blackpenredpen I feel honoured. Thanks Btw since I am talking to you, One thing I never understood, why is the gamma function defined the way it is, I mean why couldn't they just define it to be the same as the factorial (for positive integers), and that -1 in the integral always seems a bit weird. I know that there's a PI function which fits my purposes, but Gamma is so much more in use. Why would the early mathematicians (who loved it's elegance) choose to define it as such? It will probably have some physics or application related reason, but nevertheless, it is worth asking
I'm a physicist and I find this kind of integral quite often. We always say "some mathematician proved this result" but I never actually checked it. Nice to see the proof, good job BPRP
wow! i've seen integrals like your example near the end in statistical mechanics and elsewhere and never knew how they were evaluated. the textbook just gave us the value.
This was a question on my final exam of intro to real analysis (maa 5055) and I used this technique!!! I got stuck on proving the integral at 5:20 is uniformly convergent so I was unable to continue forward after that step:( but I’m glad to know I was doing the right thing lol
Man, this is very cool. As a recent ME grad, I've been out of the calc3 maths world for a little while, but this makes me want to jump right back into it.
I recall seeing this result in my statistical physics course when we dealt with bose-einstein distribution. I never saw how this mathematical derivation, though. :) Awesome vid!!!
You might also see this expressed as the product of the gamma function and the polylogarithm function Li_s(1) (which is zeta(s)). Polylogs crop up in related things like integral of x over exp(x) - 1
I'm not sure if this is true, but I think I've heard from somewhere that Fubini's Theorem (or Tonelli's) only works when the function is continuous in the given interval. But then, I don't think we could integrate something that's not continuous in a given interval (unless we break it up into continuous intervals). But very interesting video! 😯 I bet the next video is "Proof of Riemann's Hypothesis".
I think for a counterexample where the exchange of sum and int doesn't work you have to come up with some really stupid functions like f_n= a_n*x from 0 to 1/n with a_n so that f_n(1/n)=1 or something like that
AndDiracisHisProphet Excuse me man I can see you've got a great understanding of maths and physics Never thought of making some vids? (I would love to see them)
How do you know I have understanding of math and physics? Also, I would probably make them in german^^ Also, I am ugly af. You wouldn't want to see me :D
AndDiracisHisProphet Well I've read your previous comments in many other math channels and they don't seem to be written by someone who doesn't understand No problem if you do them in German Don't say that you're ugly I personally don't care about that
When you found my comments on other math channels you probably have already noticed that I make a lot of jokes. Not necessarily funny ones. The "ugly" comment was such a joke. Anyway, I indeed have planned doing math videos, but not using this user name (this is quite a private one), because I am an actual self employed tutor and I will use my companies name. Thing is I will mainly do stuff for grades 6 to 12, or so. Probably not so interesting for someone watching this kind of videos.
I was getting ready to yell at you for casually swapping sums for integrals, but you passingly referred to absolute convergence. This is one of those I'll take your word for it :-)
After doing some homework of my own on this, it seems like a much safer assumption than I would have thought that you can make that kind of interchange. According to single-variable special cases of the Fubini/Tonelli theorems, if int(sum(|f(n))|) < inf, or sum(int(|f(n)|)), then the two are interchangeable for the entire function, sans absolute values. Tonelli's hinges on Fn(x) >= 0, and I start to lose the trail after that, but I think you're really quite safe. So if you found a function who had negative values and a sum which is conditionally convergent and married the two... maybe?? Also, good job. Also also, can we get a LaTeX editor built into UA-cam Comments?
You can change the order because functions are non-negative. It doesn't need to converge. If you get infinity in one case, you get it also in the other.
If you read Planck Vorlesungen you find 1+1/2^4+1/3^4...=Zeta(4) for Stefan Boltzmann Integral. You can Take any Zeta of even Numbers of zeta, If you adapt the prefactor.
In complex numbers, if "s" is complex, e^(ns) != (e^s)^n... be aware of it because complex exponentiation is not a univocate value operation... so surely the relation is true for real "s", but for complex "s" you could find issues related to the multyvalued results of complex exponentiation
Let the equation below accept a single solution(n) specify both(a,b) in terms of (n) f(x)=X^2-(a+b+1)X+(ab)=0 since f(x)=0 is equivalent to x= (x-a)(x-b) I think in this space there are zeros of the zeta function .
