Thank you for this! I remember taking adv cal and the prof showed us a crazy integral of a function with a discontinuity. He then said, no problem, let's move it over to complex space, solve it there and bring it back to Real space. That's when I began to understand the power of mathematics.
U mean Feyman's method of integration under differential sign, i tried it like that... but it is not working out... Edit: Sorry... its differentiation under integral sign
in the fall I took my first math course with complex analysis, and we ended with evaluating these types of real integrals using complex analysis. I think I solved this exact problem as homework. it totally blew my mind how strange and powerful the technique was-and that it actually worked
Yeah, it's been a couple of years since I took the class but it was one of my favourite in the whole degree. Which is saying something, because I'm really not a fan of integrals. I need to revisit the subject, and might check out that Math Major channel.
12:55 A minor error: the integral of a constant over a complex curve is not the length of the curve! One needs to replace _z_ with _R*exp(i*t)_ and _dz_ = _i*R*exp(i*t)*dt_
There were absolute values around the integral to begin with, and he basically used the "ML-inequality", not bothering to introduce |dz| or other notation to signify that the integral is no longer a complex line integral, but rather an integral of a positive function w.r.t. path-length. He did it correctly, modulo perhaps this notational ambiguity. There was a valid upper bound for the integrand on c_R and the correct path-length for c_R.
That's pretty amazing how you can derive this answer using this method and also using a completely different method which is by feynman integration (set I(t)=integral(cos(tx)/(1+x^2) dx)) and solving the ode that we get for I(t), I''(t). Two utterly different method to the same result! That's something I love seeing in mathematics.
Define I(a) = \int cos(ax)/(x^2+1) dx. Then I(a) satisfies the differential equation d^2 I(a) /da^2 = \int cos(ax) = 2\pi \delta(a) . The unique solution to this differential equation which goes to zero at infinity is \pi exp(-|a|), which gives the integral you want at a=1.
Thank you Michael, this was awesome! It has been 30 years since I took an elective course in Complex Variable Theory. It really is a beautiful framework for solving problems like this nasty integral. How about an example using conformal mapping?
I suspect that for people who haven't seen complex analysis before, this video is going to feel like a string of steps that don't make sense to them with solutions that come out of nowhere, and that's going to turn them away from checking out your other channel because they will presume that they won't understand it (i.e. the confidence problem for math students). This means that students who might really enjoy your content on your other channel might never even check it out, and so this video will fail them as an advertisement for video series they might really benefit from. One way to remedy that would be to provide them with the steps that they don't need complex analysis for up front; this could be done by giving a sketch for the battle plan to tackle the problem. Even if you only know integration techniques and some light analysis but haven't seen complex variables, you will still understand making problems slightly more complicated so that they can be broken down into easier pieces, solving a limit one piece at a time, citing theorems with useful results to do work for you, and bounding a piece of a limit to show that it goes to zero. This way viewers can build confidence while watching the video even if they haven't learned the content you are trying to advertise. Finally, link the relevant videos and clearly cite what theorems require as hypotheses and the results that they guarantee! Honestly, this should be your bread and butter for these kinds of videos because you already have a very strong history of introducing and using theorems you don't intend to prove as fun and interesting tools to solve cool problems. Then, when you use a cool theorem from complex analysis, you can cite specifically where the relevant theorem can be found in your video series. As an example: the upper bound on the modulus of an integral can be found at 22:40 in video 12 of your complex analysis playlist (you should link the relevant video with a clear label). This is more of a youtube thing, but you are currently competing for their eyeballs with every advertised video in the sidebar, so you want to lower viewers' inertia as much as possible to check out your other channel. By specifying where they can find something specific that you have now made them interested in knowing more about, you help direct them toward your other channel. This also has the side benefit of making these videos easier to use by students, because they can go "oh I'm trying to solve a contour integral problem and I remember Michael solved one of those on his channel. Oh hey! He has even cited exactly where to find information on how to use this in a more general context. Now I have an example more information on how this thing works!" Anyways, sorry for the masters thesis worth of criticism for your video; tldr: this video has some major problems but they should all be fixable by methods and skills you've already used in previous videos. Love your work. Happy Mathing.
I'm not sure this would work, as the style of the MathMajor channel is very different to this one. I had a look at the proofs in question, and you are into a morass of punctured disks, Laurent series and the like. Not fodder for the casual browser. What would be useful over here is the basics of analytic functions and the Cauchy-Riemann equations. Followed up by a "lite" proof of the zero integral result, say on a rectangular contour with one edge varying. Then a demonstration of residues with simple poles e.g. 1/z, z^2/(z-a) etc. The MathMajor approach is different to my dim recollection of the way it was done at Cambridge. MathMajor calls up the antiderivative of the analytic function, for example if f(z) is 1/z it is log(z). This makes path independence straightforward, but opens up another can of worms with the Riemann cut. As far as I recall, our course proved the fundamental theorem straight off, by some kind of dissection argument. I'm not sure I followed it completely, but I bought it. EDIT: Michael did something like this a few weeks ago in "Complex integrals are ... different", but it wasn't for a general f(z) and he left analyticity as kind of a tease. Maybe one to revisit.
