An amazing hyperbolic integral with a surprisingly beautiful solution involving one of our favourite functions and a couple other along the way too. AND ofcourse a missing negative sign😂
How would you feel about adding some graphs to these vids (while still using standard methods of solving)? I think a lot of us want to build our geometric intuition, while simultaneously building the skills you share with us.
@@maths_505 I was thinking you could do some simple stuff in manim (the thing made by 3blue1brown) but it's a little bit more work than might be necessary.
Using integral of x^s/(1+e^x) from 0 to infinity =Eta(s+1)Gamma(s+1) for s > -1, one can deduce that your integral is equal to 1/(2sqrt(2)) * Eta(1/2)Gamma(1/2) =1/(2sqrt(2)) * Eta(1/2)*Sqrt(Pi).
I feel like in the same way that you sometimes invoke the equivalent forms of the Beta function, you could also invoke the Bose integral, and it’s cousin that you use here.
Ah knew it involved series, just differentiating totally slipped my mind 11:10 19th century mathematicians on their way to create the goofiest functions in all of the greek alphabet
I prefer the yellow result cuz I define zeta(1/2) by its relationship with the eta function, so for me we’re getting backward when expressing eta in term of zeta
Hey, so I tried using cosh definition, and I defined the whole thing as int((x^2 * e^2x^2)/(1+e^2x^2)^2) and when I tried by- parts on this separating x^2 from the rest of the function, I got a dumb answer like -2. Where am I going wrong with my method?
A me risulta sqrt(pi)/2(1-1/3sqrt3+1/5sqrt5-1/7sqrt7+.....)....(sqrtpi)/2B(3/2)... =0,766146,la mia soluzione è corretta... Nella tua soluzione non capisco cos'è Z(1/2)?
You said we can throw the k=0 term out because it's just zero but right after the next step the k=0 term appears to prove problematic with the 1/√(2k) term. What's up with that?
Can you take e^-2u and add 0 in the numerator to get I = Int(0,oo of (1 + e^-2u - 1)*u^1/2/(1+e^-2u) du) = **Int(0,oo of e^-2u*u^1/2/(1+e^-2u) du)** - ***Int(0,oo of u^1/2/(1+e^-2u)^2 du)*** Then I1 = Integral in * I2 = Integral in ** And use integration by parts in both cases. u1 -> u^1/2 dv1 -> e^-2u/(1+e^-2u) du => v1 -> -1/2 ln(e^-2u) : . I1 = Int(0,oo of u1dv1) u2 -> u^1/2 => du2 -> -1/2 u^-1/2 dv2 -> 1/(1+e^-2u)^2 du Substitute t -> e^u => u -> lnt => du -> dt/t To get u2 -> (lnt)^1/2 dv2 -> 1/(1+1/t^2)^2 * dt/t = t^2/t(t^2+1)^2 dt = t/(t^2+1)^2 dt = 2t/2(t^2+1)^2 dt => v2 = 1/(t^2+1) : . I2 = Int(0,oo of u2dv2) Use IBP ==> I = I1 - I2
I think exchanging the sum and integral is illegal here because the series is not absolutely convergent at x=0. Another way too see it, do the same problem but with cosx instead of x^2 (integral of cosx/(coshx)^2) it will lead to a divergent series. It's confusing 🤔
Have you seen that 3blue1brown is one again running the Summer of Maths Exposition for MathTubers? Google it, I think it will be great exposure for your channel to enter.
Recently found this channel mate and I'm thinking of binging your videos while eating some food. I do love me some cheeky integral calculus.
So this is what UK maths teachers do in their spare time
How would you feel about adding some graphs to these vids (while still using standard methods of solving)? I think a lot of us want to build our geometric intuition, while simultaneously building the skills you share with us.
Sounds like a nice idea. I'll see what I can do
@@maths_505 I was thinking you could do some simple stuff in manim (the thing made by 3blue1brown) but it's a little bit more work than might be necessary.
@@maths_505 Thanks! Really appreciate the work you do on here.
@@maths_505 Someone suggested 3b1b's manim program, but you could just show the graph with desmos.
