I subscribe to a few other maths channels..all of them just show the right way(s) of getting to the ans..here, we get to understand WHY a particular way won't work, & how to get around/through/above the block..much appreciated!
It's really interesting how the integration of the original function between 0 and 1 ends up being equal to the infinite discrete sum of the same function starting from 1. I'm not sure why but it just feels fascinating that something like this exists.
I got the same feeling. There must be some interesting symmetry. On the other hand, I also feel like the answer isn't any more helpful than the original question 🤣
I can think of the Riemann integral from the shape of the function and the intervals of the integral and series, but I can’t quite come up with a way to express the Riemann sum properly.
@@letao12 the power in answer is negative and it is sum rather than integration... it can be calculated by a computer easily with some approximation so I think it is very helpful answer
@@letao12 if you learn integral properly in calculus, you'll know it's not really that surprising, considering integral comes from limit to infinity of the sum of the function.. This is why we learn calculus, the study of limit, we must never forget the origin of derivative and integral, that they are all just limits..
I think it’s “those who stop learning stop living”. Not to be nitpicky but changing “those” to “if” makes it sound like you’re suggesting suicide if they stop learning😂
Man I aspire to understand math this well one day. I don’t know how to do any of this but the way you work and alter the math so intricately is beautiful
very beutifully done. i just love the way you put up an act of a few fumble here and there - fumbling like any average student would. Please do a video on tests for convergence. the answer for this integral converges to approx 1.29129. i would have preferred if you had finished off the video with a quick evaluation of the infinite sum - may be calculating 5 or 6 terms to show how quickly this converges. from a student perspective she is going to demand the value of the infinite sum at the final answer. ofcourse if i were the techer i would have said “thats left as an exercise”😅😅😅
The video felt very interactive because instead of directly showing us the solution, you walked us through the problems by showing us the various ways you tried to approach the problem.
I appreciate you all's scientific curiosity, but what is the point of computing the numerical value of a converging series if you can't prove its convergence, can't write it in simple terms usually involving natural numbers and usual constants and without having a manual method of solving the series.
Finally one that does not spend one entire minute on multiplying both sides of an equality sign by the same expression! You're beatutifully talking about the essencial stuff all the way through, without spending time on trivialities. It was a pleasure to follow you! (P.S.: A little (unimportant) tip: Use equivalence () and not implication (=>) whenever that is correct. That would be stronger.)👍
I’ll be honest I have no idea why someone would ever want to learn how to do these kind of integrals, as I don’t see a reason to use them anywhere in real life problems, but I recognise your math skills to be a thousand times better than mine, and your videos to be a lot helpful to get ready for Math exams at uni, so, kudos.
Thank you for another wonderful video and what a beautiful and fascinating result. As an alternative approach. I had the idea of trying to compute the Riemann sum for the integral: Lim as n tends to infinity of the sum from r = 1 to n of 1/n*(r/n)^~(r/n) directly. But so far I have not been able to do this. I also had the idea of proving the result you found by considering the difference between this sum, re expressed in the form: lim as n terns to infinity of the sum from r = 1 to n of r^-r and the above Riemann sum This is potentially easier I think, by establishing an upper bound, in terms of n, on the mod of this difference, and then showing that this bound is 0 in the limit as n terms to infinity. But so far I have not been able to do this either. I would be interested if you or others know if either of these alternative approaches can be made to work for this particular problem. Again, than you for your blessed and inspirational videos ❤
You just did something called discretisation. You essentially converted an integral of a continuous function 1/x^x to a sum of the same function of n+1 where n is an integer.
I have not done calculus in over 40 years .This is beyond what I am currently capable of doing.I am College level Algebra I couldn’t pass pre-calculus / Trigonometry right now .
Thank you, this was a perfect presentation, congratulations! One question still remains: Is there a closed expression for sum_(k=1)^infinity(k^(-k)) ? It can easily numerically be computet but the question would be, if this number can be expressed as a multiple of pi , e or whatever. Would be very interesting to know!
Місяць тому
from Morocco thank you for your clear wonderful explanations
4:03 😂😂😂 I'm dead. I havent laughed that hard in a math video in a long time. Hahahaha 😆 😂 😆 But for real... I hate this problem... sometimes I wish math was easier.
