It "Cannot" Be Done (Integrals)

Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

[@BriTheMathGuy]

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[@Jj-gi1sg]

Me an intellectual: "Oh its obviously e^x²/2x "

[@Yutaro-Yoshii]

Lol same, it's so frustrating that there's that one pesky term that you can't get rid of. tried throwing complex numbers to the mix, but got messier well, good old power series to the rescue I guess

[@ignantxxxninja]

Yea that would be nice if that worked like that wouldn’t it. Whoever asks for the integral for e^x^2 is an asshole

[@manojsurya1005]

I also literally thought the same

[@theimmux3034]

Don't forget +C

[@dominicstewart-guido7598]

+C

[@tomasstana5423]

A long time ago, with my friend at high school we once felt bored and unchallenged with the integrals we were computing for homework, so we decided to pick one integral we already finished and make it more difficult. The one we picked was x*e^x^2 and we dropped the x. Two hours later we gave up and admitted defeat ...

[@عَدِيُّ-م3ح]

lel, i feel you.

[@wabc2336]

I did integrals on my own in high school sometimes... never knew about non elementary functions at the time

[@IfeelFearForTheVeryLastTime]

Welp ... i know ur feeling because I've been in the exact same situation And my integral was e^x/x I've been trying for a month to solve it and i couldn't until I knew it was impossible

[@redbaron07]

Same here, and my math teacher did not tell me that it was impossible. Perhaps he was hoping I was Euler and would somehow solve it.

[@rengokukyojuro1384]

I don't see the problem to this integral. If its x*e^x^2 Do you mean x*e^(x^2) or x*(e^x) ^2. Both ways its solvable. For first one use substitution x square as t. And for second one simply use by parts.

[@Dongerd]

I just saw that the integral had the word “sex” and clicked

[@funkygawy]

reminds me of what I learned in calculus, the integral of e^x = f(u^n)

[@einsteingonzalez4336]

It's so hard because I treat the "s" differently in different forms.

[@6754bettkitty]

Noice!

[@alistairmackintosh9412]

Sex with dx.

[@donovanb8555]

@@funkygawy you killed me 🤣🤣

[@ClumpypooCP]

I like these funky integrals... do more of them

[@mohammadfahrurrozy8082]

Yeah me too!

[@BriTheMathGuy]

I'll see what I can do!

[@IngTomT]

I would keep the factorials, I think it looks nicer in the form { Sum(n=0,oo): x^(2n+1) / (2n+1)n! } + c

[@mcsyllesen5183]

same

[@hardchemist]

That was great and funny as hell at the same time lol

[@BriTheMathGuy]

Glad you thought so! Have a great day.

[@EebstertheGreat]

A more precise way of stating this idea is that exp(x²) has no _elementary_ antiderivative. Roughly speaking, elementary functions on the complex numbers consist of the rational functions and finitely many extensions by exponential and logarithmic functions. On the real line, restrictions of these are also usually included, such as the trigonometric functions (as well as their inverses). That is, a real-valued function of real numbers is elementary iff it is a composition of finitely many functions in the set {+, ×, exp, log, sin, arcsin} and constant and projection functions of the real numbers. An even more precise statement is that the function f satisfying f(x) = exp(x²) has no primitive in any elementary extension to the differential field of rational functions (i.e. not in any differential field which can be obtained by a finite chain of logarithmic, exponential, or algebraic extensions starting with the rational functions), which can be checked by applying Liouville's Theorem and solving the resulting differential equation.

[@EebstertheGreat]

Also, you mentioned the imaginary error function erfi. The exact definition is erfi(z) = -i erf(iz) = 2/√π ∫ exp(t²) dt, where the integral is taken from 0 to z. In other words, it is a scaled version of the antiderivative you supplied when c = 0, continued to the entire complex plane. (I find the 2 in the numerator irritating. I don't really know why it's there. I guess it's supposed to be easier to construct confidence intervals with erf this way.)

[@lamemelord]

of COURSE π is here

[@InzaneFlippers]

i dont understand how you have so few views

[@BriTheMathGuy]

Mystery of the universe :)

[@lial2410]

Who else is here bc of German highschool finals?

[@Zeebzz]

I really thought it was (e^x)^2 and I was like why did you make it so difficult, then realized it was e^(x^2)

[@henrytang2203]

I like how they had to invent a completely new function to describe this integral.

