You know you're an integral ninja when you can whip out power series to evaluate integrals that otherwise do not have finite elementary solutions. Watching this channel inspired me to do a detailed review of the chapter in my undergrad calculus text on power series. Sadly, it doesn't cover too many examples of integrals that can be solved using them. Thanks!
The method I used to evaluate this before watching the video was Feynman's technique, contour integration, and then geometric series. The first substitution in the video would not have occurred to me.
All is fine but I'd like to know how do we know the first substitution is the right one? What criterium is used to find the right substitution? That's the question I often ask myself when I watch people coming with the solutions for integral questions. Sure, I can go along and say that all steps are correct, but I'd like to know what idea stands for all these steps.
Yes buddy, I totally agree with you. I would love to understand the reasoning behind methods of solving. Not just the method, many times this aspect of learning is missing.
I agree. This is missing in his videos. It comes about as a bunch of unnatural tricks. It's doesn't teach one on how to be a mathematician. It's the thought process that matters - including going down wrong paths and failing at first.
I would say this is just a lot of experience in solving integrals using substitution. Sure there are some typical substitutions depending on the type of the integrand function but if u solved hundreds and hundreds of integrals u get a feeling which sub will work and of course there is some trial and error
You might find this deeply unsatisfying, but I don't care: this one, consider completely unmotivated. next time you see something where this substitution looks promising, you can be motivated by the memory of this problem. maybe by the time you've done half a dozen or a dozen of them, you'll consider it a "standard technique for these types of integrals". Maybe someone has something for this one in particular, but big picture, it doesn't all get to be deeply intuitive immediately. Train your intuition, don't just indulge it.
Always start with "If I have a function inside a function, let's first try to susbsitute u = that inside function, and see what happens". Doing some more integrals or reading about some emergent patterns will somewhat allow you to choose better susbstitution, or at the very least avoid traps in order to advance more quickly on problems that are more and more complicated.
The computation of int log(x)log(1-x)/x dx,x=0,1 can be achieved by integration by parts to obtain a result in function of the integral of log(x)^2/(1-x)dx,x=0,1 already computed.
4:41 - while it's pretty clear to accept this limit goes to 1 as a rule, a very clean way of dealing with rational limits like these in my opinion is dividing the numerator and denominator by x, leading to 1/1 + (1/x), the 1/x in the limit clearly goes to 0 leaving the limit as 1/1 = 1.
There Is a Minus sign error in the First substitution. Check It out. Not so important but valuable if you consider that it comes at the beginning of the process
@@richardheiville937 Very interesting. I played around with numbers in PARI with 500 digit accuracy in decimal fraction part and got this: int = intnum(x=0,+oo,(log(x))^k*log(1+x)/x/(1+x)) k = 0, int = zeta(2) k = 1, int = zeta(3) k = 2, int = 7*zeta(4) = 7*Pi^4/90 k = 3, int = 3!*(zeta(5)+zeta(3)*zeta(2)) k = 4, int = 279/2*zeta(6) = 31/210*Pi^6 k = 5, int = 5!*(zeta(7)+zeta(5)*zeta(2))+210*zeta(4)*zeta(3) k = 6, int = 5715*zeta(8) = 127/210*Pi^8 k = 7, int = 7!*(zeta(9)+zeta(7)*zeta(2))+9765*zeta(6)*zeta(3)+8820*zeta(5)*zeta(4) k = 8, int = 804825/2*zeta(10) = 2555/594*Pi^10 k = 9, int = 9!*(zeta(11)+zeta(9)*zeta(2))+720090*zeta(8)*zeta(3)+ 635040*zeta(7)*zeta(4)+703080*zeta(6)*zeta(5) k = 10, int = 87048675/2*zeta(12) = 1414477/30030*Pi^12 And so on. Motivation for odd k was your result for k = 3.
@@alexlee6557 Actually i have computed nothing (but i can i know how to compute this one), i have guessed what is the form of the result using lindep function of GP PARI.
Does this mean that the integral we started with is an integral representation of the Riemann Zeta function at 3? Does that mean there are similar integrals to this for different Zeta values?
increase the power of the logarithms and probably you will get results in function of zeta(k),k>=3. For example, integral of log(x)^2*log(1+x)/((1+x)*x),x=0,infinity=7zeta(4)
Probably integral of log(x)^(2n+1)*log(1+x)/((1+x)*x),x=0,infinity is a rational linear combination of zeta(2n+3) and of productS of zeta(k)zeta(k') with k+k'=2n+3
3:25 I think you missed a factor of -1 for du and I haven't seen you correct it 30ish seconds further along in the video, so I'm making a comment. (I will edit my comment when I see you correct it.) EDIT: I was wrong I forgot to consider the chain rule.
