Another, more elementary, method would be to go off of the addition formula of arctangents [i.e. arctan(a) + arctan(b)=arctan((a + b)/(1 - a b)), for ab
I(a) =arctan ( 2/(a+x^2)) dx from 0 to infinity I'(a) = - int 0 to inf, 2 /(4+a^2 + 2ax^2 + x^4) dx By Glasser's master theorem or by general long method, we can deduce that, int 0 to inf, 1/(x^4 + bx^2 +c) dx = 1/2 * c^(-3/4) pi /(2 + b/sqrt(c)) We use it here, So, I'(a)= - pi/sqrt2 1/{sqrt(4+a^2) * sqrt(a + sqrt(4+a^2)} I(1) - I(0)= int 0 to 1, - pi/sqrt2 1/{sqrt(4+a^2) sqrt(a + sqrt(4+a^2)} da Now take 2tany = a = - pi/2 int 0 to arctan(1/2), secy /sqrt(secy + tany) dy = - pi/2 int 0 to 1 arctan(1/2), secy(secy+tany) /(secy +tany)^(3/2) dy = [ pi (secy + tany)^(-1/2)] first put arctan(1/2) then 0 = pi/sqrt(golden ratio) - pi I(1) - I(0)= pi/sqrt (golden ratio) - pi I(0)= int 0 to inf, arctan (2/x^2) dx G(a) = int 0 to inf, arctan (a /x^2) dx G'(a) = int 0 to inf, x^2 /(a^2 + x^4) dx set x=1/t G'(a) = int 0 to inf, 1/(1+a^2 x^4) dx G'(a) =int 0 to inf, 1/a^2 1/{(1/a^2) + x^4} dx Use the result again again, G'(a) = pi/(2sqrt2) 1/sqrt(a) G(2) = pi I(0)=pi I(1) - I(0) = pi /sqrt(golden ratio) - pi I(1) = pi /sqrt(golden ratio) - pi + pi = pi /sqrt(golden ratio)
@@aravindakannank.s. I know you're trying to do a joke, but just so everyone else knows (because I was also confused as hell) "Anal" here actually comes from anal-retentive which means extremely or overly neat, careful, or precise, even pedantic. You could also handwave it as being an abbreviation of analytical. Now you know
i tried your approach, it results in a nasty denominator since the argument is then 1 + (k^2/(1+x^2)^2 which does not get eliminated with the remaining differential expression
@@alexander_elektronik I'(k) = integral 0 to infinity (1+x^2)dx/[(1+x^2)^2+k^2] Now notice that [(1+x^2)^2+k^2] = (1+ik +x^2)*(1 - ik + x^2). Then rewrite (1+x^2) as (1+x^2) = 0.5*(1+ik+x^2) + 0.5*(1-ik+x^2). See where this is going?
The z variable lies on the imaginary axis so z=iy. Since the radius of the circle is growing without bound, z takes on values 0 (lower limit) to i♾️ (upper limit)
That's happen when you are integration by parts hater This integral can be calculated in elementary way First integration by parts with du=dx , v = arctan(2/(1+x^2)) After integration by parts Substitution u = sqrt(5)/x add to the integral before substiturion and divide by two Another substitution u - sqrt(5)/u = v and finally we will have arctan
the phi jumpscares are getting out of hand
Hi,
12:15 : note that tan^-1 (-2) = 2 tan^-1 phi where phi is the golden ratio.
"ok, cool" : 0:17 , 2:42 , 3:07 , 4:35 , 5:40 , 6:36 , 7:01 , 8:43 , 9:49 , 12:15 ,
"terribly sorry about that" : 14:47 .
❤❤❤❤❤❤
Wake up babe, new Maths 505 dropped.
This truly is complex.
