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![[UNCUT] The Loyal Pin ปิ่นภักดิ์ EP.7 (1/4)](http://i.ytimg.com/vi/2rPGOyKs0VE/mqdefault.jpg)
34:56 Can't get over how slick you were picking up that eraser
The keyhole integral should be the mascot for complex analysis because it makes a giant ‘C.’
This is the reason I taught myself complex analysis over the past couple years from a couple books from Amazon. While I didn't memorize all the rigorous proofs, I made sure I good understanding of everything. It wasn't easy, though.
Book suggestions?
@@dhanvin4444 Visual complex analysis by tristan needham
18:35 I guess the answer to your "what happens if a is negative" question is we just pick a semicircle contour below the x-axis, and then e^(-ay) will still be less than 1, as the signs of a and y cancel out
Amazing problem solving, you really helped me a lot with great explainations on each question while i am preparing for my final exam. Thanks and keep up the good work! :)
One thing I've wondered is is there ever a case (when taking the limit to infinity of these contour integrals) where we have to worry about a pole at infinity affecting the contour?
This would imply that the denominator converges to zero as z goes to infinity, meaning the integral would be divergent anyway?
@@freyac9424 Yeah you might be right
Vaguely speaking, if you think in terms of the Riemann Sphere, I believe you can say that all polynomials have poles at infinity, so you're already familiar with exactly how divergent they are.
You can think about this like taking sum of negative integer powers, with any coefficients, and plugging in a substitution that has some power, like u=x^p. If p is a positive integer, you know the poles might shift around and change degree, but there's still there. If p is a negative integer, it's like extending that same tendency, but the poles all move to infinity.
Thinking about this in reverse, it turns out exponential functions have _lots_ of behavior at infinity, but not exactly poles. If you 'shift' this behavior around like we were doing with polynomials, you get the very nasty 'Essential Singularity.' The simplest example as such is e^1/z. In a neighborhood of an essential singularity, the function takes on every complex value, except possibly two (that is in the case of meromorphic functions, whereas for analytic functions it's all but one value), infinitely often. Not 'infinitely often' in any branching sense, but that as you get closer and closer to the singularity, you find more and more copies of all the values.
For exponentials, you can sort of find where these values were hiding by examining the regular e^z exponential: looking on the negative real side of the complex plane, as you move up and down in the imaginary direction you find copies and copies every 2pi units of the same values, obviously, with all of them getting smaller as you move further negative-real. Likewise on the positive-real side, but getting larger and larger magnitudes as you move further positive-real, with copies and copies every 2pi in the imaginary direction. Everything out at these extremes also exists when you substitute from e^z to e^1/z, but all packed in to _every_ neighborhood, especially the smallest ones, of the essential singularity (in this case at z=0).
In your squareroot example, technically the bounds on the integrals over the lines don't go from epsilon to R. It's fine to start at epsilon, but its ending is actually sqrt(R^2-epsilon^2), which clearly converges to R as epsilon goes to 0.
I hope you soon cover analytic continuity.
Really beautiful stuff. I have a question - are there *any* of these integrals that *cannot* (by any means) be proven by trig function substitutions? i.e. is this "essential" or "algorithmic"?
For anyone who wants solutions to the warm-up problems at the end, here are my solutions. Feel free to double check them, because I'm not certain.
For the first integral, I got pi/(2a^3) assuming a>0.
For the second, I got 3pi/(2e^-4).
For the third, I got (pi/12sqrt(2))*ln(3+2sqrt(2)), which I'm not sure if it's right.
For the fourth, you have to be careful with the branch cut (like Michael said), but I got pi*(cuberoot(4)-cuberoot(3))/sin(pi/3). In particular, keep in mind if you define cuberoot(re^itheta)=cuberoot(r)e^(i(theta/3)) for theta in [0,2*pi), then note cuberoot(-4) is NOT the same as -cuberoot(4).
Edit: For the second, I re-solved it and got 5pi/(2e^4). For the third, I got pi*(-16+12sqrt(2)) after I used the substitution z=2e^itheta, which apparently the correct solution is pi*sqrt(2). Still working on it.
Edit 2: I got pi*sqrt(2) for the third integral. I made a silly mistake. The reason I substituted z=2e^itheta is because when I substituted z=e^itheta, I made a mistake and thought I needed a larger radius. Either way, it worked.
1st and 4th are right (note btw you can expand sin(pi/3) out!), 2nd should have a 5 not a 3 coefficient, and 3rd should be a nicer pi*sqrt(2). All checked with wolfram alpha.
@Dan You want to integrate over a contour with a detour around the positive real axis. I used a circle of radius R>4 (centered at 0) with straight lines coming back to the imaginary axis distance epsilon
About the fourth. Do I understand correctly that in order to obtain the asked value in the real domain, I should take the module from the complex number obtained from the calculated kehole contour, i.e. the answer on the complex domain is 2*Pi*i*(2^(2/3)-3^(1/3))/(1-exp(i*2*Pi/3)) which in real domain, after applying module operator, changes to 2*Pi*i*(2^(2/3)-3^(1/3))/sqrt(3)) ?
@@sergpodolnii3962 I can't really tell what you mean by "module." It sounds like you mean either the _modulus_ (which is the length of a complex vector) or the _real part_ which is not exactly what we are going for.
The "keyhole" contour (as you describe it) is composed of 4 curves. The circular curves vanish as R goes to infinity and epsilon goes to 0. The line above the real axis approaches the desired integral, and the line below the real axis approaches the desired integral scaled by a factor of -e^(2pi*i/3) (because of the peculiarities of the discontinuity of the complex cube root function, and the reverse orientation). This allows you to solve for the desired integral.
One caveat I will mention: your solution doesn't seem to take into account how the complex cube root gets applied to -3 and -4. These have obvious _real_ cube roots, but this particular complex cube root maps it a little differently (specifically, there should be an extra factor of e^(pi*i/3).
Lastly, your answer couldn't be correct because it is imaginary, and the original integral was with real functions. This was the realization that I had which led me to correct my original solution.
@Dan My understanding is that in these problems, you are integrating over a contour, and if you choose your contour carefully, you will be able to solve for your desired integral. That’s how all these problems work.
Would you solve more difficult problems on branch cut integrals?
Good lecture. Just curious, are you going to include Laplace reverse transformation and Bromwich contour later?
I'm a bit unsure about the sin -> exponential trick. get that sin x is the imaginary part of e^ix, but there's more in the function than just sin. Wouldn't we want the real part of the denominator? How do we know it's safe to mix them like this?
... I think the idea is that is that on the real number line, the Imaginary part of the alternate function is always equal to the original function. So we can calculate the integral of the alternate function and then take the imaginary part of it.
28:13 Why are you including the origin in the region of the branch cut, since the √z is analytic at the origin, no matter which branch cut you use it will be equal to zero
"integrals that might be popular on TH-cam" *gestures at today's main channel video*
Roughly speaking how much content does each episode cover when compared to an undergraduate course? A week’s worth? Half a week’s worth? A third of a week’s worth?
Books?
How did u know on the last one it was a keyhole and not a half circle.
im not a mathematician or anything but if the question states that the integral is between 0-infinity you should do keyhole and if it states -infinity to infinity you should do half circle.
This method is useful for improper integral, but is this possible to use this method to calculate proper integrals?
You can, mostly when the bounds relate to pi. I watched a video which I can't find, which used it.
It's called "Don't be fooled! Integrating a resistant integral", also 21:27.
I misread that as "centaur integration"
39 minutes of babbling.
get to the point