WARNING: At 2:40, he replaces a function (e^xlnx) with its taylor series. This only works because the taylor series of e^x converges everywhere, so e^f(x) converges everywhere for all functions f(x). However, if you try to do this with a function whose interval of convergence does not cover the interval of integration, then you end up integrating with the a different function outside the interval of convergence than what you started with

The Bernoullis reproduced like rabbits. Should have better studied Fibonacci kind of problems

this professor never lets me stay away from mathematics.

At 5:41 when you plug in, you should put e^(-u) in a big pair of parentheses, to emphasize that you execute this exponentiation first, and then raise the result to the power (n+1). The way you wrote it, by convention, means that e is raised to the power (-u)^(n+1), and that is totally different.

TYPO 1: I got a bit silly with whether my index variable is called i or n. It doesn't matter, it just needs to be the same throughout! TYPO 2: At 6:30 I use a power rule (a^b)^c=a^(bc). However, I should have make it explicit that there is brackets around all of e^(-u). Clarification: The key trick in solving this integral was writing the integrand as a power series and then interchanging the integral and the summation. If the sum of the integral of the absolute value of a sequence of functions converges, then we can do this interchange (this is a consequence of Fubini's Theorem). In our case, the integrands are either all positive or all negative, depending on n, so taking absolute values is equivalent to removing the negative signs in the resulting series which we can show converges via the ratio test. This justifies the interchange.

I always forget you can’t always swap sums and integrals. My physics profs always just do it without comment

Back when I was a math major, we tackled this very integral in one of my advanced calculus courses. After I left college I went into programming. Sadly, I never used much of the math again. So forgive me 40 years of eroded skills. I still love this stuff. Having a strong math backgroud has always served me well.

Another clever way to evaluate the integral from 0 to 1 of x^n*(ln(x))^n w.r.t. x is to use Feynman's trick, i.e. differentiation under the integral sign, a special case of the Leibniz integral rule. Start with the integral I(t), defined as the integral from 0 to 1 of x^t w.r.t. x, with t being a nonnegative continuous variable. I(t) is the first and easiest integral anyone learns in calculus, evaluating to 1/(t+1) after applying the reverse power rule and the Fundamental theorem of calculus. Next notice that, upon differentiation w.r.t. t and exchanging the integral and derivative operators (the derivative becoming a partial derivative under the integral sign), that I'(t) is the integral from 0 to 1 of x^t*ln(x) w.r.t. x. So differentiating I(t) once chains out a single factor of ln(x). This pattern continues, meaning I''(t) is the integral from 0 to 1 of x^t*(ln(x))^2 w.r.t. x, I'''(t) is the integral from 0 to 1 of x^t*(ln(x))^3 w.r.t. x, and so on, since d/dt(x^t) = x^t*ln(x) and ln(x) is a constant with respect to t. Therefore, differentiating I(t) n times and evaluating at t = n returns the integral in question, so I^(n)(n) is the integral from 0 to 1 of x^n*(ln(x))^n w.r.t. x. To evaluate I^(n)(t), look back at I(t). Since I(t) = 1/(t+1), I^(n)(t) = d^n/dt^n(1/(t+1)). Now a seemingly difficult problem of integration has been transformed into a rather simple problem of differentiation. To evaluate this nth derivative, look at the derivative for the first few values of n and try to recognize a pattern. I^(0)(t) = 1/(t+1) = (-1)^0*0!/(t+1)^(0+1), I^(1)(t) = -1/(t+1)^2 = (-1)^1*1!*1/(t+1)^(1+1), I^(2)(t) = 2/(t+1)^3 = (-1)^2*2!*1/(t+1)^(2+1), I^(3)(t) = -6/(t+1)^4 = (-1)^3*3!*1/(t+1)^(3+1)... , the pattern continuing on up to I^(n)(t), suggesting that I^(n)(t) = (-1)^n*n!/(t+1)^(n+1), which can be proved by induction. Finally, evaluating I^(n)(t) at t = n gives the desired result: I^(n)(n) = (-1)^n*n!/(n+1)^(n+1). Some lessons to take away from this solution are 1) that recognizing and exploiting patterns can be very helpful when solving math problems and 2) starting with a similar-looking problem that you know how to solve can often provide some insight into the solution of a more difficult problem. As an aside, this method of differentiation under the integral sign can equivalently be used to prove that Gamma(n+1) = n!, giving a greater intution as to why this technique gives the same result as that of this video, but I will leave that up to you!

