Integral x^x from 0 to 1
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- Опубликовано: 11 окт 2024
- In this video, I evaluate the integrals of x^x and x^(-x) from 0 to 1. Although there is no explicit formula for this integral, I will still evaluate it as a series, and the answer is very pretty! Enjoy!
Nice. I feel like that putting the final result in the form of
int_0^1 1/x^x = sum_{n=1}^{infty} 1/n^n
would be even more beautiful.
That’s amazing, thank you!!!!
This integral is a non-elementary, since the function x^x is a non-integrable function.
People who are so happy and passionate when doing mathematics are impossible not to like. I am glad YT recommended your channel!
Sometimes i get tired of maths, but then i watch your videos and feel so motivated again. The way you present maths is just great and so much fun to watch! Keep up the great work, and thank you for making such insightful videos available for everybody.
Thanks so much!!! ❤️ I really appreciate it :)
Functions of the type f(x)^g(x) are the most mysterious, who knows how many hidden properties there are, great video love to see more like this :D
Just watch this impressive Math channel... also you can find the generalization of integral x^x ... ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
Presumably assuming f and g aren't constant.
@@xinpingdonohoe3978 Calling them functions of (x) heavily implies they are not constant.
@@dan-us6nk no it doesn't.
the pure mathematician vs the applied
One benefit of the series you've derived, is that both of them converge *very* rapidly. Just the first 4 terms get you almost 4 places' accuracy; 10 terms get you 11 places.
Fred
.
ffggddss do you do one of the convergence tests to see if the series is in fact converging?
I hope to put a spell on my previous girlfriend by sending her this.
This is called sending an x^x hex to the ex.
lol
I didn't know what to make of you at first, Dr. Peyam. I finally conclude that you simply find joy in mathematics.
Interchanging integration and sommation is easily justified using Lebesgue's convergence theorem, even when the limit of the integral is infinite!
"I'm feeling like a bad boy today " omg almost spit my lungs
the enthusiasm you have even after that much time of solving a ridiculously demanding problem is commendable . you were joyed up like a child after getting the solution .
I swear this is my favorite channel of all RUclips!!! You have the most amazing and interesting content and your videos are always funny and entertaining. Please keep doing such an amazing job Dr. Peyam!!
Mauro Castañeda His content is so interesting as it's already kinda deep into math and the problems aren't too easy what the video lengths confirms. I'd advice you studying mathematics, then it will get even more interesting
I'm actually studying physics, so I'm no stranger to some of the things doctor Peyam teaches, but still I know I'm not going to learn everything that he knows hahahahaha I just love maths
You can actually combine these results to get more integrals:
The sum of all the odd terms is equal to the integral from 0 to 1 of cosh(xln(x)) and the even terms yield the Integral from 0 to 1 of sinh(-xln(x)). There are probably many more interesting identities hidden in there.
When I saw the integral I immediately knew that there was some kind of Gamma-function involved lol.
I played a little bit around with this trick and I got a really fancy result!
Take a look at the integral from 0 to 1 of x dx. This integral isn't that hard to solve and we get the answer 1/2.
Let's try to solve this integral in the same way as the in integral in the video. You can write x as e^(lnx), so you can write x as sum_(n=1)^(inf) (lnx)^n/n!.
Now we switch the summation with the integral and deal with the integral from 0 to 1 of (lnx)^n dx.
Let u = -lnx, then dx = -e^(-u), u(0) = inf and u(1) = 0
So we get the integral from inf to 0 of (-u)^n *(-e^(-u))
After simplifying we get (-1)^n * integral from 0 to inf u^n * e^(-u) du = (-1)^n n!
So the orginial integral is equal to sum_(n=0)^(inf) (-1)^n/n! * n! = sum_(n=0)^(inf) (-1)^n = 1-1+1-1...
Comparing it with our first answer leads to this interesting result:
1-1+1-1+1-1.... = 1/2
*Ramanujan shakes in the tomb*
Of course the resulting equation is incorrect, as 1-1+1-1+1-1 is divergent and does not converge to an answer. The reason why you found this answer is falling into the fallacy that Dr Peyam mentioned at 3:00, that you cannot interchange summation and integral if the summation is not convergent (in Dr Peyam's case, it was convergent; however, in yours, it is not). An interesting result nonetheless, though.
@@berkealtparmak1699 if we call 1-1+1-1...=A
1-A =1-(1-1+1-1...)
1-A=1-1+1-1+1...=A
1=2A
1/2=A
Did anyone notice?