Great insight and suggestions. I found his way so reminiscent of my profs: Clear, quick, concise. Expects you did your homework and you may have to go back on some material after class. I love him!
Given the length and specificity of your comment, I would enjoy seeing a video produced by you that would incorporate improvements based on your criticisms and detailed review of this video.It’s quite evident your strengths are in presentation and pedagogy and many would benefit from your efforts.
Actually it is unnecessary to use the residue theorem for such integrals, applying the Cauchy integral formula to the function g(z)=e^{iz}/(z+i) on the given contour gives the result very neatly...
3:45 perhaps i would have done: cos(x) = (e^ix + e^-ix)/2, so the integral can be split into two forms, one with e^ix and one with e^-ix, and using the substitution u=-x, we get that they have identical integrands. their paths, however, are 180 rotations about the origin of one another, which means their sum is in fact the circular path parametrizable by R e^it for increasing real values of t from 0 to 2pi. Interesting how the Re() function is actually necessary this time, i think?
But dz is complex. Easily repaired, put z = R exp(i theta) dz = R i dtheta exp(i theta) |dz| = R dtheta I +infinity Strictly you should re-state the integral as the limit of a sum, but that's the bones of it. Apologies if I misunderstood
So just adding an important part in my opinion. The symmetric limit calculated here does not always have to be equal the desired improper integral. For further info a viewer can check The Caucy principle value. Here they are equal since the integrand is an even function
I believe one can combine what is shown here to show that integral 0 to infinity of exp(-x -1/x) 1/sqrt(x) = sqrt(pi)/ e^2. I never saw this latter integral, or that one can get an answer in closed form.
Replacing cos(x) by Re(e^ix) and then taking Re outside of the limit only works when x is real; otherwise the denominator can be complex too. Not hard to fix, but a bit of a mistake.
Really makes me wonder, why some integral with constants suddenly goes zero, when it covers an increasing long curve. I guess things cancel out like it's symmetric at x=0, but still not intuitive, although I somehow knew, the result will end up being zero 😅
İ think there is quite missed something, when we take abs value of inside integral we should also take abs of dz |dz|= length of curve which we are integrating İn our case thats piR And since denominator has R²,thats 0 anyway İntegral of dz, i think quite different thing
I’m wondering if that equality linking the integral in the real domain to an integral to the complex domain is valid Of course in hindsight we know that along the C_r the integral will tend to 0 as R blows up but we don’t know that preemptively. Shouldn’t we just stop. Start at the complex integral show that the curved part tends to the 0 and what reminds is what’s equivalent to the original problem
One important difference between complex and real calculus is that the semi-circle you draw isn’t a representation of the function, but only the _domain_ of the function. In real calculus, this would be like taking the integral of a function, and instead of drawing the function, you only draw the portion of the x-axis you want to consider.
@@angelmendez-rivera351 All I meant was that the diagrams you typically draw aren’t of the function, but just of the domain. If we were integrating over a circle, then a semi-circle _would_ be the appropriate drawing in real calculus.
@@angelmendez-rivera351 Of course not, I said nothing about the integral of this particular function. I don’t think I misunderstood anything, though my phrasing might not be perfectly accurate. I think you are being a little too pedantic.
I have a question: At 1:11 you mentioned that taking a single limit of the bounds is sketchy, why is that? also why does an even function bypass this issue? thank you :)
Usually, it’s because for odd functions, if you’re not careful and only use one limit, it’ll cancel out to 0 because you’re making the assumptions it’s approaching both infinities at an equal rate. That’s why usually you need to break it up into two limits. For even functions, since there is none of that area canceling bc it’s symmetric around the y-axis, it’s safer to take the one limit
you should go over probability and complex analysis! I wanna know if you can wvaluate the CDF of normal distribution using contour integrals?? Also I want to see how the gaussian distribution, Gamma distribution, and Poisson distribution relate to complex analysis - contour integrals, conformal mappings, etc.
What happens if you add TIME to all this? YOU ADD THE NUMBER ONE Because without the number One, time can't exist, why? Because Zeros go around and around, they are eternal, infinite, so all time begins with a One then followed by Infinite Zeros Thx for the video, great work!
