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
your integral is from 0 to infinity. Ideally we choose a contour which includes a path over which u wanna integrate. If it was the integral from -inf to inf we would take a semicircular contour in 1st and 2nd quadrants
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)
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)
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?
@@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
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.
„2 is an even number the last time I checked“ XD
This truly is complex.
What an amusing video, nevertougth contour integration had such elegance when solving integrals, love it
thank you for this wonderful contour integration
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
3:51 yes bro, we like it thicc
the elusive contour integration video!
ah yes, the complex realm
Hi!
Can someone explain to me why at 3:24 we're considering only the first quandart?
To decrease number of poles
@@AliAkl-un2ys ???
your integral is from 0 to infinity. Ideally we choose a contour which includes a path over which u wanna integrate. If it was the integral from -inf to inf we would take a semicircular contour in 1st and 2nd quadrants
Bro can you take a video on contour integration problems. I am confused on what contour to take,
Specially in the cases of double circle requirements
4:11 how do you know z1 is the one in that quadrant?
@@egg10104 convert it into polar form
Okay, cool!
Couldnt you solve that integral with partial fractions? Great video!
I'd assume so, but that's less fun!
@@jamiepianistexactly
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)
Beautiful result 🎉🎉🎉
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?
Kamaal what's the source do you have to get those crazy integrals 😮
§ f-¹(f') vibes
More video ideas 😂
😮 i just realised there is an symbol like this in my keyboard for more than 2+years
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
OK Cool! 😃
They weren't wrong when naming it the COMPLEX plane
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 👍🏼
If we talk about comments, then it is not so easy to solve equations of the type (z^2-z1)(z^2-z2) = 0.
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 😅