If you’re talking about the expression for dx in terms of du, then the -1 from the power rule (1-u)^-1 cancels out with the -1 from the chain rule (-u).
Whenever you do an integral in a sum, or vice versa, and switch the order, I always feel a little cheated, because you never explain under what circumstances you're allowed to do that.
I think it is still not known whether zeta(3) have a nice and neat closed form, unlike zeta(2) or zeta(4) etc. Which is why we define it as a constant on it’s own, called Apery’s constant.
@@CryToTheAngelz You’re right sorry. I got mixed up with that zeta(3) has been proven to be irrational while zeta of other odd numbers are still unknown.
@@richardheiville937 a closed form is basically just expressing the value of the infinite sum exactly using a finite expression of some numbers. For example, zeta(2) is pi^2/6. zeta(4) is pi^4/90. These are the infinite sums expressed in closed forms It was proven that zeta(2n) where n is an integer has a value of pi^2n multiplied by some rational number to be determined. but for odd inputs we know barely anything.
You know you're an integral ninja when you can whip out power series to evaluate integrals that otherwise do not have finite elementary solutions. Watching this channel inspired me to do a detailed review of the chapter in my undergrad calculus text on power series. Sadly, it doesn't cover too many examples of integrals that can be solved using them. Thanks!
The use of series expansion is for beginners.
The method I used to evaluate this before watching the video was Feynman's technique, contour integration, and then geometric series. The first substitution in the video would not have occurred to me.
The Apéry constant, as allways, appears just where you would not expect...
Nobody expects the Apéry constant.
15:49
All is fine but I'd like to know how do we know the first substitution is the right one? What criterium is used to find the right substitution? That's the question I often ask myself when I watch people coming with the solutions for integral questions. Sure, I can go along and say that all steps are correct, but I'd like to know what idea stands for all these steps.
Yes buddy, I totally agree with you. I would love to understand the reasoning behind methods of solving. Not just the method, many times this aspect of learning is missing.
I agree. This is missing in his videos. It comes about as a bunch of unnatural tricks. It's doesn't teach one on how to be a mathematician. It's the thought process that matters - including going down wrong paths and failing at first.
I would say this is just a lot of experience in solving integrals using substitution. Sure there are some typical substitutions depending on the type of the integrand function but if u solved hundreds and hundreds of integrals u get a feeling which sub will work and of course there is some trial and error
You might find this deeply unsatisfying, but I don't care: this one, consider completely unmotivated. next time you see something where this substitution looks promising, you can be motivated by the memory of this problem. maybe by the time you've done half a dozen or a dozen of them, you'll consider it a "standard technique for these types of integrals". Maybe someone has something for this one in particular, but big picture, it doesn't all get to be deeply intuitive immediately. Train your intuition, don't just indulge it.
Always start with "If I have a function inside a function, let's first try to susbsitute u = that inside function, and see what happens".
Doing some more integrals or reading about some emergent patterns will somewhat allow you to choose better susbstitution, or at the very least avoid traps in order to advance more quickly on problems that are more and more complicated.
Thank you, professor. That was very fun.
The computation of int log(x)log(1-x)/x dx,x=0,1 can be achieved by integration by parts to obtain a result in function of the integral of log(x)^2/(1-x)dx,x=0,1 already computed.
4:41 - while it's pretty clear to accept this limit goes to 1 as a rule, a very clean way of dealing with rational limits like these in my opinion is dividing the numerator and denominator by x, leading to 1/1 + (1/x), the 1/x in the limit clearly goes to 0 leaving the limit as 1/1 = 1.
There Is a Minus sign error in the First substitution. Check It out. Not so important but valuable if you consider that it comes at the beginning of the process
d( 1/(1-u) - 1) = d(1/(1-u))=-1/(1-u)^2*d(1-u) = -1/(1-u)^2 * (-1)*du = 1/(1-u)^2*du. So, the video is correct at that point.