Woah hold up when you’re integrating on this domain should you break up the boundary depending on their path? I.E. you have your curve C_r and you also have the straight line going from -R to R call it S_r So wouldn’t the integral be something like Int on boundary = int on C_r + int on S_r
He does this. He calculates the contour integral over the whole path using the residue theorem and then finds each piece of the path individually. C_R is defined as only being the top part of the semi-circle. Your S_r is the original integral
Do you ask this for a deeper reason because I am working on a breakthrough paper in mathematical physics, and this can relate, it would relate to geodesics and some obscure math
@Solar Flair Yes. I wanted an un-biased answer from someone who understands basic geometry, in a mathematics related channel, in order to screencap it and show it to someone else who is insisting that "C/2=D". Just to let that person know that his argument is wrong. I mean, can you imagine? Half the circumference equal to the diameter? Both terminated at the same 2 points, one line straight the other line curved, yet equal in length? But like I said, this person I'm talking about is very biased, so it's no biggie. It's just an internet contrarian :) Thanks again @MrPwmiles
I had never seen that amazing number π/e coming out from a relevant result before. Will we maybe someday use this or other results concerning π/e to prove it is a transcendental number?
Okay, I did something dumb calculating the residue, and I don't know what. Please help? cos(z) = Re( e^iz ) for all complex z, but cos(i) = 1/2 (e+1/e) is not the real part of e^(i^2) = 1/e.
It's because the correct definition of cos(z) for any z is cos(z) = 1/2 (e^(iz) + e^(-iz)) Whereas Re(e^(iz)) = 1/2 (e^(iz) + e^(-iz*)) In the second term you have to conjugate everything including the z. So if z is not real, Re(e^(iz)) is not the same as cos(z).
@@pwmiles56 Thanks. That makes sense, the substitution cos(x)=Re(e^ix) works up front because x is real. But that still doesn't help me understand the bigger picture, because why take the real part in the first place when this equality fails over the complex numbers? The two formulas you gave only agree over the reals. It seems to me like taking the real part is rewriting the integrand to be a different function over the complexes that only happens to agree with the original integrand on the real axis. Since computing the residue requires evaluating (z-i) times the integrand at z=i, we get different answers depending on which formula we use, so taking the real part is actually /required/ in this case. How am I supposed to know to do this for arbitrary integrands? I'm lacking intuition here.
@@davidblauyoutube We want I = integral along the real axis of cos(z)/(1+z^2) = Re[ integral along the real axis of exp(iz)/(1+z^2) ] Then everything inside [ ] is replaced by the closed contour integral, minus the integral around the semicircle. It's technically correct. It would have been clearer to split out cos(z) = 1/2 (exp(iz) + exp(-iz)) and handle the second part in the negative-imaginary halfplane. But a little more work, hence the shortcut presumably.
That has a singularity on the real line, x = -1. So no, or at least not without some fancy way to integrate through the singularity. In a simple view the integral does not exist
Take a half-circle around x=-1 with a small e>0, and compute the residue as usual. You can derive the Laurent series a shown that except the Residue term any Higher order terms are zero.
Neither. It's f(z) deltaz summed along the path, in the limit as deltaz->0 The real part of the path is given but the rest is our choice. You could complete the path as a rectangle, or a triangle, whatever. The semicircle is chosen because it simplifies the limit calculation.
@@pwmiles56 Sir can you please explain 4:50 "We Added this in in this integral, so we need to subtract it out", this region C_r is already covered by D_r boundary (first integral)?? And why need subtract it out? Dont understand this step.
Hmm, the way that it's done is a little misleading. I personally, would have started with the contour integral and got the real integral in question as part of breaking up the contour. The reason for this is that you can't always tell what bit of the integral will contribute.
Been 25 years since I studied residues. Just trying to clean up the shards of brain tissue that left on the ground as I attempted to wrench memories of that skill out of the recesses of my mind. Mostly I just remember the girl who typically sat in the front row during lectures. Such a rarity to find an attractive female in an applied math course. I definitely need a refresher. Will have to check out your other course.
Isnt the worry about the upper and lower limit only real if theres a potential problem with convergence? abs(cos(x)/(1+x^2)) is bounded by 1/(1+x^2) so no worry there.
I'm very rusty on complex calculus, but I seem to remember that a contour integral over a closed path is always zero, which does not appear to hold here. Maybe there is a problem with the function not being differentiable at the (R, 0) or (-R, 0) points, but I can't put my finger on the definition of "differentiable" in this context.