ζeta if a function can be expressed as an infinite series and if it has a radius convergence, and if f(x) is differentiable and can be integrated.... then you are allowed to perform calculus on the series which means treating terms like constants and variables of integration. Meaning you can interchange the summation and integration. However you will have to find a proof of this theorem :)
2:06 apparently you sound exactly like me according to my phone. You saying "and take a look" in the video triggered my "Ok Google" command and searched for "look". :l
Does this formula work on the critical strip? Trying to graph |zeta(s)| as a function of x + iy (3D plot) did not seem to work. Maybe it's the program I'm using. Also, when I put in the first non-trivial zero of the zeta function, this formula you have does not return zero, so I think maybe it does not work on the critical strip... sadly. The function Riemann gives defines the zeta function in terms of the zeta function of 1-s. And when a function is defined using itself (sort of)... that's where the issues arise. My opinion. I wish there were a better form of the zeta function that did not do that.
Use the integral to plot a keyhole contour over the complex plane, then use the Cauchy Residue Theorem and the Zeta Functional Equation are derived. The Analytical Continuation of Zeta are derived.
Hey nice video! But I just have one question; when saying a geometric series has the sum of a/(1-r), doesn't the sum need to start at 0 and not 1 like it did in the video?
Sadly, it's not really any better than using the series formula, though I admit it's a heck of a lot more aesthetic. In particular, you do have zeta(s) = 1/Gamma(s) int_{0...inf} u^(s-1)/(e^u - 1) du and since there's no zeta on the right, it is not circular, but the trouble is, it's no better than the series in terms of its range. The right hand integral, unfortunately, diverges for the good bit. Too bad, because as said, this is way cooler.
Hold on, x is always a variable, I think it should primarily be converted into something with u.. unless u r doin partial integration, which doean't seem to be the case, is it?
Andy Arteaga yes but because it is continuous, the value equals the limit, given that the value (inside) exists. It does exist (of course, it was calculated) so the value is the same
stamsarger only if the first term is one. It always is, however, if you evaluate the sum of (anything)^n from n=0 to infinity. In this case the sum started at n=1 so the more general formula was needed.
Hey!, That's the Bose integral right? (Or something with a similar name)
It comes up in Statistical mechanics quite some times.
Nice!!!
Yes it is!!!
blackpenredpen Wait, did you just change the title of the video or did I just not notice it before?
Gurkirat Singh
I just changed the title. Since you reminded me.
Gurkirat Singh as a thank you. I will also pin your comment now!! Thank you!
blackpenredpen I feel honoured. Thanks
Btw since I am talking to you, One thing I never understood, why is the gamma function defined the way it is, I mean why couldn't they just define it to be the same as the factorial (for positive integers), and that -1 in the integral always seems a bit weird.
I know that there's a PI function which fits my purposes, but Gamma is so much more in use. Why would the early mathematicians (who loved it's elegance) choose to define it as such?
It will probably have some physics or application related reason, but nevertheless, it is worth asking
I feel like there should be a line in a horror movie where the antagonist slowly says "we are in the u world now."
"Prepare to be integrated."
Is this better of worse than to be differentiated?
AndDiracisHisProphet - isn't that the equivalent of being dis-integrated💀
only if you are constant
@@elijahshadbolt7334 OH GOD
t = nu then tt = nut
nut. Haha
Oon Han yup! Hahha
Oon Han when you read it out: titty equals nut
Haha aren't you a bit too young for such jokes
This is not for kids
*isn't it*
This is very useful in the Stefan-Bolzmann's law (black body irradiance) to get the sigma constant
Well, for everything related to blackbody radiation you need this.
For example, the number density of photons is given by this integral too.
incredible application
Also the Sommerfeld expansion
Thanks
I'm a physicist and I find this kind of integral quite often. We always say "some mathematician proved this result" but I never actually checked it. Nice to see the proof, good job BPRP
I had a question:
Is it legal to use n first as a variable and then take the sum? It feels very sketchy.