[@nickronca1562]

His videos become a lot more impressive when you realize ... He has to write everything backwards.

[@altarius44]

Oh, actually he doesn't :D (most likely) in this video someone describes how it's done: kzbin.info/www/bejne/m4eygXeHarCMqdE basically, he writes on a glass frame and mirrors the video. That's why most people who use this technique appear to be left handed. You could check older videos of Brian where he writes on a regular paper, you'll see he's actually right handed (:

[@SunroseStudios]

wow so we're seeing his mirror image?

[@altarius44]

@@SunroseStudios yep

[@badradish2116]

@@SunroseStudios thats why he uses his "left" hand

[@MrCigarro50]

Thank you. For us statisticians this is a very important function.

[@BriTheMathGuy]

Absolutely!

[@lukedavis6711]

Do you know the name of this function?

[@Exachad]

I thought it was e^(-x^2) not e^x^2

[@lucafurlan6224]

@@lukedavis6711 Taylor function

[@rosario6217]

@@Exachad that's the error function. In the video, the one he talks about is the *imaginary* error function, which is the integral of e^x^2 multiplied by π^(-1/2) (without the constant of integration of course)

[@YoutubeUser-yl9ys]

Y’all didn’t pay attention how much he would struggle to write in reverse in front of him so we see directly.

[@squidboi9001]

Y'know he can just flip the video 😐

[@justinberdell7517]

Fascinating! I never saw it explained like that. And in 3 minutes no less! Lol I have however used the error function a lot. I'm an electrical engineer and in undergrad I took a class in thin film semiconductor fabrication and the error function is used to calculate dopant or impurity concentration in constant source diffusion. Really cool stuff but I went into ai and robotics so I never use it now

[@ezequielgerstelbodoha9492]

I just found so anti-intuitive that the integral of xe^(x^2) is fairly simple to solve, that e^x is one of the easiest ones, but e^(x^2) doesn't have an elementary answer. I was breaking my head around a couple substitutions when I searched for this

[@fatalvampire]

I have a dumb question. When you're writing on that transparent blackboard, are you writing normally? You don't have to flip your writing for us to see it correctly?

[@leoallentoff]

He probably writes on it normally then just flips the footage

[@ignantxxxninja]

Yea so we’re seeing him like how he see himself in the mirror

[@brian8507]

No he writes backwards

[@leoallentoff]

@@brian8507 he’s just built different

[@mysticdragonex815]

Better way will be to use the Gamma function I think

[@hernandofamily3118]

ohh god that integral...one of my struggles in college...wahahaha..great content

[@jaskaransingh7025]

Well if u integrate this u will get the integral of erfi. And I integrated to be as so that the integral is equal to (sqrt(pi) erfi(x))/2 +c 😊

[@charbel6677]

do you write in reverse and then flip your video horizontally?

[@colt4667]

I think he has a bad case of dyslexia. Poor guy.

[@lht001300000]

Euler: It is trivial that 2pi Exp[z^2+y^2] dzdy = 1/2 Exp[r^2] dr^2 dtheta in polar coordinate, 2pi cancels dtheta, so the integral solves to Exp[x^2]/2x + C

[@GammaFZ]

lol I thought you were kidding when you said that the video is over, it actually was true smh you are becoming an engineer

[@Gunslinger-us1ek]

so we can use this technique for the gaussian integral?

[@huwdte]

This could be a good integral for like a calc 2 final, refreshing on both series and integrals Actually nevermind it's pretty abstract. Still gonna leave this up because it could be good for something else

[@bringonthevelocirapture]

Why not write it in terms of n is the sum? Also, aren't you missing a multiplicative term for the antiderivative? And, why use x^2 and not x^-2 if you were going to mention erf(x)? MADDNESS

[@quantumphilosopher1729]

Can’t we use Feynman’s technique of differentiation under the integration sign?

[@funnydubbingclips175]

I did that before u

[@maximilianofloresguillen7265]

You should cover the gamma function in a future video!

[@BriTheMathGuy]

Nice idea - I'll see what I can do!

[@justabunga1]

Even the answer is non-elementary, some people considered the answer to be sqrt(pi)/2 erfi(x)+C. erfi(x) stands for the imaginary error function.

[@adb012]

Why didn't you write e^(x^2) as sum |i=0~inf| x^2i/i!, then move the sum out of the integral, so you are left with sum |i=0~inf| {x^(2i+1)/[(2i+1)*i!]} + C?