"Never put your X's together..."
Nice pun :))))
u v wher 2 egs can i?
At first, I thought that this absolute madman, I mean madboi, has found an elementary antiderivative for x^x. And then my head exploded as the gamma-function showed up. Now you're guilty of murder xD.And by the way: Thanks for giving my challenge a try. I didn't get a notification either as you replied to my message.
Just watch this impressive Math channel... also you can find the generalization of integral x^x ... ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
“If you’re confused, just substituted”.
Love the dude😂
What I find interesting about this, is bounds approximation formulas.
sum of 1/(n^n) from n=1 to infinity is equal to area of columns width 1 and height 1/(n^n) arranged along x axis.
then curve 1/(x+1)^(x+1) where x in [0, infinity] is whole inside that columns area -> area under curve is less than area of columns
and curve 1/(x^x) where x in [1,infinity] is whole above that area -> area under curve is bigger than area of columns except first column
I'm excluding [0,1] for 1/(x^x) because I'm unsure with its behavior on this interval.
So, we know how to calculate area under curve. It's yet again another integral, but wait!
It's integral of 1/(x+1)^(x+1) from 0 to infinity in first case, which is same as integral of x^(-x) from 1 to infinity.
And, second bound, is integral of 1/(x^x) from 1 to infinity which is same as integral of x^(-x) from 1 to infinity.
Summary: integral of x^(-x) [1, infinity] < integral of x^(-x) [0, 1] < 1 + integral of x^(-x) [1, infinity]
(extra one is made by first column which is 1/(1^1))
And that is *surprising* . Maybe I'm just too dumb, maybe something similar holds for 1/x for example? :\
OHH you forgot the dx in the 5th equation! (2:28) (or did not you?)
BTW NOICE video! Really like this integral too ^^
Next video try: Integral from 0 to 1 of x^x^x dx
dx shmee-x, haha :) That’s a great idea, I’ll try it out, although it probably won’t be as nice as x^x. I don’t think it’ll be 1/(n^n^n)
That would be nice, otherwise I will wait for your next video :-)
If you are still interested in this integral:
As Dr peyam said, the result won't be as nice as x^x, but I tried anyway. I was finally able to express it with a double sum and a limit.
The limit is needed because there are otherwise convergence problems. I think it's because the sum_{n = 0}^{m} (x^x*lnx)^n/n! blows up at 0 for any finite m (see Desmos: www.desmos.com/calculator/yc1yqbilt9)
So I tried to evaluate the integral from epsilon to 1 of x^x^x dx and then take the limit as epsilon approaches 0+.
The trick is basically the same as in the video, write the function as a series, in this case you have to use it twice. (x^x^x -> sum_{n = 0}^{infinity} (x^(nx)*(lnx)^n)/n! -> sum_{n = 0}^{infinity} (sum_{k = 0 }^{infinity} (n * x * lnx)^k/k! * (lnx)^n)/n!)
After using the trick twice and integrating and simplifying, I ended up with this expression, which also involves the Beta-function :D :
imgur.com/a/YZWkeK3
You can also take a look on it on Desmos:
www.desmos.com/calculator/pdrhazw8yq
@@maxsch.6555 De fuck??? This is awesome!
One of the most elegant integrals using power series there is!
P.S. I start reading Princeton Lectures in Analysis, and I understand why you recommended it, this is great!
Glad you’re enjoying it :)
I watched lots of Payem videos without noticing that these videos are not some old learning videos, but actually made in present (of my own lifetime). It kind of makes me happy :-)
Nice proof of sophomore's dream. I would love to see you do a proof on the convergence of infinite exponentials!
Thank you! Hahaha, it is truly a sophomore’s dream 😂
What exactly do you mean by infinite exponentials?
For those who don't get it en.wikipedia.org/wiki/Sophomore%27s_dream
Dr. Peyam's Show in the same way you can have an infinite sum, you can have an infinite exponential i.e. a^a^a^... for some number a. At first it may seem like this would be divergent but it actually converges for e^(-e) < a < e^(1/e)
@112BALAGE112:
Thank you for the link. First time I have heard this term.
Hassan Ali Husseini me too. kinda cool. 'too good to be true', haha
But, are you a bad boy, or a bad boi?
Hahaha, a bad boi, obviously 😜
AndDiracisHisProphet 😂
A bad boi!
blackpenredpen m surprised to see your comment..korean boi
krishna bhakta I m not Korean.
