best convergence test for integrals: we can switch the integration and differentiation operations because the integral converges, we knoe that because if it didnt then we wouldnt solve the integral in the first place
Sadly that doesn't work ... the integral converging is nowhere near sufficient. The usual theorem you use here is the dominated convergence theorem, which requires the integrand to be uniformly bounded (independent of the parameter, e.g. alpha in this case) by a function whose integral converges.
I calculated it with double integral Int(sqrt(x^2+2)/(1+y^2(x^2+2)),y=0..1) = arctan(sqrt(x^2+2)) Int(arctan(sqrt(x^2+2))/((x^2+1)*sqrt(x^2+2)),x=0..infinity)= Int(Int(sqrt(x^2+2)/(1+y^2(x^2+2)),y=0..1)*1/((x^2+1)*sqrt(x^2+2)),x=0..infinity)= Int(Int(sqrt(x^2+2)/(1+y^2(x^2+2)) * 1/(((x^2+1)*sqrt(x^2+2))) ,y=0..1),x=0..infinity)= Int(Int(1/((x^2+1)*(1+y^2(x^2+2))),y=0..1),x=0..infinity)= Int(Int(1/((x^2+1)*(1+y^2(x^2+2))),x=0..infinity),y=0..1) And it seems to be equivalent to the Leibniz rule
Very cool result. As a former instructor myself, I'd suggest that you include a link to the specific 'B roll' where you calculated those residues. I see you have link to the complex analysis lectures, but for beginners, the direct link can overcome a lot of frustration when you're not sure what you're looking at.
The only point where he mentioned residues was for int_{0}^{infty}dx/(x^2+1), but of course you don't actually need complex analysis for that - it's just yet another arctan integral, like all the rest in the video!
I don't understand why your limits of integration were from 0 to 1 (at the moment you integrate to find the function as it though a simple ODE)… despite the limits of integration of the first integral were from 0 to infinity. That is the only thing by which I can't sleep because of my overthinking lol
Hi Fellow overthinker here 😂 I(1) was my target case I(0) was my initial value So that's why I evaluated the integral w.r.t alpha from 0 to 1. The original integral was an integral w.r.t x from 0 to ♾️
best convergence test for integrals:
we can switch the integration and differentiation operations because the integral converges, we knoe that because if it didnt then we wouldnt solve the integral in the first place
Sadly that doesn't work ... the integral converging is nowhere near sufficient. The usual theorem you use here is the dominated convergence theorem, which requires the integrand to be uniformly bounded (independent of the parameter, e.g. alpha in this case) by a function whose integral converges.
Amazing Intregral. Congratulaions and thanks by solution.
Where do you find your integrals? Is there a textbook or set of textbooks you use?
when there is an arctan function in the numerator, fubini's theorem often works well.
Great video! Amazing as always.
This is an easy Ahmed integral.The real monster Ahmed integral has bound s from 0 to 1.The answer is (5/96)(pi)^2.
Already solved that one
Thank you for your featured effort.
Bro. If you don't write a book on integral. The gods of math won't be happy with you. 😂
It's mandatory at this point.
You can look the Cotex's integral up
oh wow i'm glad this is easier than ahmed's 😂
what program do you use for writing all of these solutions?
I calculated it with double integral
Int(sqrt(x^2+2)/(1+y^2(x^2+2)),y=0..1) = arctan(sqrt(x^2+2))
Int(arctan(sqrt(x^2+2))/((x^2+1)*sqrt(x^2+2)),x=0..infinity)=
Int(Int(sqrt(x^2+2)/(1+y^2(x^2+2)),y=0..1)*1/((x^2+1)*sqrt(x^2+2)),x=0..infinity)=
Int(Int(sqrt(x^2+2)/(1+y^2(x^2+2)) * 1/(((x^2+1)*sqrt(x^2+2))) ,y=0..1),x=0..infinity)=
Int(Int(1/((x^2+1)*(1+y^2(x^2+2))),y=0..1),x=0..infinity)=
Int(Int(1/((x^2+1)*(1+y^2(x^2+2))),x=0..infinity),y=0..1)
And it seems to be equivalent to the Leibniz rule
Very cool result. As a former instructor myself, I'd suggest that you include a link to the specific 'B roll' where you calculated those residues. I see you have link to the complex analysis lectures, but for beginners, the direct link can overcome a lot of frustration when you're not sure what you're looking at.
The only point where he mentioned residues was for int_{0}^{infty}dx/(x^2+1), but of course you don't actually need complex analysis for that - it's just yet another arctan integral, like all the rest in the video!
Ah, yet another victim to a classic-like integral, It's nice to see that we're starting to observe integral problems in the same manner.
I don't understand why your limits of integration were from 0 to 1 (at the moment you integrate to find the function as it though a simple ODE)… despite the limits of integration of the first integral were from 0 to infinity. That is the only thing by which I can't sleep because of my overthinking lol
Hi
Fellow overthinker here 😂
I(1) was my target case
I(0) was my initial value
So that's why I evaluated the integral w.r.t alpha from 0 to 1. The original integral was an integral w.r.t x from 0 to ♾️
And that's pretty much what Feynman's trick amounts to: turning an integral into an ODE
@@maths_505Got it. Tsm. Btw, I love your videos about crazy integrals!
Why is this integral famous?
mein könig
Noice
:D