[@fadihalaweh8018]

it was in a exam of mine , you scared me Bro with the Click bait solving = ln the eX2 and it will be ez

[@jrcarlyon680]

Your videos were better when you didn't wave your hands around all the time

[@khanhtran-gw3pm]

Obviously it’s 2/sqrt(pi) * erfi(x) +c 😒

[@nik_semperlotti1062]

√π erfi(x)/2 + c

[@Omar_MTH]

Ok now where is the 2 in differentiating x^2

[@Diego-qt3xy]

Me an intellectual: hehe that integral looks like sex hehe

[@MICR0N.official]

Can we replace e^x^2 by e^ln(e^x^2)? Integrating this we can realize that's special function called Gaussian Error function.

[@einsteingonzalez4336]

They actually mean that it's impossible to do it in a finite expression. :) That's an infinite expression, but it can't be simplified to a better, more elegant, and more concise expression. Knowing this fact could help prove that there is no such closed form for the sum of the reciprocals of cubes. Yes, I mean zeta(3), zeta(5), and all zeta(2n+1), where zeta(x) is the Riemann zeta function.

[@wydadiyoun]

but why can't we solve it???

[@polizario7942]

srry but this is a wast of time xD

[@maythesciencebewithyou]

When I applied for college to study biotech, I was interviewed by an IT professor and he asked me to integrate this function. Somehow I still managed to get it, but barely.

[@UdayadityaSankarDas]

I encountered this function while doing differential equations and it made me lose a question in exam. I've been both fascinated and scared of this one since then.

[@maximilianofloresguillen7265]

I think it's the incomplete error function

[@BriTheMathGuy]

You might be right!

[@ScandGeek]

The error function (incomplete or otherwise) deals with the integral of e^(-x^2). The imaginary error function equals -i times the error function of an imaginary value. So erfi(x) = -i erf(ix)

[@VincentGPT-lol]

As I was trying to make sense of the above equation, it dawned on me that he actually wrote it backwards ........ crazy 😮

[@madhavsoni2144]

Can't we just apply integration by parts Take first function to be e^x^2 and 1 as second function

[@huxleyleigh4856]

Just express that infinite sum with big sigma notation. Doesn't help much mathematically but makes the solution more satisfying

[@KishoreRana567]

What Bullshit.. The problem with these online Maths Gurus.. They make necessary out of unnecessary

[@tastypie2276]

At least it has a solution. Thanks for the video!

[@BriTheMathGuy]

You bet!

[@l.s.d.154]

This power serie of e^x is only for x approaching zero. So this primitive is only an approximation when x is close to zero isn't it ?

[@NoMan-hf6oq]

Integral ( e^x^2 dx) ( Let, u=e^x du= e^x dx du/u= x) = Integral( u^2 du/u) = Integral( u du) = u^3/3 + c = e^3x /3 +c Is it right?

[@claymusic2205]

Sir, we're going to have to exclude you from the STEM community since you don't meet the appearance requirements. You're too cute to be both smart and cute.

[@TheOnlyBootlegger]

holy interlacing

[@squashgoogolplex9392]

woah did you teach yourself to write backwards to do this? cause it looks like that

[@energyeve2152]

Lol. Simple I guess haha Thanks for sharing

[@MathZoneKH]

It’s good knowledge for me today

[@gumbyballs9312]

ah yes, the sex 2: dx

[@use-zakar]

But this expansion is only valid near zero. For higher values of x this solution well not be correct.

[@lukedavis6711]

Explain

[@use-zakar]

@@lukedavis6711 this expansion is convergent only between - 1 and 1 elsewhere it diverges (doesnt give any sesible value).

[@Exachad]

This is not true. The more terms of the series you have, the more accurate the representation is after 0. If you want, you can Google the e^x power series on Desmos and see that the 7th degree estimate is accurate for much more than from -1 to 1. The 100th degree estimate is accurate for a very long distance. The series representation of e^x converges to e^x for the whole distance as the number of terms approaches infinity.

[@matrefeytontias]

The expansion he used in the video is well-defined for every real number (every complex number actually). It's true that usually Taylor series expansions only hold in a neighbourhood of a single point, but for the exponential function it happens that the Taylor series expansion around 0 also holds everywhere. This is because the exponential function admits a power series decomposition.