Just to answer the questions:
integral 0 to 1 x^x = 0.7834305107121343
integral 0 to 1 x^(-x) = 1.291285997062663
You did engineer math!!! Yuck!!!
do you know if integral 0 to infinity x^-x converges? thx in advance. i put the integral into an online calculator and it computed some number: 1.995455957500093 so i thought maybe 2 is a limit of that (since the function has an asymptote, and the area is getting smaller)
@@Fishinator33
The integral definitely converges, since x^(-x) = e^(-x ln(x)) is less than e^(-x) for x > e, and the integral of e^(-x) from 0 to infinity is 1.
Just watch this impressive Math channel... also you can find the generalization of integral x^x ... ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
« now were gonna do something really nasty I´m feeling like a bad boy today » Dr Peyam hahahaha (2:55)
Hausuua gangstyle
Waoo!
The sum of 1/n^n converge of 1.29128599706.........
These decimals not change after of 1/10^10
Very impressing!
JUST LOVE IT!!! THE ENTHUSIASM YOU PUT INTP THIS IS JUST TOO MUCH FOR ME!!
Ur first video to hit 2 lakh views!! Ur YT channel is growing fast 🔥🔥🔥🔥🔥 keep going :)
Why the defined integral is still unknown, is it because we don’t have enough natural functions that we can express the final result in?
I mean we can work with sin and cos functions freely and interchange between them and with exponents and polynomials because all these functions are natural and can be related to real systems.
Exactly, our language of functions is limited
@@drpeyam you mean that if we let
L(x) = the integral from 0 to x of x^x
then L(x) can be expressed in terms of known function, that still unknown until now?
Well then other functions can be expressed in terms of L(x), just like other integrals can be written in terms of sin and cos
Thank you for making this! I work with its derivative for fun but had trouble with integral
Dr Peyam is a German-speaker! He said that German is his native language under the comments of his most recent video.
I've seen this integral on Wikipedia, where it was called the sophomore's dream
If this was on my test, I would just say that the function is not always well-defined on the given interval, and stop there.
This is the approach I ended up taking as well. I spent some time developing a maclaurin series representation of x^x, and mine certainly doesn't look so neat. The pattern I am getting is interesting though. f(x) = x^x, f'(x) = f(x)ln(x+1), f''(x) = f(x)/x + f'(x)ln(x+1). It seems like I may have messed up somewhere on f'''(x) because I am getting f'''(x) = -f(x)/(x^2) + 2f'(x)ln(x+1)/(x) + f''(x)ln(x+1). I can see the pattern in the increasing power of X on the denominator and the next derivative containing the previous derivatives multiplied by ln(x+1), but I'm not seeing a pattern with the coefficients just yet.
9:38 that feeling when youre solving x^x and you end up with constants of the form (n+1)^(n+1)
oof, I think I need to do alot more maths before I can understand this.
But I will come back to this!
*determined!*
You can write it as: - sum n=0 to inf of (-n)^(-n). I think it's more beautiful.
I like how in step 2 you just went for a change of variable, whenever i see a product of polynomial and ln, i think integration by parts.
Just watch this impressive Math channel... also you can find the generalization of integral x^x ... ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
I love you methods of mathematics and you cheerful way studying and teaching
What is the answer of the final result (1+1/2^2+...........) i think it must converge because its an answer of an erea from 0 to 1 so it must converge ( im right ????? )
It converges, but it cannot be written in terms of numbers that we know
@@drpeyam thank u i get it 💓
thanks man, there is nothing is the whole step that I don't know about.
this integral came in my mind when I was in BSc 1 year( now in the second year), I asked about it, but his answer discouraged me, so I moved on.
what I learn today is, there is nothing we can't do, but what we need is the right person who encourage us, not discourse ...
As a rough check, x^x is between 0 and 1 on the unit interval, so the area is less than one, but greater than zero.
2:57 that made me laugh harder than it should have
I loved watching this video! I try understanding these integrals only as a hobby of mine. Too old for college or start over again unfortunately but since two years ago I began, just for the fun of it, to take a quick internet course of my 'Highschool Mathematics. Until the point I quit school so I didn't get to the calculus-section. Since two years I try to study this. With Internets RUclips and an american Calculus book (Thomas') I have now finally reached the end of calculus 2, I am not going really fast but I sometimes veer of course by studying other kinds of math like the complex-plane, just the surface, just to get the idea. Once again, I like your way of explaining very much and although I didn't go through the gamma-function yet, I understand the implications. My question to you, or for anyone who managed to read this far, is this: what would you recommemd me to study, what courses after I finished calc 2. What would I need to study first i.g. to be able to understand courses like Complex Variables Calculus or what are the usual courses to take? Sincerely thanks for taking the time to read all this and in advance would you be willing to answer these questions. Jurgen from Vianen, Netherlands.
i like the white board its so clean and soft and cuddly!