I thought so at first as well but there is a chain rule -1 that cancels the -1 from the exponent/ power rule
11:00 If i use substitution i will get Gamma function but i prefer integration by parts directly
Now compute integral of log(x)^3*log(1+x)/(x*(1+x)),x=0,infinity (the integral does have a nice closed form)
= 18.0854
@@charleyhoward4594 It's not a closed form. You can prove that is equal to 6zeta(5)+6zeta(3)zeta(2)
@@richardheiville937
Very interesting. I played around with numbers in PARI with 500 digit accuracy in
decimal fraction part and got this:
int = intnum(x=0,+oo,(log(x))^k*log(1+x)/x/(1+x))
k = 0, int = zeta(2)
k = 1, int = zeta(3)
k = 2, int = 7*zeta(4) = 7*Pi^4/90
k = 3, int = 3!*(zeta(5)+zeta(3)*zeta(2))
k = 4, int = 279/2*zeta(6) = 31/210*Pi^6
k = 5, int = 5!*(zeta(7)+zeta(5)*zeta(2))+210*zeta(4)*zeta(3)
k = 6, int = 5715*zeta(8) = 127/210*Pi^8
k = 7, int = 7!*(zeta(9)+zeta(7)*zeta(2))+9765*zeta(6)*zeta(3)+8820*zeta(5)*zeta(4)
k = 8, int = 804825/2*zeta(10) = 2555/594*Pi^10
k = 9, int = 9!*(zeta(11)+zeta(9)*zeta(2))+720090*zeta(8)*zeta(3)+
635040*zeta(7)*zeta(4)+703080*zeta(6)*zeta(5)
k = 10, int = 87048675/2*zeta(12) = 1414477/30030*Pi^12
And so on.
Motivation for odd k was your result for k = 3.
@@alexlee6557 Actually i have computed nothing (but i can i know how to compute this one), i have guessed what is the form of the result using lindep function of GP PARI.
Does this mean that the integral we started with is an integral representation of the Riemann Zeta function at 3? Does that mean there are similar integrals to this for different Zeta values?
increase the power of the logarithms and probably you will get results in function of zeta(k),k>=3.
For example, integral of log(x)^2*log(1+x)/((1+x)*x),x=0,infinity=7zeta(4)
for integral of log(x)^3*log(1+x)/((1+x)*x),x=0,infinity the result is a bit more complicated.
Actually, i conjecture that the integral of log(x)^(2n)*log(1+x)/((1+x)*x),x=0,infinity= a rational times zeta(2n+2)
Probably integral of log(x)^(2n+1)*log(1+x)/((1+x)*x),x=0,infinity is a rational linear combination of zeta(2n+3) and of productS of zeta(k)zeta(k') with k+k'=2n+3
Very interesting result!
3:25 I think you missed a factor of -1 for du and I haven't seen you correct it 30ish seconds further along in the video, so I'm making a comment. (I will edit my comment when I see you correct it.)
EDIT: I was wrong I forgot to consider the chain rule.
d( 1/(1-u) - 1) = d(1/(1-u))=-1/(1-u)^2*d(1-u) = -1/(1-u)^2 * (-1)*du = 1/(1-u)^2*du. So, the video is correct at that point.
If you’re talking about the expression for dx in terms of du, then the -1 from the power rule (1-u)^-1 cancels out with the -1 from the chain rule (-u).
@@skvortsovalexey thank you I forgot about the chain rule.
Such an awesome problem, thank you, I learned allot!
Finally done! Thank you.
Another way to compute this integral is to use the derivative of a Euler's beta function.
Very nice video
Can this be done with a contour integral with a detour around the origin?
Whenever you do an integral in a sum, or vice versa, and switch the order, I always feel a little cheated, because you never explain under what circumstances you're allowed to do that.
Dominated Convergence Theorem. And I know that because he’s mentioned it in many previous videos.
Here you go: en.wikipedia.org/wiki/Dominated_convergence_theorem
And now you know.
When all is positve (or have a constant sign), it's allowed. See theorem of Fubini-Tonelli
I have the ominous feeling this can be done by parts... although I feel it will get hairy quite quickly
Convirgin Theorem
The forest integral
Approx 1.20206...
Would be nice if the next video was displaying a closed form for zeta(3) lol
I think it is still not known whether zeta(3) have a nice and neat closed form, unlike zeta(2) or zeta(4) etc. Which is why we define it as a constant on it’s own, called Apery’s constant.
@@CryToTheAngelz You’re right sorry. I got mixed up with that zeta(3) has been proven to be irrational while zeta of other odd numbers are still unknown.
What do you mean by closed form? zeta(3) is a closed form for sum 1/k^3,k=1,infinity
@@CryToTheAngelz What is the definition of closed-form?
@@richardheiville937 a closed form is basically just expressing the value of the infinite sum exactly using a finite expression of some numbers. For example,
zeta(2) is pi^2/6.
zeta(4) is pi^4/90.
These are the infinite sums expressed in closed forms
It was proven that zeta(2n) where n is an integer has a value of pi^2n multiplied by some rational number to be determined. but for odd inputs we know barely anything.
8 minutes ago and only 16 likes.
This was absolutely whack
I clicked because of log 🪵 ^.^