The contour need only be continuous (possibly excluding pathological cases such as fractals). The function must be differentiable i.e. df = f'dz is a meaningful identity. This is the same thing as it being analytic. Some simple functions are not analytic e.g. f(z)=z*. However polynomials and power series such as cos(z), etc, are generally analytic over some domain. And then yes, the contour integral can be "shrunk" to a tiny one round any singularity(ies), which gives us Cauchy's theorem.
But, in this case, the real part of the whole contour integral is pi/e. Both the real and imaginary parts should have been zero, then. I was wondering what is broken here.
@@juandesalgado It's the singularity. The zero result only applies if the function is holomorphic within the contour, which is a fancy way of saying there aren't any singularities. E.g. f(z) = 1/z f(z) has a simple pole (a first order pole) at z=0 Integrate round a circle of radius r, centred on the origin Put z=exp(i theta), dz = i z dtheta I = integ (theta=0 to 2 pi) iz / z dtheta = 2 pi i This illustrates the Cauchy residue theorem. The residue is the limit of z f(z) as z->0. In general if a is a simple pole the limit is (z-a)f(z) as z->a.
A ridiculous instructive video at introducing and using the residue theorem to an audience that is assumed to not have ever been introduced to it before. In that case, the viewer will be totally confused with what you just did. Why not just spend 10 minutes introducing the residue theorem and then an application of it using this example.
I agree. There seems to be a fixation on technique, but knowing a technique is not the same as understanding it. I did find this quite helpful, as I'd forgotten the details of residues, but it doesn't offer much to newbies.
In the first 20 seconds of the video he mentions he will be using methods from complex analysis and links to a channel with a 20+ part series, the title of one of which is "the residue theorem". How much hand-holding do you want?
You know, you are right. Why don’t you replace that downloaded Nancy Sinatra video on your channel with a 10 minute video introducing the residue theorem?
@@m44lshannon I shouldn't labour this, but you kind of made my point. Knowledge doesn't just lie there in some book or video course. It is something the student has inside them. Generally acquired by practice, working on problems and proofs. The buzz-phrase is "active learning". Michael is great, he knows the material, his presence and delivery are superb, but he can actually be a bit too convincing. Sometimes. Seems to me
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Then there was the math prof who named his dog Cauchy because he left a residue at every pole.
Nice one
🤣
Finally a complex analysis example! I’ve been using complex methods for years on this channel, finally glad to see an integral tackled in this way.
Did this in a complex variable class in 1967 and the complex methods are very beautiful. Very enjoyable review.
Thank you for this! I remember taking adv cal and the prof showed us a crazy integral of a function with a discontinuity. He then said, no problem, let's move it over to complex space, solve it there and bring it back to Real space. That's when I began to understand the power of mathematics.
You can also solve this using a laplace transform. If J(t) = integral from 0 to infinty of cos(tx)/(x^2 +1), the desired integral is equal to 2J(1)
But how do you actually find the exact value of J(1)?
U mean Feyman's method of integration under differential sign, i tried it like that... but it is not working out...
Edit: Sorry... its differentiation under integral sign
@@parijatsutradhar It will not. The reason is that the second derivative of J(t) does not converge
@@fartoxedm5638 exactly... but like maybe there is something else... maybe something that i didn't realise or don't know
@@parijatsutradhar no. They said laplace transform, not feynmann's technique
in the fall I took my first math course with complex analysis, and we ended with evaluating these types of real integrals using complex analysis. I think I solved this exact problem as homework. it totally blew my mind how strange and powerful the technique was-and that it actually worked
Yeah, it's been a couple of years since I took the class but it was one of my favourite in the whole degree. Which is saying something, because I'm really not a fan of integrals. I need to revisit the subject, and might check out that Math Major channel.
12:55
A minor error: the integral of a constant over a complex curve is not the length of the curve! One needs to replace _z_ with _R*exp(i*t)_ and _dz_ = _i*R*exp(i*t)*dt_
There were absolute values around the integral to begin with, and he basically used the "ML-inequality", not bothering to introduce |dz| or other notation to signify that the integral is no longer a complex line integral, but rather an integral of a positive function w.r.t. path-length. He did it correctly, modulo perhaps this notational ambiguity. There was a valid upper bound for the integrand on c_R and the correct path-length for c_R.
@@juha-mattiperkkio7646
He should've put the |dz| there. It's actually a robust limit to use |dz| instead of _dz_ actually.
"... and so we calculate the residues of singularities of these two functions."
Mmmm, yes, I know some of these words. :)
That's pretty amazing how you can derive this answer using this method and also using a completely different method which is by feynman integration (set I(t)=integral(cos(tx)/(1+x^2) dx)) and solving the ode that we get for I(t), I''(t).
Two utterly different method to the same result! That's something I love seeing in mathematics.
