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.
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!
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😂
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.
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”😅😅😅
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
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 ❤
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 😁
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 🤔
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!
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.
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.
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.
I found that you could derive a quick formula for any integral ₀∫¹ rx^(tx) dx === (k=1 to ∞)Σ (r*(-t)^(k-1))/(k^k), where r and t are any real number. For example, if you plug in 1 for r and -1 for t, the series will simplify to the final answer at the end of this video. Because that will create the problem that is presented and answered in this video ₀∫¹ 1x^(-1x) dx :)
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
This is a really surprising result. The "super sum" of 1/x^x from 0 to 1 is equal to the rest of the normal sum from 1 to infinity, fascinating! Do you have any idea why this pattern appears?
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
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}
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
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!
Could you please attempt x^(1/x) from 0 to 1? I've managed to create a series for the general case of x^x^s when s >= 0. I could share my derivation if you (or anyone else) is interested.
I regret to say I am not impressed, just as I am not impressed with so many of the make-work problems on the MIT integration Bee. The final sum is just as intractable as the original integral, so one has made no progress at all. Much better would be to go back to G. H. Hardy's classic Pure Mathematics and master the 10 or so pages in the section entitled On the Practical Problem of Integration.
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
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.
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!
"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..
Excellent teacher! It is so refreshing to experience mathematics taught well. His enthusiasm and knowledge makes the difficult easy.😊
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.
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.
It really feels like a Sophomore’s Dream!
And the sum from (k=1) to ∞ of [k^(-k)] = 1.29128599706
and thx for the great video
Clear and detailed explanation of the steps taken to tackle the problem , thank you , opera writer !
As an ex calculus private teacher i appreciate your expression so much.Your english and explanation is so clear.
You do a very good job explaining the solution.The result is really nice.
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”😅😅😅
That was spectacular! Beautifully done!
Excellent explanation! You are even better than some math professors!🎉
Haha. That's hard to prove.
Really good video! Even though I'm not that good with math, I find you videos really understandable!
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
Sloane's constant ~ 1.29...
Well the ending was a Revelation! Thanks for sharing!
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 ❤
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!
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 😁
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.
most intellectual 20 minutes & 36 seconds of my life. Thanks
Never Stop Teaching!
Not sure that I understood everything but it's awesome!
That made my Sunday evening pleasant.
I'd approach it via the lambert W function. If that pays off, then your result gives an interesting expansion for W(x)
Einfach genial!! Und so was von unterhaltsam 🙂 (Genius and best Entertainment!!)
Top quality! Grateful for this teaching!
WOW!! Outstanding!! I did not foresee a Gamma Function was going to be applied.
keep spreading the Revelation! Hail Him, The Almighty Glory.
{1/(-x+1)}.(x)^(-x+1)
The upper limit
(1/0).(x)^0=infinite
Lower limit
1.x=x=0
Infinite -0=infinite
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 .
God this is epic
Hooooo , hermoso , me quedé pegada viendo, que lindas son las matemáticas❤
Very interesting problem and clear explanation. Also you have such a lovely voice
and such perfect handwriting on blackboard ! A joy to behold.
me don't understand anything but just wants to watch it
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!
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.
U sub is so useful.
Very interesting ! Thank you for your solution
Love uuuu ❤
Love uuuu tuuu
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.
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.
I love Math.
Think you, sir.
Could you do a more elaborate video on when you can swap an integral and a sum, and when you cannot? Preferably with some examples.
Can you teach a full course on calculus from beginning through Cal III?
That is my new goal. I'm working on it
Chapeau 👌🙏👍
Beautiful
I found that you could derive a quick formula for any integral ₀∫¹ rx^(tx) dx === (k=1 to ∞)Σ (r*(-t)^(k-1))/(k^k), where r and t are any real number.
For example, if you plug in 1 for r and -1 for t, the series will simplify to the final answer at the end of this video. Because that will create the problem that is presented and answered in this video ₀∫¹ 1x^(-1x) dx :)
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
Am I a geek for liking these calculus videos?
You are not alone bud, I don't even get most of it.
This is a really surprising result. The "super sum" of 1/x^x from 0 to 1 is equal to the rest of the normal sum from 1 to infinity, fascinating! Do you have any idea why this pattern appears?
I'm not sure, but there's an even better result.
The "super sum" of nCr(α,x) dx from -∞ to ∞ is equal to the normal sum of nCr(α,x) from -∞ to ∞.
Premium sum of 0 is same like normal sum of 0💀
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
well if you replace x with -x you just have sophomore's dream 🤷♀️
Wow. My compliments!
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}
love it ❤
Wow! Great job.
LOVE YOU DUDE
دائما براهينكم رائعة
That was a juicy one! ❤
I've seen this attributed to John Bernoulli
Beatiful
What a question wow
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
you miss a minus: right side of the board,third line: minus e to the minus t. am I right?
will this hold if 'a' is a complex number?😉
👏👏👏👏
I have one doubt here, how did you write (-1)^n as 1? Shouldn't it be kept as (-1)^n only in the final answer?
❤❤
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!
Hermoso problema
Is it me or we can reach the result in one lign with riemann sum ?
Perfect
Could you please attempt x^(1/x) from 0 to 1? I've managed to create a series for the general case of x^x^s when s >= 0. I could share my derivation if you (or anyone else) is interested.
Respected Sir, Good morning.... Pls get me the solution of integration {1/(x^6+1)}, 0 to 3
Bravo
you can't add more calculations??
No way I got this right 😂
В России такое решит любой 11 классик…
Also there was one integration I still remember from Calc 2 or one if u can solve its integration of 1/(1+tan^4(×))
bad light system
I regret to say I am not impressed, just as I am not impressed with so many of the make-work problems on the MIT integration Bee. The final sum is just as intractable as the original integral, so one has made no progress at all. Much better would be to go back to G. H. Hardy's classic Pure Mathematics and master the 10 or so pages in the section entitled On the Practical Problem of Integration.
The text is covering your writing and you can’t follow what you are doing ! Why you need the text ?
I am not aware of any text. Check your settings. You may have cc turned on.
I couldn't understand. There was no good explanation.
Whats your intro music?
It's the prime Newton's signature music. I went to the studio myself.
Αυτά είναι πανεπιστήμιο.εγω δε τα ξέρω λύκειο ειμαι😢
!!!!!!!!!!
😇
Excellent !!!
1.29128599706... (I'm gonna assume there's no exact solution)
I think a beautiful power series is exact enough
asnwer=1xy isit