Fascinating video about the process but I'm not quite sure what we achieved - given the form of the solution looks so very similar to the original integral 🤔
Sir ,I am from India ,preparing for JEE exam which is an entrance exam to get into IITs which are just like MITs of India,I am currently in 12th standard and I really loved ur approach towards this problem which seems easy at first sight but is quite difficult ❤The exam for which I am preparing also asks quite difficult problems ,thanks for the Video ,Love from 🇮🇳🇮🇳🥰🥰 U got a new sub.
@@aalekhjain2682 bro I have done these kind of probs which r bit out of syllabus but only for timepass or entertainment purpose. So chill,I m jee 2025 aspirant btw 😁
Hey newtons, I’m a 10 year old learning calculus, I know a lot (not like a whole college course) I’ve started Calculus 3, So I need help and my exams are there too. everything’s to me is easy. I started in February of my advanced mathematics learning when I was 9.
No, the integral is n! since: Γ(z) = (z-1)! = ∫[0 to ∞] t^(z-1) e^(-t)dt i.e. Γ(z+1) = z! = = ∫[0 to ∞] t^z e^(-t)dt The power of the integrand is itself shifted in the definition of the Γ function
I agree with @misteribel that you replaced one riddle with another one. And solving that one, gives the first again. The only thing (okay, a great find) you showed is that some finite integral of x to the power -x can be replaced by an infinite sum of more or less the same function. What I missed in this video is what in fact is the meaning or consequence of this result.
It’s more than half a century since I last studied maths, but I’m still a bit wary of your answer. I think you need to show that that series actually exists and is well defined. Unfortunately I can’t remember the conditions for convergence.
Nice work 👍. I think you made a mostake though. At the end, you obtained the zeta function of n which is equal to (n-1)! Not n!. Edit: the gamma function of n
1 / x ^x = y 1/ y = x ^x Log ( 1/y ) = x log x 1 = x / y log x 1 / log x = x / y X ^x = x log x Int .( 0 to 1 ) 1 / x log x 1 / x logx - 1 / x^2 Don't remember it right now I think in this boundary region the function is undefined means outside boundary ( not continuous, may have divergence) ?????????? Also may Don't have proper knowledge of mine in this matter
Thanks for the great illustration, however I'm not sure what has been achieved here, all I can notice that the original integral is replaced by the sum of the similar function, which is basically the integration 🤔 Not sure if I'm seeing the full picture here!
so did the integral just convert to a more "discrete" form like earlier it was integral of all x^(-x) from 0 to 1 and in the end we are summing all k^(-k) for each natural k ... on the left we see is some summation of uncountable number of points but on the right its just some countable number of points .. am i missing something please help .. thank you
So, we go from one over x to the x, and the integral from 0 to 1 of that equals the sum of k=1 to infinity of k to the minus k, which is one over k to the k. But what does this approximate? You've rewritten a finite integral into an infinite sum (of the same function), but that's only one step.
Sir , can we indefinitely integrate the function x^-x once as the form of a^x and once in the form of x^n and sum those 2 up and plug in the limits { for 0 (the limit) we could just substitute α and make α tend to 0}
I'm a little bit confused. He started with an integral[0,1] of x^(-x) and ended up with a sum, that basically is sum[1,infinity] of x^(-x) 🤔 what's the clue?
hello, is the final solution just a Riemann sum version of the integral? The last line looks like some high school questions on the limit of some summations, which those questions require kids to transform the sum into the integral to get the final answer. Thanks!
It feels like I am watching a Mathematical opera when ever I watch your videos. The drama! The suspense! Bravo 👏
Meanwhile I feel like I'm being tied down and mathematically sodomized
I really like that sir shows it's OK to back track & re-think when we reach a road block in solving
Yes, it's very reassuring
He sees the future that we have never, as of yet, seen, then he backtracks.
Yes I agree, but I think this is very common in calculus especially.
I subscribe to a few other maths channels..all of them just show the right way(s) of getting to the ans..here, we get to understand WHY a particular way won't work, & how to get around/through/above the block..much appreciated!
It's really interesting how the integration of the original function between 0 and 1 ends up being equal to the infinite discrete sum of the same function starting from 1. I'm not sure why but it just feels fascinating that something like this exists.
I got the same feeling. There must be some interesting symmetry.
On the other hand, I also feel like the answer isn't any more helpful than the original question 🤣
I can think of the Riemann integral from the shape of the function and the intervals of the integral and series, but I can’t quite come up with a way to express the Riemann sum properly.