Also a Laplace transform.
the second derivative of I(t) does not converge even absolutely
@@fartoxedm5638 I think he meant the derivative with respect to x
Would be more amazing if you got different results, no? 🤣
@@alomirk2812 khm... What? X is constant for I(t)
I really like complex integration and I really hope that you make more videos about hard integral being tackled by contour integration
Define I(a) = \int cos(ax)/(x^2+1) dx. Then I(a) satisfies the differential equation d^2 I(a) /da^2 = \int cos(ax) = 2\pi \delta(a) . The unique solution to this differential equation which goes to zero at infinity is \pi exp(-|a|), which gives the integral you want at a=1.
Thank you Michael, this was awesome! It has been 30 years since I took an elective course in Complex Variable Theory. It really is a beautiful framework for solving problems like this nasty integral. How about an example using conformal mapping?
More visually, on e^(iz) for that semicircle, multiplying it by i rotates it anticlockwise, which means all the real values of iz are
I suspect that for people who haven't seen complex analysis before, this video is going to feel like a string of steps that don't make sense to them with solutions that come out of nowhere, and that's going to turn them away from checking out your other channel because they will presume that they won't understand it (i.e. the confidence problem for math students). This means that students who might really enjoy your content on your other channel might never even check it out, and so this video will fail them as an advertisement for video series they might really benefit from.
One way to remedy that would be to provide them with the steps that they don't need complex analysis for up front; this could be done by giving a sketch for the battle plan to tackle the problem. Even if you only know integration techniques and some light analysis but haven't seen complex variables, you will still understand making problems slightly more complicated so that they can be broken down into easier pieces, solving a limit one piece at a time, citing theorems with useful results to do work for you, and bounding a piece of a limit to show that it goes to zero. This way viewers can build confidence while watching the video even if they haven't learned the content you are trying to advertise.
Finally, link the relevant videos and clearly cite what theorems require as hypotheses and the results that they guarantee! Honestly, this should be your bread and butter for these kinds of videos because you already have a very strong history of introducing and using theorems you don't intend to prove as fun and interesting tools to solve cool problems.
Then, when you use a cool theorem from complex analysis, you can cite specifically where the relevant theorem can be found in your video series. As an example: the upper bound on the modulus of an integral can be found at 22:40 in video 12 of your complex analysis playlist (you should link the relevant video with a clear label). This is more of a youtube thing, but you are currently competing for their eyeballs with every advertised video in the sidebar, so you want to lower viewers' inertia as much as possible to check out your other channel. By specifying where they can find something specific that you have now made them interested in knowing more about, you help direct them toward your other channel.
This also has the side benefit of making these videos easier to use by students, because they can go "oh I'm trying to solve a contour integral problem and I remember Michael solved one of those on his channel. Oh hey! He has even cited exactly where to find information on how to use this in a more general context. Now I have an example more information on how this thing works!"
Anyways, sorry for the masters thesis worth of criticism for your video; tldr: this video has some major problems but they should all be fixable by methods and skills you've already used in previous videos. Love your work. Happy Mathing.
I'm not sure this would work, as the style of the MathMajor channel is very different to this one. I had a look at the proofs in question, and you are into a morass of punctured disks, Laurent series and the like. Not fodder for the casual browser.
What would be useful over here is the basics of analytic functions and the Cauchy-Riemann equations. Followed up by a "lite" proof of the zero integral result, say on a rectangular contour with one edge varying. Then a demonstration of residues with simple poles e.g. 1/z, z^2/(z-a) etc.
The MathMajor approach is different to my dim recollection of the way it was done at Cambridge. MathMajor calls up the antiderivative of the analytic function, for example if f(z) is 1/z it is log(z). This makes path independence straightforward, but opens up another can of worms with the Riemann cut. As far as I recall, our course proved the fundamental theorem straight off, by some kind of dissection argument. I'm not sure I followed it completely, but I bought it.
EDIT: Michael did something like this a few weeks ago in "Complex integrals are ... different", but it wasn't for a general f(z) and he left analyticity as kind of a tease. Maybe one to revisit.
Great insight and suggestions. I found his way so reminiscent of my profs: Clear, quick, concise. Expects you did your homework and you may have to go back on some material after class. I love him!
Given the length and specificity of your comment, I would enjoy seeing a video produced by you that would incorporate improvements based on your criticisms and detailed review of this video.It’s quite evident your strengths are in presentation and pedagogy and many would benefit from your efforts.
Cauchy's biography is also worth reading.
This one is a classic!
Classic blooper! dz is complex, this simply wasn't addressed
I very much like Complex Analysis.
Thank you, professor.
