When I have a child I'll tell them the natural numbers go: 1, one over pi squared times the integral from 0 to infinity of the natural log squared of x over squared root x times one minus x all squared, 3, 4, and so on.
@@orionspur exactly, the question isn't very well-defined, the lemma, imo, might go like this: for all { x ; x belongs to Z), (X is even xor X is odd). can be done with basic first-order, or even just propositional.
@@orionspur I think zero is fairly even though, it can’t be written in the form, 2k+1... but then again that depends on the set of axioms that define odd-ness. Edit: It really depends on the field we’re working in.
1. Replacement in the original integral t=lnx. We get (the given multiplier 1/π^2 will be taken into account at the end): ∫ (from -∞ to ∞) t^2*e^t*dt/(1-e^t)^2. (1) We divide this integral into two (from -∞ to 0) and (from 0 to ∞). In the first one, we do the replacement t→-t. After simple transformations (1) is reduced to the form ∫ (from 0 to ∞) t^2*[e^(-t/2) +e^(-3t/2)]*dt/[1-e^(-t)]^2 = (*) 1/[1-e^(-t)]^2 = - e^t * d/dt [1/(1- e^(-t))] =- e^t*d/dt[1+∑(n from 1 to ∞) e^(-nt)]= =∑(n from 1 to ∞) n*e^[-(n-1)t]. (*) = ∑(n from 1 to ∞) n*∫ (0 to ∞) t^2*[e^(-(n-1/2)*t)+e^(-(n+1/2)*t)]. (2) 2. Using Feynman's trick, we obtain that if J(a)=∫ (from 0 to ∞)e^(-at)dt =1/a, a>0, then ∫ (from 0 to ∞)t^2*e^(-at)dt = d2J(a)/d2a = 2/a^3. (3) From (2) taking into account (3) (for a= n-1/2 and a=n+1/2) we get 2*∑(n from 1 to ∞) n*[1/(n-1/2)^3+1/(n+1/2)^3]= = 8*∑(n from 1 to ∞) {[(2n-1)+1]/(2n-1)^3+[(2n+1)-1]/(2n+1)^3} = 8*∑(n from 1 to ∞) {[1/(2n-1)^2+1/(2n+1)^2 +1} =16*∑(n from 1 to ∞) 1/(2n-1)^2 =16*(π^2/8)=2*π^2. Answer: 2*π^2/π^2=2.
In the last step we could not use the geometric series, instead we can visualize the integral as a second derivative of beta function which involves some digamma function values.
Hahaha. Reminds me of times in high school when I would write out a full page of work on solving an integral, only to end up with the value of Zero. I remember always remarking to myself : "Gosh, I did all of that work for Nothing..." 🙂
I did the same thing except I substituted sqrt(x) before the integral because I didn't want to deal with fractions in the denominator (It doesn't change much at all).
3:45 it might be my minor color blindness, but the yellow and orange are hard to tell apart EDIT: Later on it was easy to tell, must have been a lighting issue in that shot
Agree. Michael's yellow looks like a lightgreen (yellowish) to me. But that could be a color difference between the video cam and the final edited video.
In my attempt before watching the video, I used contour integration twice. The first time was a half circle in the upper half plane (and avoiding the origin) to put this integral in terms of a different integral, but quite similar. The second was a proper evaluation of that with a keyhole contour. I think the square root in the denominator made this easier to evaluate.
13:18 forgot to write the integral sign for the sum term, which made it really hard to follow that step and the next, the rest is very well put together
Heres a nice question- What irrational numbers have a continuous simple fraction expansion {a0,a1,a2,a3,...} So that their decimal representation b0+b1/10 + b2/100+.... Is such that b0=a0, b1=a1, b2=a2,.... What about any other representation? (Binary, hex,...)
Gradshteyn and Ryzhik anyone? I wonder if the Integral Suggester knows this book. Bought it for my grad level modeling class and barely cracked it since. Some of the integrals had no closed-form solution as indefinite integrals, only as definite. Really interesting stuff!
This maybe provides a bit of clarity: go to like 8:17 and imagine substituting x=t into the left integral and y=t into the right integral. Now you can add them together, and even just work the whole thing in t if you want to. The variable of integration can be changed to anything, since it never appears in the final result. It is like the k in summations you often see, simply used to index some set, a "dummy variable"
Hiya Doc! It always fascinates me, but how do you know which lemma, or tool each integral requires? I mean, not you specifically, but how does one know?