[@lukedavis6711]

@@matrefeytontias thank you!

[@alexplastow9496]

Thanks for being casually excited about mathematics

[@drhubblebubble7]

Me, a physicist: ah yes. e^(x^2) = 1/(1-x), so Solution = -ln(1-x)

[@josephcohen734]

Was like "wow that looks so hard how will he do it?" - "there is no closed form answer" BOOM! Instantly understood what his solve was gonna be.

[@samegawa_sharkskin]

it's imaginary error function erfi(x)

[@graemehunter4395]

I thought in my head: "it must just be (e^(x^2+1))/(x^2+1) +c Then I noticed the dx.

[@violintegral]

If only it were de instead of dx lol

[@bryanbartlett5637]

simplifies the entire thing to i* x * gamma(1/2,-x^2)/(2*x)

[@maxhenderson1890]

Can’t you let y=e^(x^2) then dy/dx=2xe^(x^2) then say that the integral = (1/2x)e^(x^2) ?

[@user-wx8bm1pg1d]

That's not how it works

[@gytzgytz7262]

It could be done easily if x is different from 0

[@halchen1439]

If you think about it its actually not that surprising that you "only" get a series like that. Why? Because the exponential function itself is defined like such a series and each value is just the limit of the series for what you put in

[@isilder]

..no . .because almost everything can be a power series, eg trif functions, yrt they have nicer integrals . Being able to define as power series is not a good inducator to difficulty integrating.

[@halchen1439]

@@isilder mmmmhh good boy

[@yadavvikas8957]

It can be easily solved by taking e^x =t

[@amaanabbasi9443]

√lnx is it possible to intrigate it ??

[@ambativenkatesh7170]

Can you tell me solution for integral o to π/2 dx/sin^10x+cos^10x dx

[@ambativenkatesh7170]

Question is integral 0 to π/2dx/sin^10x+cos^10x

[@user-wx8bm1pg1d]

You can just write cos^10x and sin^10x in terms of cos(nx) and sin(nx) and integrate. Though it would be tedious

[@alperenerol1852]

You won't be wrong if you say there is pi hidden somewhere in the sum.

[@lukedavis6711]

Sqrt(pi)😏

[@Alnakera]

Use polar coordinates

[@mathunt1130]

erf function...

[@zouzouleloup675]

Thats a Taylor Serie

[@OptimusPhillip]

This is the basis for the Gauss error function, but it's not complete. The exponent of the integrand should be negative, and the integral should be multiplied by 2/sqrt(pi)

[@maythesciencebewithyou]

The Gauss error function is just this function normalized, so that the area under the curve becomes 1.

[@modolief]

Cool!

[@TLohr]

Wow this is just like what was on my AP calc ab test. It was something like integral from -1 to 1 of e^x^2. Wish I would’ve seen this 8 hours ago

[@CrisDFF30917]

Simpson :/

[@mathswithatifaslam9741]

Nice

[@BriTheMathGuy]

Thanks! Have a great day!

[@BigMarser]

this looks to me more as the integral of x*(e^x) ...

[@ddamuliraali4621]

your the best

[@BriTheMathGuy]

You are!

[@kqp1998gyy]

You are great

[@BriTheMathGuy]

You are!

[@iamtrash288]

Nice

[@BriTheMathGuy]

Thanks!

[@CTJ2619]

Why not use a substitution

[@axelnils]

Try it!

[@HDitzzDH]

It's not gonna work.

[@criskity]

It would work if it was xe^x^2, but not plain e^x^2.

[@herbcruz4697]

If you substituted for the x^2 (i.e., let u=x^2), then du=2x*dx. While we differ by a constant factor of 2, that isn't the issue, here. We also differ by a factor of x that we don't have inside of our integral, and we can't pull variables outside of integrals, so doing a U-Substitution is out of the question.

[@nithin1729s]

Thanks sir

[@BriTheMathGuy]

All the best

[@wowZhenek]

I'm pretty sure that the "I just replace" at 01:25 is an incorrect way of doing it, even though, ironically, it yields the correct answer.

[@benschmitt7035]

If you use the guassian formula that Integral e^-x^2 dx = root pi, you get that this is equal to root pi over i

[@user-om3ej7rl3e]

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[@Code4You1]

Substitution 2 does the trick, I dont know what you're talking about