Actually it is unnecessary to use the residue theorem for such integrals, applying the Cauchy integral formula to the function g(z)=e^{iz}/(z+i) on the given contour gives the result very neatly...
Complex analysis is definitely the best area of mathematics. It's super powerful and shows up just about everywhere
No it isn't. Algebraic geometry is the best area of mathematics.
Complex analysis is great
elegant solution
Please could you consider doing a series on topology/ algebraic topology on your mathmajor channel? Thank you for your videos!
Excellent! I like it a lot!!
That was just beautiful. Thanks
3:45 perhaps i would have done: cos(x) = (e^ix + e^-ix)/2, so the integral can be split into two forms, one with e^ix and one with e^-ix, and using the substitution u=-x, we get that they have identical integrands. their paths, however, are 180 rotations about the origin of one another, which means their sum is in fact the circular path parametrizable by R e^it for increasing real values of t from 0 to 2pi. Interesting how the Re() function is actually necessary this time, i think?
If only my complex analysis professor taught like you...Thanks.
But dz is complex. Easily repaired, put
z = R exp(i theta)
dz = R i dtheta exp(i theta)
|dz| = R dtheta
I +infinity
Strictly you should re-state the integral as the limit of a sum, but that's the bones of it. Apologies if I misunderstood
So just adding an important part in my opinion. The symmetric limit calculated here does not always have to be equal the desired improper integral. For further info a viewer can check The Caucy principle value. Here they are equal since the integrand is an even function
More contour integration please!
I also do the C_R part using polar coordinates, but this is so much quicker!!
I believe one can combine what is shown here to show that integral 0 to infinity of exp(-x -1/x) 1/sqrt(x) = sqrt(pi)/ e^2.
I never saw this latter integral, or that one can get an answer in closed form.
Replacing cos(x) by Re(e^ix) and then taking Re outside of the limit only works when x is real; otherwise the denominator can be complex too. Not hard to fix, but a bit of a mistake.
I like this video a lot, really really well explained
The Re part wouldn't be necessary since the integral of sin(x)/(1+x^2) is zero due to antisymmetry, so replacing cos(x) by exp(ix) should suffice.
Thank you Sir
Have a nice day
When you are computing an integral of moduli over the contour, dx should be
|dz|
Just did a course on this, good stuff. Want to explore the analysis more pure mathematically
I found that modulus trick far more powerful and eye opening than the contour integral trick😅
Cool! (:
Thx for sharing!
I believe more correct record for integral in right part of inequality at 12:30 must use |dz| instead of dz.
The link to his second channel can be found in the description if you’re having trouble searching for it.
but it's called "Course videos" there!
Residue theory is always the king
Wow, thats some heavy stuff.
Contour Integration is the best.
B I G P O W E R .
Really makes me wonder, why some integral with constants suddenly goes zero, when it covers an increasing long curve.
I guess things cancel out like it's symmetric at x=0, but still not intuitive, although I somehow knew, the result will end up being zero 😅
Take the plunge
flammable maths did one on this
What I can't understand is why we chose the region in such a way. Can someone explain why we almost always integrate over the half circle?
İ think there is quite missed something, when we take abs value of inside integral we should also take abs of dz
|dz|= length of curve which we are integrating
İn our case thats piR
And since denominator has R²,thats 0 anyway
İntegral of dz, i think quite different thing
I’m wondering if that equality linking the integral in the real domain to an integral to the complex domain is valid
Of course in hindsight we know that along the C_r the integral will tend to 0 as R blows up but we don’t know that preemptively.
Shouldn’t we just stop. Start at the complex integral show that the curved part tends to the 0 and what reminds is what’s equivalent to the original problem
Goodness, I miss this so much.
Majestic.
One important difference between complex and real calculus is that the semi-circle you draw isn’t a representation of the function, but only the _domain_ of the function. In real calculus, this would be like taking the integral of a function, and instead of drawing the function, you only draw the portion of the x-axis you want to consider.
@@angelmendez-rivera351 All I meant was that the diagrams you typically draw aren’t of the function, but just of the domain. If we were integrating over a circle, then a semi-circle _would_ be the appropriate drawing in real calculus.
@@angelmendez-rivera351 Of course not, I said nothing about the integral of this particular function. I don’t think I misunderstood anything, though my phrasing might not be perfectly accurate. I think you are being a little too pedantic.
I have a question: At 1:11 you mentioned that taking a single limit of the bounds is sketchy, why is that? also why does an even function bypass this issue? thank you :)
Usually, it’s because for odd functions, if you’re not careful and only use one limit, it’ll cancel out to 0 because you’re making the assumptions it’s approaching both infinities at an equal rate. That’s why usually you need to break it up into two limits. For even functions, since there is none of that area canceling bc it’s symmetric around the y-axis, it’s safer to take the one limit
We would normally say the integral of 1/x from -inf to inf is divergent, despite the limit as r->inf of the integral of 1/x from -r to r being 0.