This boi is a really nice one as it appears everywhere in quantum statistics in cases where the fugacity is equal to one.
wow! i've seen integrals like your example near the end in statistical mechanics and elsewhere and never knew how they were evaluated. the textbook just gave us the value.
Finally I'm able to correct you. At 4:24 you say "x to the nth power" :D
But it's still a great video =)
I was about to say that! :-(
This was a question on my final exam of intro to real analysis (maa 5055) and I used this technique!!! I got stuck on proving the integral at 5:20 is uniformly convergent so I was unable to continue forward after that step:( but I’m glad to know I was doing the right thing lol
I think i asked for this integral time ago, while i was studying statistical mechanics, but i only saw the video today! great!
Sure wish these videos were around when I did my degree!
These are great!
390 likes and 0 dislikes... that’s the most likes I’ve ever seen on a video with 0 dislikes. Keep up the good work! (:
This is equivalent to showing that the Mellin transform of the function 1/(e^x -1) is the product of the Gamma function with the Riemann zeta function
Man, this is very cool. As a recent ME grad, I've been out of the calc3 maths world for a little while, but this makes me want to jump right back into it.
Dean Congrats! :) also, out of curiosity, how much math were you required to take as an ME major?
Thanks Jeff. After doing my calc series, I also had to take a linear algebra and an ODE differential equations course.
I recall seeing this result in my statistical physics course when we dealt with bose-einstein distribution. I never saw how this mathematical derivation, though. :) Awesome vid!!!
Your best video so far. Truly interesting!
This is like the first or second page of Riemann’s paper on primes.
You might also see this expressed as the product of the gamma function and the polylogarithm function Li_s(1) (which is zeta(s)). Polylogs crop up in related things like integral of x over exp(x) - 1
oh... 我只在乎你 nice choice for an ending song ! :) Thanks for the uploading !
Thanks!!
that outro music had a Scott Joplin vibe, v-much appreesh
I'm not sure if this is true, but I think I've heard from somewhere that Fubini's Theorem (or Tonelli's) only works when the function is continuous in the given interval. But then, I don't think we could integrate something that's not continuous in a given interval (unless we break it up into continuous intervals).
But very interesting video! 😯 I bet the next video is "Proof of Riemann's Hypothesis".
and if you replace the zeta function by the alternate eta function, you change the sign in the denominator in the integral, from e^u - 1 to e^u + 1.
I think for a counterexample where the exchange of sum and int doesn't work you have to come up with some really stupid functions like f_n= a_n*x from 0 to 1/n with a_n so that f_n(1/n)=1 or something like that
AndDiracisHisProphet Excuse me man
I can see you've got a great understanding of maths and physics
Never thought of making some vids? (I would love to see them)
How do you know I have understanding of math and physics?
Also, I would probably make them in german^^
Also, I am ugly af. You wouldn't want to see me :D
AndDiracisHisProphet Well I've read your previous comments in many other math channels and they don't seem to be written by someone who doesn't understand
No problem if you do them in German
Don't say that you're ugly
I personally don't care about that
When you found my comments on other math channels you probably have already noticed that I make a lot of jokes. Not necessarily funny ones. The "ugly" comment was such a joke.
Anyway, I indeed have planned doing math videos, but not using this user name (this is quite a private one), because I am an actual self employed tutor and I will use my companies name.
Thing is I will mainly do stuff for grades 6 to 12, or so. Probably not so interesting for someone watching this kind of videos.
AndDiracisHisProphet But if one day you decide to do some calc, ODE, Linear álgebra. Abstract algebra vids
You've got a subscriber!
I was getting ready to yell at you for casually swapping sums for integrals, but you passingly referred to absolute convergence. This is one of those I'll take your word for it :-)
After doing some homework of my own on this, it seems like a much safer assumption than I would have thought that you can make that kind of interchange. According to single-variable special cases of the Fubini/Tonelli theorems, if int(sum(|f(n))|) < inf, or sum(int(|f(n)|)), then the two are interchangeable for the entire function, sans absolute values.
Tonelli's hinges on Fn(x) >= 0, and I start to lose the trail after that, but I think you're really quite safe. So if you found a function who had negative values and a sum which is conditionally convergent and married the two... maybe??
Also, good job.
