@@circuitcraft2399 (x,y): (0,0) -> (1,1) (r,θ): (1,0) -> (secθ,π/4) (This is just half the region, but the integrand is symmetric over y=x, so just double.)
When you put y=x and integrate from 0 to x, can you use polar coordinates x^2+y^2=r^2 and substitute dxdy by rdrdtheta. Anyway, Thank you for your featured effort. This channel is the best. 😍
6:20 You can get rid of this nasty complex number mess just by using the logarithm properties: log(t²-1) - log(-2) = log((t²-1)/(-2))= log((1-t²)/2)=ln(1-t²) - ln 2 what is real number because 0
Calculation of integral of log(1-t²)/(1+t²) may be simplified by Feynman's trick: log(2-x•(1+t²))/(1+t²) Differentiating by 'x' will get -1/(2-x•(1+t²)) What can be easily integrated
im not really sure if im being dumb (i tend to be dumb) but is the function even defined in the reals over the interval of integration? Or are we just addressing the integral entirely in the complex domain
If we do polar coordinates onto that double integral, doesn’t that make our integral equal to -pi/4*ln(2)? Is there something in this problem that prevents this from working?
This integrand is not continuous over the entire domain of integration: when x=y=1, the denominator is zero. It seems you can still use Fubini per the measure-theoretic description on WP, but you may have wanted to say something about measurability or just used Tonelli's theorem, instead of claiming continuity.
@@maths_505 Whoa, you remember me from my other comment, don't you? How nice of you to ask! I'm doing okay, overall, but I do have big news: I'm about to present my MS project next week! With regard to the math: The one-sided limit is definitely easier, and indeed you can use Fubini then if I'm not mistaken. Also, @circuitcraft2399, that is a good point and I didn't even notice that. However, this function is _always_ negative. A typical trick then is to just negate it, use Tonelli's theorem proper, then negate the result by linearity of Lebesgue integral.
Once again my boy πlog(2) loses the battle against the big G. Will he give up and surrender? Or will he come back stronger to defeat the devil constants? I guess we will have to wait for the next chapter of the monster integrals
The given interval is not in the domain of the square root. How does integration work with this in mind? You get a real answer to this integral without consideration of the square roots domain. That's a flaw in my view...
Mfw you didn't use Feynman. Also this log2 popping up with fractions of pi/2, pi/4 etc. makes me think there is some relation to the information integral which I mentioned a while back. Int(0,oo) - 1/(1 + exp(-x)) dx = - [log(1+exp(-oo)) - log(1+exp(0))] = log2 So all these integrals involving ratios of numbers to arc-trig functions seem to have something to do with the rate of binary information?
In base alle relazioni iperboliche,la funzione integranda è -arcth(1/√(2-x^2))/√(2-x^2)...il problema è che,nell'intervallo di integrazione 0-->1,la funzione arth non esiste...
@@takemyhand1988 Then explain how it is defined on -R keeping in mind that e^(iπ)=e^(-iπ). I think that it isn't continous neither defined (on -R) because the e^(ix) is not injective because e^(iπ)=e^(-iπ).
![ค้างคืนโรงเก็บกระสวยอวกาศร้างกลางทะเลทราย [Around The World Ep. 3]](http://i.ytimg.com/vi/E5bWZQhABJE/mqdefault.jpg)
"Thank you, the Netherlands for Virgil van Dijk" 😄
Am not a Liverpool fan myself, but my brother is, and he says the same.
Your integral content is very nice....and most difficult...only few have easy level...
I really thought you were going to go to polar coordinates when I saw the x^2+y^2
You basically just skip directly to the 11:00 mark if you do that
You're integrating over a square, so that wouldn't really work.
@@circuitcraft2399 (x,y): (0,0) -> (1,1) (r,θ): (1,0) -> (secθ,π/4)
(This is just half the region, but the integrand is symmetric over y=x, so just double.)
@@circuitcraft2399You can do it using Jacobian
You can integrate over a square in polar coordinates
If x is between 0 and 1, then x^2-2 is less than -1 so sqrt(x^2-1) is imaginary
Amazing broooo, great to see this on the tubes ❤❤
“just cancel out the sqrt(x^2 - 2) on both sides”
For the integral on 9:40, you can use Weierstrass substitution.
As soon as I saw the case f(g(x))/g(x) in an integral I instantly thought of Feynman's Technic.
In fact, I was wrong.
You can probably do use Feynman'S Technique by some adaptation
His starting structure is basically what you would get if you applied Feynman’s trick I believe.
