Ah it's so awesome that you actually updated the video! Thanks for putting in the extra effort to make a new video for us, just goes to show your commitment. Gonna settle in for a nice watch now. Bravo!
Very interesting technique for integrating these types of integrals. Keyhole Contour and Cauchy’s Residue Theorem. The world of complex analysis is fascinating.
@ 10:46 The numerator should be epsilon^(4/3). The process for the parameterization for the smaller circle is the same as that for the larger circle, but with R replaced with epsilon.
Lots of thanks for you work, my friend! It helps me a lot! Actually, your Sigma isn't a branch cut, right? You have a branch point at the origin and a branch cut going from it to infinity. Your sigma is just a curve that doesn't cross the branch cut so your function stays single-valued.
For L2 integral you explained that z becomes |z|exp(2pi/3) due to the z^1/3 in the numerator. How do we account for the Z^2+1 in the denominator? What was the justification to ignore?
Yeah boi! 😎 But on a serious note myers and zunaid parker added a couple of notes that motivated me to remake the video and then upload it. You'll notice the changes toward the end of the video where I evaluate the integrals over the lines l1 and l2
@@maths_505 oww! lemme check, but what I don't get is how do we choose our contour properly? is it like a u-sub thing where it's just intuition? or we check where the function is analytic and check for singularities and places where the laurent doesn't converge?
@@manstuckinabox3679 checking for convergence is always necessary. However it mostly comes down to the particular integral you're evaluating. For many log integrals, the keyhole contour works wonders; you'll get a feel for this with more practice. Remember that box contour? The motivation behind it was the difficulty in studying the behaviour of the integrand on that contour. So don't worry about it....you'll get the feel for what contours will work and you'll be able to picture in your mind exactly what could work where.
If you draw the vector representing any complex number on l2, that vector lies in the 4th quadrant (3pi/2 to 2pi). You see that as delta goes to zero, the argument approaches 2pi
@@maths_505 right.. got it! Thank you! ☺️ Btw can you please explain a little more about the branch cut at the origin? Is it because it is z^(1/3) so will it have 3 branches? Also why didn't you draw contours around the singularities i.e. z=+/-i? Sorry to ask so many questions. 🙏
@@imonkalyanbarua z^(1/n) is a multivalued function so we need only the principle branch. The principle branch is defined for all z other than zero. That's why we branch around z=0
I have a doubt, at 6:10, roots of a complex number are generally multivalued, so at first glance i would prefer taking -i ^1/3 as e^(-ipi/6). So this integration value is multivalued then? But the question cannot be multivalued since its real
@@maths_505 Actually I am fairly new to complex numbers, so I don't understand why we are taking a branch cut. The only place where our function isn't holomorphic is probably at +i and -i, so why the positive real line?
@@xtremeplayz3606 it restricts the argument of z⅓ between 0 and 2pi. That way you avoid traversing the entire circle centred at the origin. If you dont restrict the argument, the function is multivalued as complex numbers have 3 cube roots.
@@xtremeplayz3606 technically yes but bruh pi to 3 pi is just plain weird 😂 Start from the positive x axis and move clockwise and you'll get 2 pi -pi to pi is a pretty common branch cut too (along the negative real axis that is)
Ah it's so awesome that you actually updated the video! Thanks for putting in the extra effort to make a new video for us, just goes to show your commitment. Gonna settle in for a nice watch now. Bravo!
Not boring, necessary! Keeps us on our toes!
Very interesting technique for integrating these types of integrals. Keyhole Contour and Cauchy’s Residue Theorem. The world of complex analysis is fascinating.
@ 10:46 The numerator should be epsilon^(4/3). The process for the parameterization for the smaller circle is the same as that for the larger circle, but with R replaced with epsilon.
Oh btw I forgot to mention great work going on! Wish you all the best! 😇👍
Wolfram Mathematica gives different answer: Pi/sqrt(3) and this is 1/2 of Pi/sin(Pi/3).