@@letao12 the power in answer is negative and it is sum rather than integration... it can be calculated by a computer easily with some approximation so I think it is very helpful answer
@hammondkakavandi7738 the power in the original integral is negative also man.
1/(x^x)
=
x^(-×)
@@letao12 if you learn integral properly in calculus, you'll know it's not really that surprising, considering integral comes from limit to infinity of the sum of the function..
This is why we learn calculus, the study of limit, we must never forget the origin of derivative and integral, that they are all just limits..
"Never stop learning.If you stop learning. Stop living..." I appreciate you very much.. Nice explanation and nice question..
I think it’s “those who stop learning stop living”. Not to be nitpicky but changing “those” to “if” makes it sound like you’re suggesting suicide if they stop learning😂
@@ranae6566 learning means not only studies..
@@ranae6566 I literally liked his version better. Bro was like if you stop learning I will be personally looking for you
Excellent teacher! It is so refreshing to experience mathematics taught well. His enthusiasm and knowledge makes the difficult easy.😊
I just loved this and your whole approach. As a 74 year old UK guy who took his BSc in 1971, I am indeed still learning. Thank you!
I thought your comment was good to see that at your age you are still learning . I am sure the education system has changed since you were at uni.
Wow, did not expect that is so complicated.
You do a very good job explaining the solution.The result is really nice.
As an ex calculus private teacher i appreciate your expression so much.Your english and explanation is so clear.
Clear and detailed explanation of the steps taken to tackle the problem , thank you , opera writer !
Man I aspire to understand math this well one day. I don’t know how to do any of this but the way you work and alter the math so intricately is beautiful
It really feels like a Sophomore’s Dream!
Excellent explanation! You are even better than some math professors!🎉
Haha. That's hard to prove.
That was spectacular! Beautifully done!
never in my life did i think i would sit and watch such a crazy integral be solved yet here i am. amazing
Hey PN, you’re getting a lot of attention from other channels lately. It’s well deserved brother, god bless you and your work.
Really good video! Even though I'm not that good with math, I find you videos really understandable!
very beutifully done. i just love the way you put up an act of a few fumble here and there - fumbling like any average student would. Please do a video on tests for convergence. the answer for this integral converges to approx 1.29129. i would have preferred if you had finished off the video with a quick evaluation of the infinite sum - may be calculating 5 or 6 terms to show how quickly this converges. from a student perspective she is going to demand the value of the infinite sum at the final answer. ofcourse if i were the techer i would have said “thats left as an exercise”😅😅😅
This teacher is absolutely awesome. I am really a fan of his way to explain. Perfect !
One of your BEST videos that I have seen of yours. I really enjoyed it.
The video felt very interactive because instead of directly showing us the solution, you walked us through the problems by showing us the various ways you tried to approach the problem.
Watching this video was mesmerizing! Now, I want to know what that infinite series converges to.
1.29129
@@PrimeNewtons Thanks! I'm 79 and still learning!
do a spreadsheet, it converges very fast, after some six terms to 1.291286
I appreciate you all's scientific curiosity, but what is the point of computing the numerical value of a converging series if you can't prove its convergence, can't write it in simple terms usually involving natural numbers and usual constants and without having a manual method of solving the series.
@@BartBuzz Same here..i am 70.
Well the ending was a Revelation! Thanks for sharing!
Finally one that does not spend one entire minute on multiplying both sides of an equality sign by the same expression! You're beatutifully talking about the essencial stuff all the way through, without spending time on trivialities. It was a pleasure to follow you! (P.S.: A little (unimportant) tip: Use equivalence () and not implication (=>) whenever that is correct. That would be stronger.)👍
most intellectual 20 minutes & 36 seconds of my life. Thanks
Top quality! Grateful for this teaching!
Great, yet easy presenting approach.I like your channel.
This guy explained math in a very detailed way.
That made my Sunday evening pleasant.
I’ll be honest I have no idea why someone would ever want to learn how to do these kind of integrals, as I don’t see a reason to use them anywhere in real life problems, but I recognise your math skills to be a thousand times better than mine, and your videos to be a lot helpful to get ready for Math exams at uni, so, kudos.
Pleasant videos! I'll have to spend some hours to understand all the details here, but I think I'll set aside an evening for just that!
