A beautifully unusual integral calculation

Рет қаралды 5,021

Күн бұрын

Пікірлер: 56

@CM63_France 5 ай бұрын

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!

@GeraldPreston1 5 ай бұрын

I'm beautifully unusual, that's what my mum told me

@xizar0rg 5 ай бұрын

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.)

@kingzenoiii 5 ай бұрын

those fractional equations triggered my anxiety

@EtienneSturm1 5 ай бұрын

In this case, you don't need to calculate the constant of integration as the result is the difference of 2 values

@aravindakannank.s. 5 ай бұрын

whatever I read from ur comment feels like illegal but it is correct 😅😂

@trelosyiaellinika 5 ай бұрын

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 5 ай бұрын

@@trelosyiaellinika thanks mate. The Urdu was on point. Where ya from?

@trelosyiaellinika 5 ай бұрын

@@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...

@beautyofmath6821 5 ай бұрын

Not me binging every integral on this channel

@aravindakannank.s. 5 ай бұрын

today is the day bro u finally proven ur not scared of pesky arithmetic which involves fractions. it is very inspiring 😅😂

@patricius6378 5 ай бұрын

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 :)

@premdeepkhatri1441 5 ай бұрын

Once I start to watch I can not stop understanding completely. Thank You for this Video.

@slavinojunepri7648 5 ай бұрын

Gorgeous parametrization!

@maxvangulik1988 5 ай бұрын

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.

@maths_505 5 ай бұрын

Yes but there is a merit to solving the general problem.

@maxvangulik1988 5 ай бұрын

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

@leroyzack265 5 ай бұрын

The fractions that popped out where quite a hell of calculation.

@PyarMatKaro 5 ай бұрын

It can also be solved using dilogarithms and taking the logarithm of a complex number. Scary stuff! I prefer Kamaal's method here

@xanterrx9741 5 ай бұрын

I love watching your videos , thanks for effort you put in this videos to create them for us viewers

@MrWael1970 5 ай бұрын

Very smart solution. Thanks.

@worldnotworld 5 ай бұрын

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?!

@jejnsndn 5 ай бұрын

1:28 why did you put the sin alpha?

@premdeepkhatri1441 5 ай бұрын

This adding another variable is called " Feynman Method "

@jejnsndn 5 ай бұрын

@@premdeepkhatri1441 Why putting sin alpha not alpha?

@trelosyiaellinika 5 ай бұрын

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!

@sandyjr5225 5 ай бұрын

@@jejnsndn in order to endure lesser difficulty while solving the problem.

@abc-nd2xt 5 ай бұрын

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

@Samanthipamuduni-l6k 4 ай бұрын

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!

@TheMartinbowes 5 ай бұрын

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!

@yoav613 5 ай бұрын

Nice!! 15:31💥💥💥💥😂

@jinkela7295 5 ай бұрын

05:29 It should be (x-sin(α))^2

@vitorbordini5246 5 ай бұрын

Actually you didn't need to calculate the initial condition of I(a),because you wanted a difference

@zunaidparker 5 ай бұрын

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?

@maths_505 5 ай бұрын

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.

@vancedforU 5 ай бұрын

I believe that you don’t have to calculate the constant, since you’re taking a difference

@maths_505 5 ай бұрын

@@vancedforU I didn't want to skip it as it solves the more general problem of not having a difference of logs.

@lakshay3745 5 ай бұрын

Also if you have the chance to invoke the Basel problem why would any mortal avoid it🥰

@orionspur 5 ай бұрын

28🌜

@lyyplayscube 5 ай бұрын

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)

@Calcprof 5 ай бұрын

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

@premdeepkhatri1441 5 ай бұрын

Yes this is the good method to make quick answer

@sozeran 5 ай бұрын

Love your videos 😊

@maths_505 5 ай бұрын

Thanks

@jejnsndn 5 ай бұрын

Share in the community the desire to post geomtry videos

@theblainefarm3310 5 ай бұрын

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.