Lots of thanks for you work, my friend! It helps me a lot! Actually, your Sigma isn't a branch cut, right? You have a branch point at the origin and a branch cut going from it to infinity. Your sigma is just a curve that doesn't cross the branch cut so your function stays single-valued.
Nice!!! Good work...Which program or app do you use to write in that black panel ?
17:03 is okay to forgot the /2
bcos it will canel with the sin pi/3 in ghe denominator.
The final anwser should be
pi/sqrt3
For L2 integral you explained that z becomes |z|exp(2pi/3) due to the z^1/3 in the numerator. How do we account for the Z^2+1 in the denominator? What was the justification to ignore?
I dont understand this either
@@quite_unknown_1if you understood now, you may explain me
Yeah get to see it twice :D!
Yeah boi! 😎
But on a serious note myers and zunaid parker added a couple of notes that motivated me to remake the video and then upload it. You'll notice the changes toward the end of the video where I evaluate the integrals over the lines l1 and l2
@@maths_505 oww! lemme check, but what I don't get is how do we choose our contour properly? is it like a u-sub thing where it's just intuition? or we check where the function is analytic and check for singularities and places where the laurent doesn't converge?
@@manstuckinabox3679 checking for convergence is always necessary. However it mostly comes down to the particular integral you're evaluating. For many log integrals, the keyhole contour works wonders; you'll get a feel for this with more practice. Remember that box contour? The motivation behind it was the difficulty in studying the behaviour of the integrand on that contour. So don't worry about it....you'll get the feel for what contours will work and you'll be able to picture in your mind exactly what could work where.
@@maths_505 Ah I see! yeah most stuff concerning integrals really boils down to experience lol.
@@maths_505 Excellent! Now, all complex men shall gather around and drink from the fountain of knowledge. 😎
One aspect I didn't understand is why is the argument for the path l2 2π? Should it not be just π? If you could kindly elaborate. Thank you.
If you draw the vector representing any complex number on l2, that vector lies in the 4th quadrant (3pi/2 to 2pi). You see that as delta goes to zero, the argument approaches 2pi
@@maths_505 right.. got it! Thank you! ☺️ Btw can you please explain a little more about the branch cut at the origin? Is it because it is z^(1/3) so will it have 3 branches? Also why didn't you draw contours around the singularities i.e. z=+/-i? Sorry to ask so many questions. 🙏
@@imonkalyanbarua z^(1/n) is a multivalued function so we need only the principle branch. The principle branch is defined for all z other than zero. That's why we branch around z=0
@@imonkalyanbarua the singularities didn't bother us since we use them to calculate residues
@@maths_505 thank you so much for your kind reply and resolving my doubts.. 😇🙏
Nice bro
I have a doubt, at 6:10, roots of a complex number are generally multivalued, so at first glance i would prefer taking -i ^1/3 as e^(-ipi/6).
So this integration value is multivalued then? But the question cannot be multivalued since its real
That's why there's the branch cut [0, infty)
That makes z⅓ holomorphic
@@maths_505 Actually I am fairly new to complex numbers, so I don't understand why we are taking a branch cut.
The only place where our function isn't holomorphic is probably at +i and -i, so why the positive real line?
@@xtremeplayz3606 it restricts the argument of z⅓ between 0 and 2pi. That way you avoid traversing the entire circle centred at the origin. If you dont restrict the argument, the function is multivalued as complex numbers have 3 cube roots.
@@maths_505 Okay so will we get the same answer if we define a branch cut from pi to 3*pi ?
@@xtremeplayz3606 technically yes but bruh pi to 3 pi is just plain weird 😂
Start from the positive x axis and move clockwise and you'll get 2 pi
-pi to pi is a pretty common branch cut too (along the negative real axis that is)
x=tgu... Diventa l'integrale di sqrt3(tgu) molto semplice, ma immagino tu abbia voluto insegnare altri metodi
😀