Also also, can we get a LaTeX editor built into UA-cam Comments?
proud to be a bose
Wow...these things fascinate me...I don't know much about them...but the time will come I promise!
You should record in 1080p 60fps. Quality res videos for quality maths 👌🏼
TheOnePath he records on mac
60fps sucks
You are simply awesome Prof. 🙏
You can change the order because functions are non-negative. It doesn't need to converge. If you get infinity in one case, you get it also in the other.
If you read Planck Vorlesungen you find 1+1/2^4+1/3^4...=Zeta(4) for Stefan Boltzmann Integral. You can Take any Zeta of even Numbers of zeta, If you adapt the prefactor.
Hi! I love your videos, and your math is on point Keep it up 😉
Hi, I'm from Bangladesh. I very much like your videos. Thanks for making video on Bose integral.
U just solved the Riemann hypothesis
Amazing and beautiful result! I always snapshot these kinds of things bc that are so interesting! Thanks for sharing as always!!!!!!!!!! 😇😇😇😇😇
1 fact-oreo!
Bringing back the fun
This. Is. Awesome.
Wow, an amazing proof. The ancient Greeks would be surprised to see Gamma, Zeta, Pi in one equation.
Awesome video! You can't swap integration and addition with this thing: ∫ ∑ x²(1-x²)ⁿ dx for x∈[-1; 1] when you sum over n c:
But why?
That was a great one. How can we solve zeta(3)
10:35 Bravo!!!
thank very much you are a genius
What a great video 😍😍😍
Just like usual 😊
WOW!! This is so cool!
Now, evaluate the integral using the residue theorem, thereby deriving the reflection formula for zeta!
We need more videos about this,,, BPRP!!!!!!!!!!!!!
:V
Small speaking mistake at 4:24, but still an awesome and creative video!
Great work, really interesting as always!
I wonder what would happen if you did this with the eta function you just made a video on
4:17 I’m confused, if T(x) is defined in terms of n how can you pull it out of the summation?
In complex numbers, if "s" is complex, e^(ns) != (e^s)^n... be aware of it because complex exponentiation is not a univocate value operation... so surely the relation is true for real "s", but for complex "s" you could find issues related to the multyvalued results of complex exponentiation
This is awesome!!❤
Let the equation below accept a single solution(n) specify both(a,b) in terms of (n)
f(x)=X^2-(a+b+1)X+(ab)=0
since f(x)=0 is equivalent to
x= (x-a)(x-b)
I think in this space there are zeros of the zeta function .
you are very cool and very educational
Each and every day you look like Dr Peyam
High level maths!
😘😘ابداع يا استاد .احسنت 👏
Blackpenredpen, I am pretty sure that you can always switch the order of summation and integration
this video very good. I want that you share such a this video. I'm wacthing from Turkey.
zeta(3) = 1/2 integrate 0 to infinity u^2/(e^u-1) by u
Archik4 a close approximation is pi^3/26
I do not really understand when and why you are allowed to change summation and integration. Can anyone explain is to me?
ζeta if the integral is absolutely convergent (the integral of the absolute value is finite)
Thank you, but I was rather talking about why you're not allowed to change if the sum is divergent
ζeta if a function can be expressed as an infinite series and if it has a radius convergence, and if f(x) is differentiable and can be integrated.... then you are allowed to perform calculus on the series which means treating terms like constants and variables of integration. Meaning you can interchange the summation and integration. However you will have to find a proof of this theorem :)
Because the integral and sum operators are linear and are operating on distinct variables.
ζeta www.maths.manchester.ac.uk/~mdc/old/341/not7.pdf
Are you going off HM Edwards' "Riemann's Zeta Function"? I'm reading it now and this is in one of the first few chapters.
Bose condensate. is winter coming ? :)
Wow Bose Integral🤩
This is fucking amazing
Bernhard Riemann!
4:11 How were you able to take the gamma function out since it contains n terms?
It doesn't contain any n.
Math is divine. Bose is a great scientist from India. In physics some elementary particles are called bosons.
To permutate the infinite sum and the proper integral, shouldn't the sum converge uniformely and not only absolutely??
2:06 apparently you sound exactly like me according to my phone. You saying "and take a look" in the video triggered my "Ok Google" command and searched for "look". :l
I forgot, where do you teach again?