Your presentation of this problem is really motivating me to try on my own such incredible integrals !

Thanks for this beautiful solution to the Bernoulli integral ∫₀¹xˣdx. Just a small point about the value of 0⁰. At the start of the video you mentioned the apparent contradiction between the rules x⁰=1 and 0ˣ=0, and wondered which rule should take priority when x=0. As far as I am concerned, the x⁰=1 rule definitely takes priority. This is because it is valid for all real x≠0, whilst the 0ˣ=0 rule, on the other hand, only applies for x>0, with 0ˣ being undefined for x0 it is 0, for x0. If we let a→0 and take the pointwise limit of these functions, we get 0 for x>0, 1 for x=0 and ∞ for x

If u look up cambridge step 3, 2009 question 8 it is on this exact integral and its a very excellent question i remember doing it when studying. Very fun to figure out and pleasing result :)

Never stop makin videos professor u are an absolute gem of a person

Absolutely loved this video. Thank you Trefor for bringing this gorgeous integral ^^

Excellent presentation, professor. ⭐️

Hello everyone! I might be kinda late, but here's my generalized approach for limits from a to b, being both real numbers. When Dr.Treffor said right before trying to get the gamma function for the answer that you can integrate by parts x^n*ln(x)^n, you can and the formula for the series that will appear is actually easy but a headache to get: Σ (0=k to n) of (x^(n+1)ln(x)^(n-k)*(-1)^k*n!)/((n+1)^(k+1)*(n-k)!). You guys can plug in the values for the summatory and will see how they work. Finally, we can use the Taylor series of e^x to get a generalized integral which can be evaluated from a to b. We get Σ (n=0 to inf) of (1/(n!)*(the aforementioned summatory)). Then just use Barrow's law. Despite it is theoretically correct and perfectly usable, for a computer to evaluate it the summatory will have an error because, obviously, the limit will be a real number and not infinity, not to mention evaluating it manually. Thus, I'd like that the ones that have an extended knowledge in the field can tell me whether it is plausible for the formula to be simplified in some way or method.

What might be fun is a short video discussing what I was taught to call "Indeterminate forms" such as your example - zero to the power of zero - but including: - anything divided by zero, - infinity (or at least the cardinality of the continuum) to the power of zero, - anything to the power of infinity, and so forth, with attention to the way some make even less sense than others.

Really trending video, great job!

Oh neat, the tetrate x^^2. Wonderfully done. I throughly enjoyed your process and pacing.

すばらしい。どうしても解けなかった問題を解いてくれた。

Brilliant, beautiful derivation. Thank you

Excellent and enjoyable presentation.

Love your content !!

To finish it off split the series into the odd and even parts to get rid of the (-1)^n, then re index the two sums to get the Basel problem and plug in pi^2/6

Great video, you're an excellent professor :)

This beautiful solution by Johann Bernoulli is justly shown here by the professor, involving e/ln, integration of x, lnx b/w 1 and 0; IBP was not used then, the same Johann who was stumped with Basel problem, and got to admire Euler's solution, (sadly w/o Jakob). You are a teacher I am looking for.

I admit I like to think about the way interections within the function suggest limitations on them. For example consider this, the existence of the power rule for derivatives seems to suggest that for whole number all odd slope that are on a x of odd power will have come from a fraction and that for a whole number c that could represent any number, there is potentiolly a limit on how low it can actually be if all numbers are whole due to how power will keep multiplying up. I have a 44 page paper on the collatz conjecture, but my record of passing math classes is admittily not perfect due to several issues with hand writing and expression of knowledge on tests themselves... your channel is always a good reminder for fun. Sometimes difficult to find less known rules that you discover yourself but that makes it interesting until you want to confirm they exist already lol XD

Excellent explanation - great👌

that's one great piece of math right there!! thank you

When he said "i will leave this as an exercise for you" i got severe PTSD of my maths professors at uni. Very rarely did they show us how results came about lol

x^x is such a bad boy. You've gotta be a math nerd to understand, but I just like that function. I am curious however whether it actually represents something in the real world, in the same way that n! expresses the number of permutations of n objects.

nice video, thank you!