@@jashunpaluru887 good answer
@@stephenbeck7222 wouldn’t the larger issue be that we are integrating over a discontinuity (0)?
Wasn’t that just using Jordan’s Lemma?
WHAT IS ROOT
EXPONENTS NEED
TO PROVE
Fourier transform of Cauchy distribution...
Sweet!
you should go over probability and complex analysis! I wanna know if you can wvaluate the CDF of normal distribution using contour integrals?? Also I want to see how the gaussian distribution, Gamma distribution, and Poisson distribution relate to complex analysis - contour integrals, conformal mappings, etc.
Could be fun idea 🤷♂️
@@mitchellverhelle3986 on his second channel he has complex analysis course.
This video does the Gaussian integral using a contour:
ruclips.net/video/aw9BfO0Whic/видео.html
@@Alex_Deam thanks!
What happens if you add TIME to all this?
YOU ADD THE NUMBER ONE
Because without the number One, time can't exist, why?
Because Zeros go around and around, they are eternal, infinite, so all time begins with a One then followed by Infinite Zeros
Thx for the video, great work!
I love it when years of math you've forgotten just slaps you in the face randomly. Thanks dude.
There's a burned raisin at x = i.
Can you do an example of fractional calculus?
The i-th derivative of something? Or the e-th integral of something?
Woah hold up when you’re integrating on this domain should you break up the boundary depending on their path?
I.E. you have your curve C_r and you also have the straight line going from -R to R call it S_r
So wouldn’t the integral be something like
Int on boundary = int on C_r + int on S_r
He does this. He calculates the contour integral over the whole path using the residue theorem and then finds each piece of the path individually. C_R is defined as only being the top part of the semi-circle. Your S_r is the original integral
Does the complex analysis prove everything in it ?
شكرا لكم.رائع جدا.ذكرتني بأيام كلية العلوم بأكادير في السنة الثانية .الموسم الجامعي 1996/1997مع الأستاذ المناوي اطال الله في عمره
نعم شرحه للرياضيات جد ممتع
Was so thrown off by your use of legit 😂
Well, we have studied this on lessons of complex analysis. Btw, i am a second year student in Russia👍
Can a straight line equal the length of a curved line, both terminated at the same 2 points?
Not in Euclidean space. Euclid actually defines the straight line as the shortest distance between two points.
@@pwmiles56 Thanks for the simple and concise reply!
Do you ask this for a deeper reason because I am working on a breakthrough paper in mathematical physics, and this can relate, it would relate to geodesics and some obscure math
@Solar Flair
Yes.
I wanted an un-biased answer from someone who understands basic geometry, in a mathematics related channel, in order to screencap it and show it to someone else who is insisting that "C/2=D".
Just to let that person know that his argument is wrong.
I mean, can you imagine? Half the circumference equal to the diameter?
Both terminated at the same 2 points, one line straight the other line curved, yet equal in length?
But like I said, this person I'm talking about is very biased, so it's no biggie. It's just an internet contrarian :)
Thanks again @MrPwmiles
@@Sarugakil0l Hahaha gold !!
"Bound DDR" = playing arcade games in a gimp costume
I had never seen that amazing number π/e coming out from a relevant result before. Will we maybe someday use this or other results concerning π/e to prove it is a transcendental number?
Linear algebra please
😭😭😭😭 i hate it but yes it needs
Okay, I did something dumb calculating the residue, and I don't know what. Please help?
cos(z) = Re( e^iz ) for all complex z, but cos(i) = 1/2 (e+1/e) is not the real part of e^(i^2) = 1/e.
It's because the correct definition of cos(z) for any z is
cos(z) = 1/2 (e^(iz) + e^(-iz))
Whereas
Re(e^(iz)) = 1/2 (e^(iz) + e^(-iz*))
In the second term you have to conjugate everything including the z. So if z is not real, Re(e^(iz)) is not the same as cos(z).
@@pwmiles56 Thanks. That makes sense, the substitution cos(x)=Re(e^ix) works up front because x is real.
But that still doesn't help me understand the bigger picture, because why take the real part in the first place when this equality fails over the complex numbers? The two formulas you gave only agree over the reals.
It seems to me like taking the real part is rewriting the integrand to be a different function over the complexes that only happens to agree with the original integrand on the real axis. Since computing the residue requires evaluating (z-i) times the integrand at z=i, we get different answers depending on which formula we use, so taking the real part is actually /required/ in this case. How am I supposed to know to do this for arbitrary integrands? I'm lacking intuition here.