Does this formula work on the critical strip? Trying to graph |zeta(s)| as a function of x + iy (3D plot) did not seem to work. Maybe it's the program I'm using. Also, when I put in the first non-trivial zero of the zeta function, this formula you have does not return zero, so I think maybe it does not work on the critical strip... sadly. The function Riemann gives defines the zeta function in terms of the zeta function of 1-s. And when a function is defined using itself (sort of)... that's where the issues arise. My opinion. I wish there were a better form of the zeta function that did not do that.
As Gamma of negative integer values produces poles, I am wondering how one can define zeta there.
Use the integral to plot a keyhole contour over the complex plane, then use the Cauchy Residue Theorem and the Zeta Functional Equation are derived. The Analytical Continuation of Zeta are derived.
Just amazing...
Hey nice video! But I just have one question; when saying a geometric series has the sum of a/(1-r), doesn't the sum need to start at 0 and not 1 like it did in the video?
Suchetan Dontha it depends. That's why I put down "first"/(1-r)
4:25 x to the nth power
Can we not do something similar with the eta function
roses are red
violets are blue
now i can do maths
that my teachers can't do
Please solve integration root Sin(x) dx by udv ?
Very good!
At 7:57 didnt he forget the nth power or did i not understand something
David Wiedemann It is the sum to infinity as the summation size is added.So the nth power didn't miss
he took the infinite geometric sum with r = 1/e^u
Now derive the same equation starting from Zeta and starting from Bose
Very nice!
Forgetting Zeta function is the worst thing that can happen in your life
Could we use this connection as a definition of the zeta function ?
Jens Waelkens no we can't define zeta by this integral because of the poles of the gamma function
Also this integral converges only for Re (s) >1
Thanks for clarifying that!
Sadly, it's not really any better than using the series formula, though I admit it's a heck of a lot more aesthetic. In particular, you do have
zeta(s) = 1/Gamma(s) int_{0...inf} u^(s-1)/(e^u - 1) du
and since there's no zeta on the right, it is not circular, but the trouble is, it's no better than the series in terms of its range. The right hand integral, unfortunately, diverges for the good bit. Too bad, because as said, this is way cooler.
Is there any application of the beta and gamma functions in physics?
Quantum Mechanics are plenty used.
Love it!
Steve did you forget umlaut on the name of this integral ?
Just thinking how he hold 2 pens together and switch them while writing.....
Hold on, x is always a variable, I think it should primarily be converted into something with u.. unless u r doin partial integration, which doean't seem to be the case, is it?
Does anybody know the feeling of Depression i have it at the Ende of the Video it is so good why it has to be so good thanks for showing
Just to be sure, since *u* has to be greater than 0, should the integral be from *a* to *infinity* and then take the limit of *a -> 0* ?
Andy Arteaga yes but because it is continuous, the value equals the limit, given that the value (inside) exists. It does exist (of course, it was calculated) so the value is the same
Isn't the formula for an infinite geometric sum 1/1-r instead of first term / 1 - r?
stamsarger only if the first term is one. It always is, however, if you evaluate the sum of (anything)^n from n=0 to infinity. In this case the sum started at n=1 so the more general formula was needed.
I wonder if zeta is defined on the complex plane and your integral is defined on the real line can this be somehow accorded...
We know that zeta(2) = π^2/6, but can you show us how to calculate zeta(3) or zeta(4) maybe? I always wonder how to find the value of it..
Also when s > 2
Zeta(4) is equal to pi^4/90 ( The great Euler found them all)
David Rheault i already know the value. But i want to know how to find the pi^4/90 hahah
Fourier and parseval identity on f(x) = x^2, big formula, and you get zeta(4) as a function of zeta(2)
didnt he basically prove all the even values of zeta
t=n*u then -t= -1(n*u) or = -t=-n*-u
Plz, define the limit of x.
what if you have gamma(x)*zeta(y)
or even maybe gamma(x)*zeta(2x) or something like that.
Can you do anything?
Amazing
What math topic is this from?
The cool topic
The best topic
Histograms
algebra 2
Complex analysis
I realized you were holding 2 markers 10 minutes into the video