Me too!!!
It's official!
Dr. Peyam claims to be clean, soft, & cuddly...
Fred
. . exploiting grammatical ambiguities everywhere . .
.
Hi I am in college and RUclips recommended you. I feel blessed to see you sir
So Dr.peyam you have assumed the integrand is uniformly convergent right?
You see integral and summation
Can be interchanged if {fn} is uniformly convergent though the converse is not true.
But I like your approach
To your knowledge, does x^x model anything in nature?
Keep up the good work, I didn’t think of the series expansion.
Here the switch integral sum is indeed justified since we have an uniform convergence of the exponential series.
Interesting how one function gets you a series and the reciprocal of the function gets you the alternating version of the series. It's like e^x.
Very good content! I'm studying to be a math teacher in Brazil and your channel is an inspiration
Just watch this impressive Math channel... also you can find the generalization of integral x^x ... ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
Hello, first of all thanks a lot for your wonderfull video. Your enthousiasm is great ! Hope you'll go on that way.
May I ask a question ? How would you show that it is in fact correct to interchange the sum and the integral sign ? Do you know an exemple where one cannot interchange the sum and the integral ? Might be very interesting to study the question. Thanks in advance
Dominated convergence theorem
Interchanging the summation and integral is not so tricky if you use the uniform convergence of power series in its disk of convergence, together with |x lnx | < 1/e whenever x in (0,1].
Nice. Sorry to be a perfectionist but it would be good to know what the series evaluates to. Now I'm going to have to write a python script to find out....
Great !!! You are The Best Mathematican .. 👍👍👍
Dr. Peyam video did not calculate the end result , a Chinese simple trick gives the alternate series :
= Π²/6 - ½.Π²/6 = Π²/12
S=1+1/2² + 1/3² + 1/4² + ...= π²/6
1-1/2² + 1/3² - 1/4² + ...
= (1+1/2² + 1/3² + ) - 2(1/2² + 1/4² +1/6²+... )
= (1+1/2² + 1/3² + ) - 2*¼ (1+ 1/2² + 1/3²)
= S - ½S
Great video as always!!! I would like to know how can we prove that all the cosines and sines are a "complete" base. It is, there's no other function you need to "hug" (convolute?) to get the equality.
At the beginning you say that all are ortogonal among them therefore any element of the space (any continuous function) is a linear combination of the base. I would say that orthogonality tells you there's only one solution to that equation, only one set of coefficients. But you still need to prove that the linearly independent set is a base to prove there IS a solution.
Maybe I missed another video?
Wow amazing, can you believe I was actually discussing this with a friend the other day? I´ll tell him about your channel. Is there a closed expression for the sum from 1 to n of 1/k^k? (Your last sum but for any n, not just from 1 to infinity). Many thanks Dr. Peyam!
Thanks so much, what a coincidence! :) And Sadly I don’t think there’s a closed form, because even for the whole series there is no closed form expression!
I´ll try to work on that haha. Thanks again.
Dr. Peyam, I hope this message finds you well. I was a bit confused with the interchanging of integral and summation. Integral itself is defined as the sum of all infinitesimal points in the domain so whilst interchanging the two Shouldn't we start the summation from n=1 to infinity rather than n=0 to infinity?
Nice videos as always dr. Peyam. You're a math whiz indeed. I hope you can make videos about ALGEBRAIC TOPOLOGY.
Thank you for the videos Dr. Peyam.
2:57 “I’m feeling like a bad boy today.”- Dr Peyam
Is it proven that the antiderivative of x^x cannot be represented using elementary functions; or is it that we haven't discovered it yet?
Is your thesis about Asymptotic PDE Models for Chemical Reactions and Diffusions? I read that on Internet but I'm not sure.
By the way, beautiful and amazing video :D
Yeah! :)
I think you can exchange the sum and integral because n and the terms are all non-negative.
Oh, I don’t think positivity is sufficient, there has to be some convergence assumption, but I forgot what it is
Keep it up i really liked the way u explain the expressions
Lot of respect from INDIA sir ...