After playing around and reaching the point where for example you have int{ x^a ln(x) dx} you calculate it then and once you find the answer you rewrite the problem in a more presentable fashion. It's rare to know a priori what you need to solve a problem like this :b
For the first integral (the lemma), I first thought of Leibniz' Rule for integration, as d/da (x^a) = x^a ln(x), but that wouldn't work, as it would just increase the power of ln(x). However, then I integrated the expression in the lemma w.r.t. a two times, and I got the integral of x^a from 0 to 1, equaling 1/(a+1). Differentiating w.r.t. a two times again gives the correct result. My question is, how valid is this? I think we could argue that integration is just a fancy limit, meaning we could actually do this. But I'd love to hear a second opinion.
For heavens sake if YOU SONT KNOW the lemma or thst fsct can't you still solve the original with trying integration by parts..and or maube some u substitution like u is x minus 1 or u is ln x ..surely thst will work
When I have a child I'll tell them the natural numbers go: 1, one over pi squared times the integral from 0 to infinity of the natural log squared of x over squared root x times one minus x all squared, 3, 4, and so on.
What about 0?
@@jadonjones4590 It's not natural
@@farfa2937 Depends. In some parts of maths, it's considered to be a natural number, in others, it isn't.
@@bjornfeuerbacher5514 Yeah there're exceptions to every rule, but I didn't feel it was necessary since it was just a joke...
dont
0:40 Lemma
5:03 Main result
18:15 Good Place To Stop
14:03 Fake homework
Thanks!
Really nice integral, it's always cool when you can use sums to help solve integrals.
16:20 "That's because all numbers are either even or odd." Homework lemma. :)
Beautiful Timing just as I'm taking real analysis, plus I've never heard of an even irrational, emphasis on never heard.
@@orionspur exactly, the question isn't very well-defined, the lemma, imo, might go like this:
for all { x ; x belongs to Z), (X is even xor X is odd).
can be done with basic first-order, or even just propositional.
@@orionspur I think zero is fairly even though, it can’t be written in the form, 2k+1... but then again that depends on the set of axioms that define odd-ness.
Edit: It really depends on the field we’re working in.
1. Replacement in the original integral t=lnx. We get (the given multiplier 1/π^2 will be taken into account at the end):
∫ (from -∞ to ∞) t^2*e^t*dt/(1-e^t)^2. (1)
We divide this integral into two (from -∞ to 0) and (from 0 to ∞). In the first one, we do the replacement
t→-t. After simple transformations (1) is reduced to the form
∫ (from 0 to ∞) t^2*[e^(-t/2) +e^(-3t/2)]*dt/[1-e^(-t)]^2 = (*)
1/[1-e^(-t)]^2 = - e^t * d/dt [1/(1- e^(-t))] =- e^t*d/dt[1+∑(n from 1 to ∞) e^(-nt)]=
=∑(n from 1 to ∞) n*e^[-(n-1)t].
(*) = ∑(n from 1 to ∞) n*∫ (0 to ∞) t^2*[e^(-(n-1/2)*t)+e^(-(n+1/2)*t)]. (2)
2. Using Feynman's trick, we obtain that if J(a)=∫ (from 0 to ∞)e^(-at)dt =1/a,
a>0, then ∫ (from 0 to ∞)t^2*e^(-at)dt = d2J(a)/d2a = 2/a^3. (3)
From (2) taking into account (3) (for a= n-1/2 and a=n+1/2) we get
2*∑(n from 1 to ∞) n*[1/(n-1/2)^3+1/(n+1/2)^3]=
= 8*∑(n from 1 to ∞) {[(2n-1)+1]/(2n-1)^3+[(2n+1)-1]/(2n+1)^3} = 8*∑(n from 1 to ∞) {[1/(2n-1)^2+1/(2n+1)^2 +1}
=16*∑(n from 1 to ∞) 1/(2n-1)^2 =16*(π^2/8)=2*π^2.
Answer: 2*π^2/π^2=2.
In the last step we could not use the geometric series, instead we can visualize the integral as a second derivative of beta function which involves some digamma function values.
As others said, it's always nice when geometrical series and stuff like that shows up in integrals.
I just love it when after a long trek through math you end up with 2.
Thank you, professor.
Hahaha. Reminds me of times in high school when I would write out a full page of work on solving an integral, only to end up with the value of Zero.
I remember always remarking to myself : "Gosh, I did all of that work for Nothing..."