And the sum from (k=1) to ∞ of [k^(-k)] = 1.29128599706
and thx for the great video
You draw your xs so perfectly
Thank you for another wonderful video and what a beautiful and fascinating result.
As an alternative approach. I had the idea of trying to compute the Riemann sum for the integral:
Lim as n tends to infinity of
the sum from r = 1 to n of
1/n*(r/n)^~(r/n)
directly. But so far I have not been able to do this.
I also had the idea of proving the result you found by considering the difference between this sum, re expressed in the form:
lim as n terns to infinity of the sum from r = 1 to n of r^-r
and the above Riemann sum
This is potentially easier I think, by establishing an upper bound, in terms of n, on the mod of this difference, and then showing that this bound is 0 in the limit as n terms to infinity. But so far I have not been able to do this either.
I would be interested if you or others know if either of these alternative approaches can be made to work for this particular problem.
Again, than you for your blessed and inspirational videos ❤
amazing!! beautiful result!
Your voice is soothing
This Professor is genius
Amazing solutions. I felt like I was watching the crucial scene of John Wick. (Last sentence translated.)
This man is like the bob ross of calculus!
Not sure that I understood everything but it's awesome!
WOW!! Outstanding!! I did not foresee a Gamma Function was going to be applied.
Never Stop Teaching!
You know your stuff Man.
Keep on.
Thx for the dominate convergence theorem
You earned a subscriber bro. Hats off
I'd approach it via the lambert W function. If that pays off, then your result gives an interesting expansion for W(x)
You just did something called discretisation. You essentially converted an integral of a continuous function 1/x^x to a sum of the same function of n+1 where n is an integer.
Very interesting problem and clear explanation. Also you have such a lovely voice
and such perfect handwriting on blackboard ! A joy to behold.
Thanks a lot for you vidéo from France 🇫🇷
Well explained 👌🏼
God this is epic
I have not done calculus in over 40 years .This is beyond what I am currently capable of doing.I am College level Algebra I couldn’t pass pre-calculus / Trigonometry right now .
Thank you, this was a perfect presentation, congratulations! One question still remains: Is there a closed expression for
sum_(k=1)^infinity(k^(-k)) ? It can easily numerically be computet but the question would be, if this number can be expressed as a multiple of pi , e or whatever. Would be very interesting to know!
from Morocco thank you for your clear wonderful explanations
so int 0 -> 1 x^(-x) = sum 1 -> inf x^(-x) *mindblowing*
Pretty funny and pretty beautiful.
Hooooo , hermoso , me quedé pegada viendo, que lindas son las matemáticas❤
"now, can this be easily integrated?... no :("
4:03 😂😂😂
I'm dead. I havent laughed that hard in a math video in a long time. Hahahaha 😆 😂 😆
But for real... I hate this problem... sometimes I wish math was easier.
Well, you need to prove the uniform convergence of that series to be able to switch integral ans series sum.
Einfach genial!! Und so was von unterhaltsam 🙂 (Genius and best Entertainment!!)
{1/(-x+1)}.(x)^(-x+1)
The upper limit
(1/0).(x)^0=infinite
Lower limit
1.x=x=0
Infinite -0=infinite
Sloane's constant ~ 1.29...
Fascinating video about the process but I'm not quite sure what we achieved - given the form of the solution looks so very similar to the original integral 🤔
In terms of computing the integral's value? Not much (if anything at all). But the result looks very interesting _because_ of the similarity.
I love Math.
Think you, sir.
Sir ,I am from India ,preparing for JEE exam which is an entrance exam to get into IITs which are just like MITs of India,I am currently in 12th standard and I really loved ur approach towards this problem which seems easy at first sight but is quite difficult ❤The exam for which I am preparing also asks quite difficult problems ,thanks for the Video ,Love from 🇮🇳🇮🇳🥰🥰
U got a new sub.
JEE Advanced doesn't ask this level of calculus imo
@@aalekhjain2682 bro I have done these kind of probs which r bit out of syllabus but only for timepass or entertainment purpose. So chill,I m jee 2025 aspirant btw 😁
@@THESHAURYASHUKLA oh nice, i am JEE 2026 aspirant 😁
Hey newtons, I’m a 10 year old learning calculus, I know a lot (not like a whole college course) I’ve started Calculus 3, So I need help and my exams are there too. everything’s to me is easy. I started in February of my advanced mathematics learning when I was 9.