Which means you can take
Int(x^x + x^(-x))dx = Sum((-1/N)^N + ((1/N)^N) = Sum(1/1^1 - 1/2^2 + 1/3^3 - 1/4^4 + ... + 1/1^1 + 1/2^2 + 1/3^3 + 1/4^4 + ...) = Sum(2/1^1 + 2/3^3 + 2/5^5 + ...) = 2*Sum(1/(2N+1)^(2N+1)) from 0 to infinity
Let S = Sum(1/(2N+1)^(2N+1)) from 0 to infinity
Then Int(x^x + x^(-x))dx = 2S
Int(x^(-x)(x^2x + 1))dx = 2S
From this we can use IBP to find the integral of x^2x and then by induction that of x^(kx + b)
I just want to relax before sleeping but how did I scroll down on the video list and end up with being trapped by a giant picture showing strange integral that I can’t solve. And I clicked in for 20min and got excited at 12:00pm.
Exactly the same but it's now 11:50pm here.
Impressing video!
Your explanation was easy to understand this for even highschool student like me!!! thank you!!!
Just watch this impressive Math channel... also you can find the generalization of integral x^x ... ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
Find the following primitive?
The primitive of: arctan(x)*ln(x)?
Ótima resolução, parabéns e obrigado pelo trabalho.
Muito Obrigado!!! 😄
yes...as you said, it is really beautiful.
Just watch this impressive Math channel... also you can find the generalization of integral x^x ... ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
i like this guy teaching
Nice! This is putnam exam question.
Okay. So you rewrote the original integral of x^x as an infinite alternating series. But the question is, does that series converge? The integral from 0 to 1 of x^x should be a number, so intuitively the series should converge right? If so, what does it converge to?
Is switching the integral and sum because Taylor's theorem satisfies uniform convergence or is it because of dominated convergence?
It’s because of dominated convergence
you can still express the indefinite integral, using the incomplete gamma function and a sum
Dr. Peyam, I'm just a chemistry student so this might be a dumb question, but why is pulling out the summation symbol usually not allowed? Isn't integration just linear? Which would then mean that:
J(f(x)+g(x))=J(f(x))+J(g(x))
where J represents the integral sign.
For finite sums it’s ok, but basically for infinite sums things become dangerous and illegal :)
Ah ok, thank you :)! So does this break down for specific series? Because I saw you use it on a Taylor series. So does this rule depend somewhat on the 'type' (don't know if you call it that, I mean like Taylor or geometric etc.) of series you are using?
This should be rated R for interchanging the integral and sum signs like that.
Hahaha
Rated pi
@@drpeyam That's great, I'd hoped you would like that! Your videos are amazing I hope you are making more of them.
Beautiful result. Well done
めっちゃ楽しそうでかわいい
日本人だ
Great presentation! The voicing made it all the better!
when you try to watch math videos to fall asleep but then you get really invested in the topic
You are a true genius
I know I am late, but I am really confused about the 2. substitution (step 3) shouldn‘t be the integral from 1 to infinity?
Or why not?
Is it valid to evaluate this integral as a Cauchy Riemann sum? because then we can faster get a corresponding series even though I could only find that the value of it was betwwen 0 & 1. If my path is correct then the integral I= lim (sum(1/N(k/N)^[1/N])
I don't know English.
しかし、とてもわかりやすかったです。
im a beginner... why can't we just do the integral of e^xln(x) as e^xln(x)/ln(x) and thats our answer? 1:42
Sir I'm a beginner in calculus, I'm having a question, that why didn't you directly used a power rule and integrated it as x^x+1/(x+1){excluding limits)?
You can only use the power rule if the exponent is constant
@@drpeyam okay, thanks a lot sir !!
BEAUTIFUL!
Sir you are great Allah give you health and you do lot for us
I've been waiting for something like this!
I used a TI-84 to calculate this integral and I got 0.7834304192, which is what I assume to be the solution when you run the sum to infinity
I have a question.
How we can indetify functoons that doesnt have a antiderivative?
Wooow que hermoso!!!
Nice presentation. Thanks .DrRahul Rohtak Haryana India
Can you do some physics problems[mechanics] ?
Don’t know any physics, but Flammable Maths has lots of mechanics problems!
Ok, thank you for the answer, sir.
I felt happy when saw that you solve nice problem sir
What about the integral of x^-x from 0 to infinity? Could one write the result (~1.995) as a infinite sum...?
You are too amazing teacher ❤️❤️❤️