Maths 505 sounds like he's in a really good mood today, and that makes me happy. :) Even though I watched this video on qn's channel, I'll watch the whole thing again, in the hopes that the algorithm gods bless Maths 505. So yeah, devious integral as always, bro. Keep it up. 💪😎👌
Watched it at Q's house, None the less, I feel the combination of both feynmann and contour was the most elegant, but both of you made a tremendous job! keep it up.
this is cool, I'm not happy with the argument at 13:13 though. The idea is correct, but it's argued weirdly I think. The integral doesn't immediately say we have a limit of I. But it says that limsup=-pi/2, in other words it doesn't blow up to infinity. So c2=0 (and now we can conclude that I has a limit as a bonus)
That's a nice argument as well. However, I invoked Dirichlet's convergence theorem to provide a basis of convergence. The rest followed due to the cosine term which ensured convergence didn't break down over the t-domain.
Smart choice of tools used to implement a strategy that results in an elegant solution to this integral. You have an impressive toolbox of skills & techniques to solve these exotic integrals.
I found another approach to this integral: using Fourier transform. The integral can be seen as the inverse Fourier transform, taken at 0, of sin(x)/x ⋅ sin(x)/x ⋅ 1/(1+x²) = ℱ{½ rect(x)} ⋅ ℱ{½ rect(x)} ⋅ ℱ{exp(-|x|)}. Because of the duality of product and convolution this leads to the result basically be ½ rect * ½ rect * exp(-|•|) taken at 0. The leftmost convolution gives a "witch hat" like function, which should then be convolved with with exp(-|•|). That we only need the value at 0 of this makes the calculation easier (we don't have to consider several cases).
Similarly, you could use Parseval's identity, with one function being the squared sinc and the other function being 1/(x^2+1), which turns into a similarly simple integral of e^-|x| times a triangle function, constants not withstanding.
German guy called flammable math caculated this integral and if i remember correctly also with Leibniz rule for differentiation under integral sign It is good that you recorded it beause it is better to watch you than that German guy Integrand is even so we have int(2sin^2(x)/(x^2+1),x=0..infinity) Now define function (1-cos(2tx))/(x^2(x^2+1)) and calculate Laplace transform It will probably lead us to double integral
@@maths_505 I dont like his style of recording videos He used complex analysis and Laplace transform and Leibniz rule for integration under integral sign
Orange on black and it is not even Halloween! Even with all that trickery.
Maths 505 sounds like he's in a really good mood today, and that makes me happy. :)
Even though I watched this video on qn's channel, I'll watch the whole thing again, in the hopes that the algorithm gods bless Maths 505.
So yeah, devious integral as always, bro. Keep it up. 💪😎👌
Watched it at Q's house, None the less, I feel the combination of both feynmann and contour was the most elegant, but both of you made a tremendous job! keep it up.
love the content, love the integrals!
For the seccond integral you can use Lobachevsky's formulae if you multiply and devide by x^2
Yes, I was going to mention that. I'm glad that somebody noticed
this is cool, I'm not happy with the argument at 13:13 though. The idea is correct, but it's argued weirdly I think. The integral doesn't immediately say we have a limit of I. But it says that limsup=-pi/2, in other words it doesn't blow up to infinity. So c2=0 (and now we can conclude that I has a limit as a bonus)
That's a nice argument as well.
However, I invoked Dirichlet's convergence theorem to provide a basis of convergence. The rest followed due to the cosine term which ensured convergence didn't break down over the t-domain.
Smart choice of tools used to implement a strategy that results in an elegant solution to this integral. You have an impressive toolbox of skills & techniques to solve these exotic integrals.
Clever and thorough.
Thanks for your knowledge and time.
To show c_2 = 0 say that the integral is BOUNDED, not just convergent.
10:18
I'(0)=-pi/2
I(0)=pi/2
I(t)=Ae^-t +Be^t
I'(t)=-Ae^-t +Be^t
pi/2=A+B
-pi/2=-A+B
B=0,A=pi/2
I(0)=pi/2 e^-t
I found another approach to this integral: using Fourier transform.
The integral can be seen as the inverse Fourier transform, taken at 0, of sin(x)/x ⋅ sin(x)/x ⋅ 1/(1+x²) = ℱ{½ rect(x)} ⋅ ℱ{½ rect(x)} ⋅ ℱ{exp(-|x|)}. Because of the duality of product and convolution this leads to the result basically be ½ rect * ½ rect * exp(-|•|) taken at 0. The leftmost convolution gives a "witch hat" like function, which should then be convolved with with exp(-|•|). That we only need the value at 0 of this makes the calculation easier (we don't have to consider several cases).
Similarly, you could use Parseval's identity, with one function being the squared sinc and the other function being 1/(x^2+1), which turns into a similarly simple integral of e^-|x| times a triangle function, constants not withstanding.
unreal .much appreciated my g
For the first part, wouldn't it have been much quicker to write sin^2(x) = 1/2*(1-cos(2x)), which could bypass the integration by part?
What is this writing board software in the video?
It seems very helpful for a teacher.
German guy called flammable math caculated this integral and if i remember correctly also with Leibniz rule for differentiation under integral sign
It is good that you recorded it beause it is better to watch you than that German guy
Integrand is even so we have int(2sin^2(x)/(x^2+1),x=0..infinity)
Now define function (1-cos(2tx))/(x^2(x^2+1)) and calculate Laplace transform
It will probably lead us to double integral
Yup...I didn't like his solution so I came up with this much better one
@@maths_505 I dont like his style of recording videos
He used complex analysis and Laplace transform and Leibniz rule for integration under integral sign
@@holyshit922 oh I enjoyed his presentation but the solution put me off
Wanna do the integral of the general quadratic √(ax^2 + bx +c)? It's more of an excercise in algebra, at least the way I did it.
This is too hard....