Professor you blow my mind

With the first graph, if you plug it in in Desmos, even though desmos doesnt understand imaginary numbers, when you click home it sometimes shows things such as -1/3 as a value, although you cant click on it because buggy desmos

I Liked too much this solution, I hope more videos

Very good computation ❤

You are my inspiration Sir!

That was pretty creative

Great video thanks, In the challenge examples at the end, the first one should be : Int 0 to 1 (1/(x^x)) dx = sum (n=1 to inf) 1/(n^n) and not 1/(n^2)

Video: Ahh yes. Bernoullis contribution to math. Me: The guy who did the triangles. Nice. Video: Sorry no. The Bernoullis are to math what the Mathews family is to football. Different one

Sorry if I'm the 80th person to point this out, but your final summation is indexed by i while the terms are indexed by n. It was correct when you first introduced the Maclaurin series for the exponential function but, after you computed the integrals and put the whole series back together to finish the problem, that was when the i showed up.

Fantastic method.

Love the t-shirt!

Lovely integral bud

0^0 is obviously 1 since it is the cardinality of the set of functions which map the empty set to itself, namely the empty function an no others

Thanks 🙏🏻 teacher . Me from Cambodia 🇰🇭

That was pretty crazy.

Fascinating.

Well we could make it even more cleaner by assuming that (-1)^n =-(-1)^(n+1) than make a shift by one in the serie expression to fin the resulat equal to the sum from 1 to infinity of -(-1/n)^n

This was a fun one. Made a good break after my finance class. Question, do you think a Fuchsian group can be rendered in 3-dimensional hyperbolic space? Since hyperbolic space is weird, and Fuchsian groups are just projections into the hyperbolic space. (Sorry, doing research on Fuchsian groups and the question has been niggling at my brain for a minute.)

great answer thankj you

clicking on your video i already figured out that 0^0 = 1 from knowing the behavior of root n of x function tend to grow toward a step function at n=infinity (power of 0). and 1^1 = 1.

i once actually did the n-fold integration by parts cus i thought it was interesting. honestly im surprised i haven't seen anyone cover that method in a video

The numerical approximation strongly reminds me Pi/4. Is it true for a longer series sum?

good job

When you switched the integral with the sum in the beginning even before you said it I was thinking dominated convergence theorem or monotone/bounded CT

It has some real values for negative xs, so like the gamma function. 😀

i think i need to learn more algebra absolutely love it.

Great integral

Using the Basal problem and the steps to solve it, you can get a final solution to this as n goes to infinity of pi^2/12

Could you tell me where you purchased that beautiful T-shirt? Plz :)

Thanks for this showing this beautiful piece of mathematics btw I am so tiny to praise you sir 😅but I have a question which I want to ask Which shape (2d,3d) covers most area for least perimeter. I think it should be circle for regular shapes since it has highest area perimeter ratio but don't know if any complex shape in 3D space ( some sort of bent circular shape, like Pringles or anything) does so.

in 9:31 the "challenge examples" in the first integral of 1/x^x the result is not the sum of 1/n^2 it's the sum of 1/n^n

Simple Math, but beautiful! Congratulation!

can you make video about bernoulli number , with some contour integration behind it ? thank you so much.

What is the indefinite triple integral of x raised to the power of y, and then y raised to the power of z, with respect to dx, dy, and dz? Thanks and love you!

integral from 0 to 1 of x^x dx = - sum of n from 1 to infinity of (-1)^n * n^-n

I remember a similar integral being called Sophomore’s Dream, the integrand is x^-x instead of x^x

Could a similar method be used to integrate x^e^x?

great explanations! Could it be that you you messed up the variables i and n since the sum counted the variable i but you used n?