🙂
رائع كالعادة. واصل يا أستاذ.احتراماتي.
I did the same thing except I substituted sqrt(x) before the integral because I didn't want to deal with fractions in the denominator (It doesn't change much at all).
The most beautiful integral I've ever seen!
A very nice demonstration.
3:45 it might be my minor color blindness, but the yellow and orange are hard to tell apart
EDIT: Later on it was easy to tell, must have been a lighting issue in that shot
Agree. Michael's yellow looks like a lightgreen (yellowish) to me. But that could be a color difference between the video cam and the final edited video.
In my attempt before watching the video, I used contour integration twice. The first time was a half circle in the upper half plane (and avoiding the origin) to put this integral in terms of a different integral, but quite similar. The second was a proper evaluation of that with a keyhole contour. I think the square root in the denominator made this easier to evaluate.
Makes me think of Rube Goldberg machine. Very nice solution.
13:18 forgot to write the integral sign for the sum term, which made it really hard to follow that step and the next, the rest is very well put together
Should have put 1/(2 п^2) before the integral ig.
Heres a nice question-
What irrational numbers have a continuous simple fraction expansion {a0,a1,a2,a3,...} So that their decimal representation b0+b1/10 + b2/100+.... Is such that b0=a0, b1=a1, b2=a2,....
What about any other representation? (Binary, hex,...)
Nice video!
Today this question appeared in my test
Great video sir
Love it!
Gradshteyn and Ryzhik anyone? I wonder if the Integral Suggester knows this book. Bought it for my grad level modeling class and barely cracked it since. Some of the integrals had no closed-form solution as indefinite integrals, only as definite. Really interesting stuff!
Spoiler alert : the answer is the only prime number that can be written as the sum of two cubes ;)
I guess that generalizes all the way up right? 1^n + 1^n = 2?
Wolframalpha said that answer is 1.57.....
It's a very "easy" way to calculate 1+1.
😂😂😂
Clumsy! To prove the lemma just differentiate inegral of x^a twice.
5:01 no1 can tell me that that isnt a good transition
So technically the goal integral is 2pi^2? Can this be related to Gaussian functions?
I dont get why if x=1/y, then later y turns back to an x
Dummy variable
This maybe provides a bit of clarity: go to like 8:17 and imagine substituting x=t into the left integral and y=t into the right integral. Now you can add them together, and even just work the whole thing in t if you want to. The variable of integration can be changed to anything, since it never appears in the final result. It is like the k in summations you often see, simply used to index some set, a "dummy variable"
Thumbnail lacks pi-squared.
Then the answer is 2π 😂
@@goodplacetostop2973 Or rather 2π². ;)
@@bjornfeuerbacher5514 no, there's a pi in the thumbnail; it's just not squared
You're great
Hiya Doc! It always fascinates me, but how do you know which lemma, or tool each integral requires? I mean, not you specifically, but how does one know?
After playing around and reaching the point where for example you have int{ x^a ln(x) dx} you calculate it then and once you find the answer you rewrite the problem in a more presentable fashion. It's rare to know a priori what you need to solve a problem like this :b
Para cuando un problema de Ramanujan ?
To prove the lemma, note that x^a(log(x)))^2=d^2/da^2(x^a).
phương pháp tích phân từng phần chọn u, dv. Cảm ơn.
The lemma is easy by taking d^2/da^2 of the integral from 0 to 1 of x^a.
I want a T-shirt that says "... and that's a good place to stop"
:D
Check his shop, he sells them helpfully translated into many languages.
For the first integral (the lemma), I first thought of Leibniz' Rule for integration, as d/da (x^a) = x^a ln(x), but that wouldn't work, as it would just increase the power of ln(x).
However, then I integrated the expression in the lemma w.r.t. a two times, and I got the integral of x^a from 0 to 1, equaling 1/(a+1). Differentiating w.r.t. a two times again gives the correct result.
My question is, how valid is this? I think we could argue that integration is just a fancy limit, meaning we could actually do this. But I'd love to hear a second opinion.
The Maple Calculator gives the definite integral as 2π^2, so twice the value you got.
I ve sended u many requests of integrals
Please can u do the one of the result zeta(3)
Thanks in advance
For heavens sake if YOU SONT KNOW the lemma or thst fsct can't you still solve the original with trying integration by parts..and or maube some u substitution like u is x minus 1 or u is ln x ..surely thst will work
Or just use complex analysis...
This is most likely none of my business, but is everything ok? You look very tired in the video.