„2 is an even number the last time I checked“ XD
What an amusing video, nevertougth contour integration had such elegance when solving integrals, love it
Another, more elementary, method would be to go off of the addition formula of arctangents [i.e. arctan(a) + arctan(b)=arctan((a + b)/(1 - a b)), for ab
thank you for this wonderful contour integration
ah yes, the complex realm
I(a) =arctan ( 2/(a+x^2)) dx from 0 to infinity
I'(a) = - int 0 to inf, 2 /(4+a^2 + 2ax^2 + x^4) dx
By Glasser's master theorem or by general long method, we can deduce that,
int 0 to inf, 1/(x^4 + bx^2 +c) dx = 1/2 * c^(-3/4) pi /(2 + b/sqrt(c))
We use it here,
So,
I'(a)= - pi/sqrt2 1/{sqrt(4+a^2) * sqrt(a + sqrt(4+a^2)}
I(1) - I(0)= int 0 to 1, - pi/sqrt2 1/{sqrt(4+a^2) sqrt(a + sqrt(4+a^2)} da
Now take 2tany = a
= - pi/2 int 0 to arctan(1/2), secy /sqrt(secy + tany) dy
= - pi/2 int 0 to 1 arctan(1/2), secy(secy+tany) /(secy +tany)^(3/2) dy
= [ pi (secy + tany)^(-1/2)] first put arctan(1/2) then 0
= pi/sqrt(golden ratio) - pi
I(1) - I(0)= pi/sqrt (golden ratio) - pi
I(0)= int 0 to inf, arctan (2/x^2) dx
G(a) = int 0 to inf, arctan (a /x^2) dx
G'(a) = int 0 to inf, x^2 /(a^2 + x^4) dx
set x=1/t
G'(a) = int 0 to inf, 1/(1+a^2 x^4) dx
G'(a) =int 0 to inf, 1/a^2 1/{(1/a^2) + x^4} dx
Use the result again again,
G'(a) = pi/(2sqrt2) 1/sqrt(a)
G(2) = pi
I(0)=pi
I(1) - I(0) = pi /sqrt(golden ratio) - pi
I(1) = pi /sqrt(golden ratio) - pi + pi
= pi /sqrt(golden ratio)
@@SussySusan-lf6fk nice work
Wow! Very nice Susan 😊
And friend 505 how are you doing btw?
the elusive contour integration video!
3:51 yes bro, we like it thicc
§ f-¹(f') vibes
More video ideas 😂
😮 i just realised there is an symbol like this in my keyboard for more than 2+years
Couldnt you solve that integral with partial fractions? Great video!
I'd assume so, but that's less fun!
@@jamiepianistexactly
Hi!
Can someone explain to me why at 3:24 we're considering only the first quandart?
To decrease number of poles
Okay, cool!
because I like being anal about maths there is no residue of f(z) it's the residue of the 1-form f(z)dz, plus is sounds cooler to call it that
you mean analysis right? 😂
right? 😅
@@aravindakannank.s. I know you're trying to do a joke, but just so everyone else knows (because I was also confused as hell)
"Anal" here actually comes from anal-retentive which means extremely or overly neat, careful, or precise, even pedantic. You could also handwave it as being an abbreviation of analytical. Now you know
@@stefanalecu9532 👍🏼
Beautiful result 🎉🎉🎉
Can we also use OP Feynman's trick for this one> I(k) = integral 0 to infinity arctan(k/1+x^2) dx and differentiate.
i tried your approach, it results in a nasty denominator since the argument is then 1 + (k^2/(1+x^2)^2 which does not get eliminated with the remaining differential expression
@@alexander_elektronik
I'(k) = integral 0 to infinity (1+x^2)dx/[(1+x^2)^2+k^2]
Now notice that [(1+x^2)^2+k^2] = (1+ik +x^2)*(1 - ik + x^2).
Then rewrite (1+x^2) as
(1+x^2) = 0.5*(1+ik+x^2) + 0.5*(1-ik+x^2).
See where this is going?
I have solved this with Feynman's technique... Do you want to see?? I comment??
@@SussySusan-lf6fk I've solved it too, thanks
@@SussySusan-lf6fk yes please
If we talk about comments, then it is not so easy to solve equations of the type (z^2-z1)(z^2-z2) = 0.
4:11 how do you know z1 is the one in that quadrant?
@@niom-nx7kb convert it into polar form
Kamaal what's the source do you have to get those crazy integrals 😮
They weren't wrong when naming it the COMPLEX plane
8:04 how do you determine the limits of integration of I2 from the contour integral ?
The z variable lies on the imaginary axis so z=iy. Since the radius of the circle is growing without bound, z takes on values 0 (lower limit) to i♾️ (upper limit)
OK Cool! 😃
What, why and how?! - Anyway. Love your videos!
@@F-S. Thanks bro but what exactly were the questions 😂?
Why am I already giving up at 1:13 😭
That's happen when you are integration by parts hater
This integral can be calculated in elementary way
First integration by parts with du=dx , v = arctan(2/(1+x^2))
After integration by parts
Substitution u = sqrt(5)/x
add to the integral before substiturion and divide by two
Another substitution u - sqrt(5)/u = v
and finally we will have arctan
Please stop saying 'OK cool'
@@jpf119 i can't
Maths 505
Heritage
Please never stop saying 'OK cool'
@@maths_505 Missed opportunity to reply with "OOOkay cool"
there is a t shirt available which is written in it bro 😂
i just checked it is missing 😅