Hi, 2:41 : this integral is well known to be equal to zeta(2), 5:29 : quantity squared, 5:55 : fixed , 7:32 : ok, I have to proof that as home work :) 15:57 : yes, verified with Fraction module of Python . "ok, cool" : 1:51 , 2:58 , 4:01 , 5:59 , 11:04 , 12:58 , 13:34 , 14:20 , "terribly sorry about that" : 3:37 , 4:22 , 5:44 , 6:15 . PS : I realize that your first name is very famous now!
Given the frequency of use, you should present proofs of the Leibniz integral rule. (preferably posted not solely posted to instagram.) (Maybe also dominated convergence thm, too.)
There certainly are many other approaches to solve this integral, some maybe simpler or more straightforward as some have commented, but I love it when you play with the integral like a cat with the mouse... Not just eat it immediately!😂 It's an exercise that broadens one's vision. I enjoyed it greatly. Thanks a lot! Bahat shukriya!
@@maths_505 Mmm... That's a difficult question 🤣 I am Armenian, born and bred in Lebanon, living in Armenia already for 37 years. I know several languages, including Arabic, Russian, Turkish, Greek, Farsi etc. Unfortunately, Urdu is not one of them. But I have many friends, ex colleagues, from Pakistan, Afghanistan, India etc 😄 So, I can often catch phrases in related languages...
Loved this one. Especially the arithmethicc and your inability to write the number 8. Don't worry Kamaal, in 1st grade I was really scared I wouldn't learn how to properly write it, but I eventually got the hang of it :)
You can prove zeta(2) = pi^2/6 with this general integral. f(a)=Integral ln(x^2+2ax+1)/x dx from 0 to 1. f'(a)=2*Int 1/(x^2+2ax+1) dx = 2/(sqrt(1-a^2))*(arctan((a+1)/sqrt(1-a^2))-arctan(a/sqrt(1-a^2))). Using arctan(x)+arctan(y)=arctan((x+y)/(1-xy)). f'(a)= 2/sqrt(1-a^2) * arctan(sqrt((1-a)/(1+a))) So f(a) = 2 Int arctan(sqrt((1-a)/(1+a)))/sqrt(1-a^2) da + C sqrt((1-a)/(1+a))=t, (1-t^2)/(1+t^2)=a, da =(-4t)/(1+t^2)^2 dt, 1/sqrt(1-a^2)= (1+t^2)/(2t) f(a)= -4 Int arctan(t)/(1+t^2) dt = -2 Int 2 * arctan(t) * (arctan(t))' dt. Using (f^2)' = 2 * f * f', f=f(x) f(a)=-2*arctan(t)^2+C = -2 * (arctan(sqrt((1-a)/(1+a)))^2 + C Now f(1) = C = 2 * Int ln(1+x)/x dx from 0 to 1, using series for ln(1+x), f(1)=2*Sum (-1)^(k+1)/(k^2), k 0 to 1, = zeta(2)=C, f(-1)= 2 * Int ln(1-x)/x dx = -2*zeta(2)= -2arctan(infinity)^2 + zeta(2), arctan(infinity)=pi/2, solving for zeta(2), -3*zeta(2)=-pi^2/2, zeta(2)=pi^2/6
That's the beauty of the trick. It suits well with the x√2 & x√3 for values of sin(α) when α is π/4 & π/3... It needs an eye to see it and a little bit of imagination... I certainly enjoyed this integral although it is relatively simple thanks to Feynman!
Mathematica (assuming alpha is between 0 and pi/2) computes the integral I(alpha) in terms of two polylog sub 2's: -PolyLog[2,-I Cos[\[Alpha]]+Sin[\[Alpha]]]-PolyLog[2,I Cos[\[Alpha]]+Sin[\[Alpha]]]. It gets your answer for pi/3 and pi/4
When I see fun/fascinating problems/solutions like this, I always wonder, do such cases ever "arise" in physics or any other quantifiable domain of human experience? This integral is so _cool..._ Does it "correspond" to anything we might run across? That "root 2" vs. "root 3" thing going on -- that must correspond to some physical or metaphysical struggle somewhere, no?!
can someone explain how to choose which parameter we use to make feyman's technique work? eg;- how to decide that we should use sin(alpha) as the parameter in 1:12 , he could,ve used something else. thank you!