19:10 Isn’t the integral equal to (n-1)!, because it is gamma(n). But previously you established gamma(n+1) as equal to n! and not (n+1)!
I was wondering the same thing
No, the integral is n! since:
Γ(z) = (z-1)! = ∫[0 to ∞] t^(z-1) e^(-t)dt
i.e. Γ(z+1) = z! = = ∫[0 to ∞] t^z e^(-t)dt
The power of the integrand is itself shifted in the definition of the Γ function
I agree with @misteribel that you replaced one riddle with another one. And solving that one, gives the first again. The only thing (okay, a great find) you showed is that some finite integral of x to the power -x can be replaced by an infinite sum of more or less the same function.
What I missed in this video is what in fact is the meaning or consequence of this result.
It's the non-closed form solution for the definite integral thats much easier to evaluate than the integral by itself.
Very interesting ! Thank you for your solution
A bit disappointing, How is the series better than the integral?
It’s more than half a century since I last studied maths, but I’m still a bit wary of your answer. I think you need to show that that series actually exists and is well defined. Unfortunately I can’t remember the conditions for convergence.
INTEGRAL
The mathematical delinquency in me wants to just set u=x^x even tho i know that is one of the worst things you could do lmao
me don't understand anything but just wants to watch it
Love uuuu ❤
Love uuuu tuuu
Wow! Great job.
Beautiful
U sub is so useful.
Can you teach a full course on calculus from beginning through Cal III?
That is my new goal. I'm working on it
Wow. My compliments!
I've seen this attributed to John Bernoulli
keep spreading the Revelation! Hail Him, The Almighty Glory.
Nice work 👍.
I think you made a mostake though. At the end, you obtained the zeta function of n which is equal to (n-1)! Not n!.
Edit: the gamma function of n
No, no mistake. The value of the Euler integral used in the video is in fact Γ(n+1) which is equal to n!.
@@Grecks75 oh shoot, you're right. It happened, I started to forget basic math knowledge from school. Never thought it could be the gamma function tho
1 / x ^x = y
1/ y = x ^x
Log ( 1/y ) = x log x
1 = x / y log x
1 / log x = x / y
X ^x = x log x
Int .( 0 to 1 ) 1 / x log x
1 / x logx - 1 / x^2
Don't remember it right now
I think in this boundary region the function is undefined means outside boundary ( not continuous, may have divergence)
??????????
Also may Don't have proper knowledge of mine in this matter
LOVE YOU DUDE
Magic!
Chapeau 👌🙏👍
Thanks for the great illustration, however I'm not sure what has been achieved here, all I can notice that the original integral is replaced by the sum of the similar function, which is basically the integration 🤔
Not sure if I'm seeing the full picture here!
I would rather memorize it than solving.
Beatiful
so did the integral just convert to a more "discrete" form like earlier it was integral of all x^(-x) from 0 to 1 and in the end we are summing all k^(-k) for each natural k ... on the left we see is some summation of uncountable number of points but on the right its just some countable number of points .. am i missing something please help .. thank you
love it ❤
So, we go from one over x to the x, and the integral from 0 to 1 of that equals the sum of k=1 to infinity of k to the minus k, which is one over k to the k. But what does this approximate? You've rewritten a finite integral into an infinite sum (of the same function), but that's only one step.
But the infinite sum is always defined as a limit, which is in this case a certain (finite) constant, I think.
@@zzambezi1959Wolfram Alpha gave the finite answer of:
≈1.29128599706266
Sir , can we indefinitely integrate the function x^-x once as the form of a^x and once in the form of x^n and sum those 2 up and plug in the limits { for 0 (the limit) we could just substitute α and make α tend to 0}
Amazing
That was a juicy one! ❤
I'm a little bit confused. He started with an integral[0,1] of x^(-x) and ended up with a sum, that basically is sum[1,infinity] of x^(-x) 🤔 what's the clue?
Does sum from (k=1) to ∞ of [k^(-k)] converge ? If yes, could you demonstrate it please ?
well if you replace x with -x you just have sophomore's dream 🤷♀️
hello, is the final solution just a Riemann sum version of the integral? The last line looks like some high school questions on the limit of some summations, which those questions require kids to transform the sum into the integral to get the final answer. Thanks!