos.

[@richardturietta9455]

nice! I remember doing all sorts of math gymnastics back in college...

[@henrykwieniawski7233]

Have my babies Papa Flammy

[@nikhilnagaria2672]

0:01 the best intro of Papa flammy ever

[@danielcardenas2607]

I really enjoy your videos, you're so good doing this tricks and solve that crazy problems

[@AlEx-ro7kd]

i dont understand much of this but your videos are entertaining

[@CornishMiner]

Whoa! Retro Flammy.

[@duckymomo7935]

That was quite the adventure for that integral The indefinite integral is scarier

[@duckymomo7935]

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))

[@孙林可]

Wow, what can I say? I guess if Feynman had watched this he would open IMU's locker and award you with Fields Medal.

[@joshuawood1082]

Thank you Papa, very cool!

[@anderrafaellinaresrojas3772]

Nobody: Flammable: :D

[@brayanrojas7754]

:v?

[@TheSam1902]

This is exactly why I watch this channel !!

[@charliepilgrim1065]

Woop Woop! That's the sound the police makes

[@TheNachoesuncapo]

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]

Nacho He lifts heavy infiniti bois for about 3 hours a day.

[@TheNachoesuncapo]

No pi-in no gain

[@Bjowolf2]

Chalk gymnastics, blackboard lifting, extreme integration without limits, bending the rules... 😂

[@visheshmangla2650]

this sucker is a must for board exams(class 12) in India.

[@ИванРодькин-ю7ч]

Now, try this: integral from -inf to +inf ln(x)/(x^2+1) Pretty nice answer

[@duckymomo7935]

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

[@Mr_Academic98]

Great inspiration

[@PapaFlammy69]

[@46pi26]

1:08 Bprp=Bruce Lee Makes sense I guess

[@dhoyt902]

This video is a year old but I still clapped.

[@chaos4785]

Math is fun!!! You make things look a lot easier than they really are😂 I always enjoy watching your videos👍

[@zeroleader]

Whoo! Excellent work, dude

[@lesnyk255]

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)

[@arjunv7597]

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.

[@slavinojunepri7648]

Impressive solution from an alternate universe

[@46pi26]

Gotta chow down on some infinity bois before work for protein

[@charllsquarra1677]

badass way to integrate

[@moslemasultana9388]

balance the denominator by 1/2[(x^2+1)-(x^2-1)], then separate and divide up and downside by x^2.

[@DerekBamonte]

After just having finished calculus 2 in the US, that shit was wild.

[@DarioCasarotti-qv7vv]

I love Contour integration and Residue theorem. Much faster

[@Zzznmop]

I came back to this video, I realized what zero is. I am smart boy like papa says

[@seegeeaye]

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

[@cytyy8531]

we just had an exam today with this kind of integrand.

[@williamtachyon2630]

Why could he wrote 1 + x^2 in the numerator (in about 4:55) ? Where this +1 came from?

[@MarioFanGamer659]

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).

[@paparapiropip87]

Man, you’re the God of integrals

[@atrimandal4324]

It was kinda doable till around 10:15,after that it became typical Flammable 🔥🔥😂 Btw, that A4 paper stuff in your website was great ❤️

[@bensolomons4299]

Now we can use some sexy little symmetry tricks which will turn this into something really spicy

[@אדוארדששון-ע8ס]

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.

[@twakilon]

Could have gone for complex analysis, but yeah, this works too.

[@shubhamaryan8202]

Really studying with you maths has become easier

[@si48690]

excellent video

[@derpsalot6853]

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

[@pituitlechat3807]

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!

[@MathIguess]

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

[@EAtheatreguy]

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.

[@__shubham__]

*dx is Nuthin' but a G Thang* . If you forget to put a +C ( in indefinite boi) then the gangster's gonna rude.

[@JohnSmith-iu3fc]

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.

[@pritishmoharir3830]

Amazing stuff

[@JohnSmith-iu3fc]

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?

[@allowfire14]

Very nice! A simpler approach would of course be to write 1/(x^4 + 1) as (i/2)/(x^2 +i) - (i/2)/(x^2 - i), which can be integrated directly. 1/2 * [ sqrt(i)atan(x/sqrt(i)) + sqrt(-i)*atan(x/sqrt(-i)) ] | 0 to inf = pi/4 * (sqrt(i) + sqrt(-i) ) = pi/2*cos(pi/4) = sqrt(2)*pi/4

[@tonyng0714]

How did pi/4 * [sqrt(i) + sqrt(-i)] become pi/2 * cos(pi/4) ??? Stuck in this part

[@allowfire14]

@@tonyng0714 well i is nothing but exp(i*pi/2), or 90 degree counterclockwise rotation in the complex plane. Sqrt(i) = i^(1/2) = exp(i*pi/4). And likewise sqrt(-i)=exp(-i*pi/4). Now by euler's identity cos(z) = 1/2 * (exp(z) + exp(-z)) ; in this case replace the argument z with pi/4. I think the rest should be straight forward

[@dnranjit]

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 ?

[@giannicolombo3993]

Amazing! Thanks a lot.

[@giannicolombo3993]

@@PapaFlammy69, where did you study Mathematics? Have you some suggestions about books?

[@ashuthoshbharadwaj6703]

Hear me out here, @ 10:16, the first integral is super easy to solve because the addition of the two expressions in the denominator is twice the numerator and after that a bit of algebraic manipulation should give us an sec(a)*arctangent((x+sin(a))/cos(a)) and the other integral also becomes sec(a)*arctangent((x-sin(a)/cos(a)) and now you just have to plug in the values of the lower and upper bounds and multiply by one-eighth ( because we stated that sum of expressions in Denominator is 2*(numerator)). I just think that adding the odd fuction integral is absolutely ingenious :) but unnecessary :/

[@guillaumedeplus7727]

What about some generalisation, the integral from 0 to t of 1/(x^n + 1) dx ? :)

[@dugong369]

Dr Payam has a good video that shows the integral from 0 to infinity is pi/n/sin(pi/n)

[@neo_tsz]

Huh...my mind is satisfied.

[@thexabi123]

What a lovely "2" at 13:27

[@kelvinella]

Let t=1/(x^4+1), play around with it a little you should get 1/4*Beta(3/4,1/4) = 1/4*Gamma(3/4)Gamma(1/4) = 1/4*pi/sin(pi/4) = sqrt2*pi/4

[@ryankoepke5525]

You could also factor and use partial fractions to solve that.

[@zhangray663]

I think this can be solved via residue theorem which won’t require so many tricks.

[@Czeckie]

plug the the new information in, fellas!

[@junya4284]

Note: My first language isn't English. 12:45 Perhaps, 1/4cos(a) is mistake. I believe correct is 1/4cos^2(a) so Answer is π/2 maybe. But I know you are a professional of course. I'm very confused... please help me!

[@junya4284]

Finally I'v noticed my answer, which is not anything, is mistake. It's so simple and seem often. Differential of tan^-1(~) has 1/cos(a) in denominator.