Lol I actually plotted the graph in geogabra before starting the video. This seems like a great idea
Using integral of x^s/(1+e^x) from 0 to infinity =Eta(s+1)Gamma(s+1) for s > -1, one can deduce that your integral is equal to 1/(2sqrt(2)) * Eta(1/2)Gamma(1/2) =1/(2sqrt(2)) * Eta(1/2)*Sqrt(Pi).
Love your content dude! Can’t wait to see you at 100k subscribers
I’m a third year math major and just for a second my first thought when I saw this integral was to cancel x^2😭😭
Lamo
5:05 thanks to this damn exponential factor 🤣
I feel like in the same way that you sometimes invoke the equivalent forms of the Beta function, you could also invoke the Bose integral, and it’s cousin that you use here.
Amazing and exciting integral. You are talented for hunting such integrals to solve. Thank you very much for your fascinating videos.
Ah knew it involved series, just differentiating totally slipped my mind
11:10 19th century mathematicians on their way to create the goofiest functions in all of the greek alphabet
I'd love to see an integral where dominated convergence DOESN'T apply and you have some other creative solution.
As in a divergent integral???
@Maths 505 There's gotta be a convergent integral where the exchange of sum and integral isn't allowed. That's what I meant, sorry.
man u r grown so much
but still couldn't master writing that peski summation symbol (capital sigma) 😂
ur viewer from future ❤
I prefer the yellow result cuz I define zeta(1/2) by its relationship with the eta function, so for me we’re getting backward when expressing eta in term of zeta
My wife is getting really jealous considering that I am more excited by integrals than her.
a wild hyperbolic integral has appeared!
Oh wow I was genuinely confused about the series for some time. I didn't understand how you didn't get a minus from the exponential and the rational.
Hey, so I tried using cosh definition, and I defined the whole thing as int((x^2 * e^2x^2)/(1+e^2x^2)^2) and when I tried by- parts on this separating x^2 from the rest of the function, I got a dumb answer like -2. Where am I going wrong with my method?
A me risulta sqrt(pi)/2(1-1/3sqrt3+1/5sqrt5-1/7sqrt7+.....)....(sqrtpi)/2B(3/2)... =0,766146,la mia soluzione è corretta... Nella tua soluzione non capisco cos'è Z(1/2)?
great video, keep it up!
You said we can throw the k=0 term out because it's just zero but right after the next step the k=0 term appears to prove problematic with the 1/√(2k) term. What's up with that?
The one in yellow. Much simpler.
Can you take e^-2u and add 0 in the numerator to get
I = Int(0,oo of (1 + e^-2u - 1)*u^1/2/(1+e^-2u) du) = **Int(0,oo of e^-2u*u^1/2/(1+e^-2u) du)** - ***Int(0,oo of u^1/2/(1+e^-2u)^2 du)***
Then I1 = Integral in *
I2 = Integral in **
And use integration by parts in both cases.
u1 -> u^1/2
dv1 -> e^-2u/(1+e^-2u) du => v1 -> -1/2 ln(e^-2u)
: . I1 = Int(0,oo of u1dv1)
u2 -> u^1/2 => du2 -> -1/2 u^-1/2
dv2 -> 1/(1+e^-2u)^2 du
Substitute t -> e^u => u -> lnt => du -> dt/t
To get
u2 -> (lnt)^1/2
dv2 -> 1/(1+1/t^2)^2 * dt/t = t^2/t(t^2+1)^2 dt = t/(t^2+1)^2 dt = 2t/2(t^2+1)^2 dt => v2 = 1/(t^2+1)
: . I2 = Int(0,oo of u2dv2)
Use IBP ==> I = I1 - I2
I think exchanging the sum and integral is illegal here because the series is not absolutely convergent at x=0. Another way too see it, do the same problem but with cosx instead of x^2 (integral of cosx/(coshx)^2) it will lead to a divergent series. It's confusing 🤔
bruh it is legal
3:57 where did this come from?! It just appeared without justification, nvm! 10:18
just to annoy the purists i vote each video ends with the decimal expansion :)
Will you ever do an antiderivative problem
They're quite rare on the channel. Unless it's an integration bee problem or a DE
Guess: contour integration?
Nope
Have you seen that 3blue1brown is one again running the Summer of Maths Exposition for MathTubers? Google it, I think it will be great exposure for your channel to enter.
Ofcourse!
How could one miss it😍
Oh and I started teaching myself analytic number theory cuz I think that would be fun to cover on the channel too