QQ: how is using 2sin(alpha) better than just alpha in the Feynman trick? I don't see where it specifically made the problem any easier. You still end up with arctan integrals do I think you just get to the answer anyway no? Am I missing something in the middle of the derivation?
Nah I just wanted to try something different cuz of weird values of the parameter. I've never demonstrated something like this in previous videos so I wanted to do so here. More detail in a couple upcoming videos.
This is solvable using Feynman’s technique by replacing sqrt(2) with a second variable (I usually use “t”). This is probably not the best method though due to the extremely nasty integrals that come from it.
Hi,
2:41 : this integral is well known to be equal to zeta(2),
5:29 : quantity squared, 5:55 : fixed ,
7:32 : ok, I have to proof that as home work :)
15:57 : yes, verified with Fraction module of Python .
"ok, cool" : 1:51 , 2:58 , 4:01 , 5:59 , 11:04 , 12:58 , 13:34 , 14:20 ,
"terribly sorry about that" : 3:37 , 4:22 , 5:44 , 6:15 .
PS : I realize that your first name is very famous now!
I'm beautifully unusual, that's what my mum told me
those fractional equations triggered my anxiety
Given the frequency of use, you should present proofs of the Leibniz integral rule. (preferably posted not solely posted to instagram.) (Maybe also dominated convergence thm, too.)
In this case, you don't need to calculate the constant of integration as the result is the difference of 2 values
whatever I read from ur comment feels like illegal but it is correct 😅😂
There certainly are many other approaches to solve this integral, some maybe simpler or more straightforward as some have commented, but I love it when you play with the integral like a cat with the mouse... Not just eat it immediately!😂 It's an exercise that broadens one's vision. I enjoyed it greatly. Thanks a lot! Bahat shukriya!
@@trelosyiaellinika thanks mate. The Urdu was on point. Where ya from?
@@maths_505 Mmm... That's a difficult question 🤣 I am Armenian, born and bred in Lebanon, living in Armenia already for 37 years. I know several languages, including Arabic, Russian, Turkish, Greek, Farsi etc. Unfortunately, Urdu is not one of them. But I have many friends, ex colleagues, from Pakistan, Afghanistan, India etc 😄 So, I can often catch phrases in related languages...
Once I start to watch I can not stop understanding completely. Thank You for this Video.
Not me binging every integral on this channel
today is the day bro u finally proven ur not scared of pesky arithmetic which involves fractions. it is very inspiring 😅😂
Very smart solution. Thanks.
Loved this one. Especially the arithmethicc and your inability to write the number 8. Don't worry Kamaal, in 1st grade I was really scared I wouldn't learn how to properly write it, but I eventually got the hang of it :)
Gorgeous parametrization!
I love watching your videos , thanks for effort you put in this videos to create them for us viewers
it's unnecessary to pick a third input for alpha, because you can just integrate from pi/3 to pi/4. C just cancels anyway.
Yes but there is a merit to solving the general problem.
int[pi/3,pi/4](-pi/2-a)da=(-a•pi/2-a^2/2)[a=pi/3,pi/4]
=-pi^2/8-pi^2/32+pi^2/6+pi^2/18
=pi^2/2•(1/3+1/9-1/16-1/4)
=pi^2/2•(4/9-5/16)
=pi^2/2•((64-45)/144)
=19pi^2/288
The fractions that popped out where quite a hell of calculation.