When you put y=x and integrate from 0 to x, can you use polar coordinates x^2+y^2=r^2 and substitute dxdy by rdrdtheta. Anyway, Thank you for your featured effort. This channel is the best. 😍
6:20 You can get rid of this nasty complex number mess just by using the logarithm properties:
log(t²-1) - log(-2) = log((t²-1)/(-2))=
log((1-t²)/2)=ln(1-t²) - ln 2
what is real number because 0
Calculation of integral of
log(1-t²)/(1+t²)
may be simplified by Feynman's trick:
log(2-x•(1+t²))/(1+t²)
Differentiating by 'x' will get
-1/(2-x•(1+t²))
What can be easily integrated
im not really sure if im being dumb (i tend to be dumb) but is the function even defined in the reals over the interval of integration? Or are we just addressing the integral entirely in the complex domain
No your not dumb I noticed this too!
No you're not dumb and I was pretty surprised too....the imaginary parts cancel out
Your not dumb and because of the domain issue, I don't even think this integral is Riemann integrable.
If we do polar coordinates onto that double integral, doesn’t that make our integral equal to -pi/4*ln(2)? Is there something in this problem that prevents this from working?
Never mind my region was wrong, we are integrating over the line y = x, not that circular region you drew. Very cool problem!
This integrand is not continuous over the entire domain of integration: when x=y=1, the denominator is zero. It seems you can still use Fubini per the measure-theoretic description on WP, but you may have wanted to say something about measurability or just used Tonelli's theorem, instead of claiming continuity.
Tonelli's theorem doesn't apply, since some values are negative (e.g. x = t = 0.5)
Yeah you're right...another option is to invoke one sided limits.
Btw how you been bro? How's life going?
@@maths_505 Whoa, you remember me from my other comment, don't you? How nice of you to ask!
I'm doing okay, overall, but I do have big news: I'm about to present my MS project next week!
With regard to the math: The one-sided limit is definitely easier, and indeed you can use Fubini then if I'm not mistaken. Also, @circuitcraft2399, that is a good point and I didn't even notice that. However, this function is _always_ negative. A typical trick then is to just negate it, use Tonelli's theorem proper, then negate the result by linearity of Lebesgue integral.
@@MyOldHandleWasWorse woah! Good luck with the project mate....you got this 😎😎🔥🔥
Once again my boy πlog(2) loses the battle against the big G. Will he give up and surrender? Or will he come back stronger to defeat the devil constants? I guess we will have to wait for the next chapter of the monster integrals
Due to the way you are trying to write θ, I can tell that you are a fan of mathematics mi
Fan of mathematics what?
The given interval is not in the domain of the square root. How does integration work with this in mind? You get a real answer to this integral without consideration of the square roots domain. That's a flaw in my view...
😁😁😁😁😁 11:05-11:15
Mfw you didn't use Feynman.
Also this log2 popping up with fractions of pi/2, pi/4 etc. makes me think there is some relation to the information integral which I mentioned a while back.
Int(0,oo) - 1/(1 + exp(-x)) dx = - [log(1+exp(-oo)) - log(1+exp(0))] = log2
So all these integrals involving ratios of numbers to arc-trig functions seem to have something to do with the rate of binary information?
Perhaps.
It's not exactly a topic I have much knowledge of. But your argument does seem enticing.
That's in our board's syllabus
In base alle relazioni iperboliche,la funzione integranda è -arcth(1/√(2-x^2))/√(2-x^2)...il problema è che,nell'intervallo di integrazione 0-->1,la funzione arth non esiste...
Can there be a another method rather going through double integral ??
In my view the integral doesn't exist in the real number system. The square root isn't defined on the given interval of integration
Having looked at the options, this is the easiest path in all honesty
Make a video on borwein integrals
Wow
happy pi day bhaiya
What is C, Caplan's constant?
Catalans Constant
Please put a vedio about who can participate in mit bee . I am interested to participate in it from India
Bro I think it's pretty much exclusive to MIT students
Wait: cotan x = tan (1/x) ? or is it rather equal to 1/tan (x)?
No bro
Arctan(x)=arccot(1/x)
somehow its not imaginary and somehow everything vanished i dont know
Happyy π dayy
I really can't get it why don't we use polar coordinates at some point
Oh we can do so here but I just wanted to raw dog this one using symmetry.
Let’s goo
I wanted to substitute u = 1/sqrt(x^2-2), then realized u would be imaginary… but the answer ended up real?!
The Log isn't defined on -R
In complex plane, it is
@@takemyhand1988 Not on -R because it ain't continuous even on the complex plane and even the principle log. Because e^(iπ)=e^(-iπ).
@@Dany161-w1i complex logarithm exists. The problem is it isn't injective so you can't inverse the function
@@takemyhand1988 Then explain how it is defined on -R keeping in mind that e^(iπ)=e^(-iπ). I think that it isn't continous neither defined (on -R) because the e^(ix) is not injective because e^(iπ)=e^(-iπ).
@@Dany161-w1i the function exists but isn't injective. Just like how arc trigonometry exists. You can limit the domain to principal argument