The integral of x^-x from 0 to infinity cash be solved basically the same way and turns out to be 1 + 1/2^2 + 1/3^3 +1/4^4…. (Even more beautiful) but here is my Question: is there a (similar) way to calculate the Integral from 1 to inf of f(x)=x^-x ? (It definitely has to be a finite number because this function is getting smaller than any f(x)= 1/x^n for any n >1 and and all integrals from 1 to inf of f(x)/x^n are finite for n>1 )

Nice!!

Hehe I have another trick to calculate this integral and is equal to exp(-1/4), but your work is really cool

i'm preparing for JEE Advanced, sir.....one small query....what were the ratio tests etc u talked to check if its converging or not, coz lot of times I misunderstand it to be a part fo taylor's expansions

So so elegant

Is there any important link between this integral and the integral of x TO THE -X FROM 0 TO INFINITY? [Sorry keyboard is broken>}

Can you make a playlist on series and sequences

I would still ask for proof of speed of convergence. What if converges really slowly and terms appear to stabilize only to change a lot later on in the expansion?

Can this integral be solved using Faradays method?

6:30 it looked like n+1 was the exponent on -u alone so I didn't think you could do that step like that

Thank you for this journey down memory lane. Once we realized that Calculus is Algebra with limits, mostly Algebra, it became much easier. Just remember that the endpoints of an integral are not included in the value, because it's ultimately a limit operation. Using a program to evaluate something like 0^0 is just accepting a definition, they simply define 0^0 to be 1. Computers are detrimental to understanding.

The answer is 0.5

Could you write out the reduction of e^-u^(n+1) to e^-(n+1)u

5:44 The brackets should enclose the entire e-function, not only the exponent. There are already a couple of other comments wondering how you simplified further using the power rule, which was correct, but didn't make sense if the brackets are only about the -u in the exponent.

This, along with an even nicer integral, is collectively known as the "Sophomore's Dream"! The other integrand is x^-x and the result is even nicer: integral(x^-x) = sum(n^-n)

Eeeeh waitaminute. @8:40 The condition of absolute convergence is not a thing (it has other properties, though, such as the re-ordering of the terms). In order to swap the integral and the summation signs, you need either: 1) uniform convergence; this is the case, because the exponential series Σ y^n /n! has a infinite radius, therefore UC, and even normal convergence is guaranted on [-1,0] or whatever interval (x ln x) spans; 2) dominated convergence; on the interval [0,1], | x^x | = | exp(x lnx) | < 1 and ∫ 1 dx is well defined on [0..1] so it's ok; 3) monotonous convergence; (x lnx)^n / n! < 0 so the series is strictly decreasing as n goes to ∞, so it's ok. Forgive me if I'm completely mistaken, because it's 1 AM, I just came back from a festival, I'm exhausted and I'm doing math, ah ah.

Can you tell me how to create png images of these beautiful mathematics equations

Why can we switch 0 and infinity on the integral?

if you enjoy format of maple, you should definitely try wolfram mathematica

How you rearrangeed integral and summation

I don't follow the rule of exponents thing at around 6:35 where you bring down the (n+1)...I mean isn't it like saying exp(X**3) = exp(3x)?

I've got a print of a painting of two brothers Bernoulli deep into some maths hanging above my desk. They wore curly hair wigs like judges used to do.

7:13 Animation of power n is strange when left and right are swapping.

I’m 73 and my math knowledge stays at the level of sophomore in college. Still I find your topics are always interesting and intriguing.😊😊😊

He reminds of my maths professor who used to gesticulate a lot, in funny manner. 😄

Serie de fonctions qui converge normalement sur [0,1], on peut inverser l'integrale et la somme.

You are very very smart and remarkably incredible

Is it possible to get a closed form of the final result?

The retroactive justification based on absolute convergence works, but can't you justify the swap on the basis that the Taylor series for the exponential function has an infinite radius of convergence (meaning it is analytic on the whole complex plane thus complete)? You can't swap the sum and integral if the bounds of integration extend beyond the radius of convergence since this pretty much guarantees the resulting solution will not converge.

Sir Pls make a Video series On Beta and Gamma Function Seperately 🙏.

Can someone explain why when converting x^n in terms of u, you add 1 to n? In other words why does the exponent become n+1 and not just n?