It can also be solved using dilogarithms and taking the logarithm of a complex number. Scary stuff! I prefer Kamaal's method here
You can prove zeta(2) = pi^2/6 with this general integral. f(a)=Integral ln(x^2+2ax+1)/x dx from 0 to 1. f'(a)=2*Int 1/(x^2+2ax+1) dx = 2/(sqrt(1-a^2))*(arctan((a+1)/sqrt(1-a^2))-arctan(a/sqrt(1-a^2))). Using arctan(x)+arctan(y)=arctan((x+y)/(1-xy)). f'(a)= 2/sqrt(1-a^2) * arctan(sqrt((1-a)/(1+a)))
So f(a) = 2 Int arctan(sqrt((1-a)/(1+a)))/sqrt(1-a^2) da + C
sqrt((1-a)/(1+a))=t, (1-t^2)/(1+t^2)=a, da =(-4t)/(1+t^2)^2 dt, 1/sqrt(1-a^2)= (1+t^2)/(2t)
f(a)= -4 Int arctan(t)/(1+t^2) dt = -2 Int 2 * arctan(t) * (arctan(t))' dt. Using (f^2)' = 2 * f * f', f=f(x)
f(a)=-2*arctan(t)^2+C = -2 * (arctan(sqrt((1-a)/(1+a)))^2 + C
Now f(1) = C = 2 * Int ln(1+x)/x dx from 0 to 1, using series for ln(1+x), f(1)=2*Sum (-1)^(k+1)/(k^2), k 0 to 1, = zeta(2)=C,
f(-1)= 2 * Int ln(1-x)/x dx = -2*zeta(2)= -2arctan(infinity)^2 + zeta(2), arctan(infinity)=pi/2, solving for zeta(2), -3*zeta(2)=-pi^2/2, zeta(2)=pi^2/6
1:28 why did you put the sin alpha?
This adding another variable is called " Feynman Method "
@@premdeepkhatri1441
Why putting sin alpha not alpha?
That's the beauty of the trick. It suits well with the x√2 & x√3 for values of sin(α) when α is π/4 & π/3... It needs an eye to see it and a little bit of imagination... I certainly enjoyed this integral although it is relatively simple thanks to Feynman!
@@jejnsndn in order to endure lesser difficulty while solving the problem.
Mathematica (assuming alpha is between 0 and pi/2) computes the integral I(alpha) in terms of two polylog sub 2's: -PolyLog[2,-I Cos[\[Alpha]]+Sin[\[Alpha]]]-PolyLog[2,I Cos[\[Alpha]]+Sin[\[Alpha]]]. It gets your answer for pi/3 and pi/4
Yes this is the good method to make quick answer
Way Cool! I've been working on my 8s as well recently moving to the two stacked balls approach. So stick to your balls and work on your 4s!
When I see fun/fascinating problems/solutions like this, I always wonder, do such cases ever "arise" in physics or any other quantifiable domain of human experience? This integral is so _cool..._ Does it "correspond" to anything we might run across? That "root 2" vs. "root 3" thing going on -- that must correspond to some physical or metaphysical struggle somewhere, no?!
can someone explain how to choose which parameter we use to make feyman's technique work?
eg;- how to decide that we should use sin(alpha) as the parameter in 1:12 , he could,ve used something else.
thank you!
Since the result is the difference of two integrals, the C cancel out and you didn’t need to find out the C (easier evaluation is always a W)
QQ: how is using 2sin(alpha) better than just alpha in the Feynman trick? I don't see where it specifically made the problem any easier. You still end up with arctan integrals do I think you just get to the answer anyway no? Am I missing something in the middle of the derivation?
Nah I just wanted to try something different cuz of weird values of the parameter. I've never demonstrated something like this in previous videos so I wanted to do so here. More detail in a couple upcoming videos.
Love your videos 😊
Thanks
Nice!! 15:31💥💥💥💥😂
28🌜
05:29 It should be (x-sin(α))^2
Actually you didn't need to calculate the initial condition of I(a),because you wanted a difference
This is solvable using Feynman’s technique by replacing sqrt(2) with a second variable (I usually use “t”). This is probably not the best method though due to the extremely nasty integrals that come from it.
I believe that you don’t have to calculate the constant, since you’re taking a difference
@@vancedforU I didn't want to skip it as it solves the more general problem of not having a difference of logs.
Also if you have the chance to invoke the Basel problem why would any mortal avoid it🥰
Today I realized I'm a nerd, I laugh with some guy solving integrals, who's strugling to write numer 8. And i have absolutly no problem with that.
BABE WAKE UP SNOWMAN 8 ARC IS OVER
Share in the community the desire to post geomtry videos
19π²/28🌜
For my thesis, somehow 1 + 1 = 3. Sigh
15:30 Pacman
Actually that's close to how we write 5 in Bengali, on eof the many languages spoken in India. It's also the national language of Bangladesh.
almost half of the video is just the simple algebra lol
It was the biggest struggle I've faced so far
Love you
Will you stop saying "Ok, cool" repeatedly. The solution, as usual, is unnecessarily complex