@@davidblauyoutube We want
I = integral along the real axis of cos(z)/(1+z^2)
= Re[ integral along the real axis of exp(iz)/(1+z^2) ]
Then everything inside [ ] is replaced by
the closed contour integral, minus the integral around the semicircle.
It's technically correct. It would have been clearer to split out
cos(z) = 1/2 (exp(iz) + exp(-iz))
and handle the second part in the negative-imaginary halfplane. But a little more work, hence the shortcut presumably.
@@pwmiles56 Thank you.
Does anyone know where I can find Micheal Penn doing math in 4K 60fps? Wait, nevermind...
I’m such a dumbass. So interesting to see what I’m clueless about though.
Can you do something like that with the function cos(x)/(1+x^3) ?
That has a singularity on the real line, x = -1. So no, or at least not without some fancy way to integrate through the singularity. In a simple view the integral does not exist
You could find the principle value by taking a small 'ε-bump' around that point and setting ε -> 0
@@user_2793 That wouldn't be the integral along the real line. Nice thinking!
Take a half-circle around x=-1 with a small e>0, and compute the residue as usual. You can derive the Laurent series a shown that except the Residue term any Higher order terms are zero.
Good
So is the output of the integral interpreted as the length of the path or the area enclosed by the path?
Neither. It's f(z) deltaz summed along the path, in the limit as deltaz->0
The real part of the path is given but the rest is our choice. You could complete the path as a rectangle, or a triangle, whatever. The semicircle is chosen because it simplifies the limit calculation.
@@pwmiles56 Sir can you please explain 4:50 "We Added this in in this integral, so we need to subtract it out", this region C_r is already covered by D_r boundary (first integral)?? And why need subtract it out? Dont understand this step.
@@solarflair3613 Hi Solar Flair.
Best take a step back. The trick is to take the integral we want, along the real line, and then >>complete
@@pwmiles56 Aaah I kind of understand it now :D It does makes sense. Thank you so so much !! All the best. You're awesome
Hmm, the way that it's done is a little misleading. I personally, would have started with the contour integral and got the real integral in question as part of breaking up the contour. The reason for this is that you can't always tell what bit of the integral will contribute.
💓
The function is odd so of course the limit from - infinity to +infinity is 0. QED
a calculator
:D
Mlp
Been 25 years since I studied residues. Just trying to clean up the shards of brain tissue that left on the ground as I attempted to wrench memories of that skill out of the recesses of my mind. Mostly I just remember the girl who typically sat in the front row during lectures. Such a rarity to find an attractive female in an applied math course. I definitely need a refresher. Will have to check out your other course.
There's a nice account on the web, Contour integration and Cauchy's theorem by Cosgrove. (I just tried to paste the link but it got deleted)
She was hawt, I remember her. Top 5 for most attractive females I have ever seen, you just could not help hoping to see her.
👍🏻
Hi??
Cool, but this method made me a bit dz.
grubo
First like
good good
Show taped letters🔠 in high resolution
It's very good for⭐✨ eyes👀👀👀👀👀👀👀👀👀
If it's an even function and the bounds are in the form (a, -a), how is the result anything other than 0?
This holds for odd functions.
@@Eismann1 oh duh, i'm stupid lol
pi/e
Integrand is even so it looks like Laplace transform is possible but does Laplace transform really help to calculate this integral
Isnt the worry about the upper and lower limit only real if theres a potential problem with convergence? abs(cos(x)/(1+x^2)) is bounded by 1/(1+x^2) so no worry there.
Not if x is complex! | cos | goes to infinity for both positive and negative Im(x)
@@pwmiles56 This was a question about the part of the path thats on the real axis. The other part is only auxilliary.
Yes, here both limits converge separately so it's fine to combine them. Don't know what the function being even has to do with anything.
@@vizart2045 But you need the integral around the semicircle to get the final result. So it does need to be bounded, at least.
@@pwmiles56 Yes thats true but you could just as well have chosen a box shaped path. Its only auxilliary and only there to make a closed path.
I'm very rusty on complex calculus, but I seem to remember that a contour integral over a closed path is always zero, which does not appear to hold here. Maybe there is a problem with the function not being differentiable at the (R, 0) or (-R, 0) points, but I can't put my finger on the definition of "differentiable" in this context.
Integral of an *analytic* function over a closed path is zero (Cauchy's thereom). The function is analytic everywhere except for z=±i.
The contour need only be continuous (possibly excluding pathological cases such as fractals). The function must be differentiable i.e. df = f'dz is a meaningful identity. This is the same thing as it being analytic. Some simple functions are not analytic e.g. f(z)=z*. However polynomials and power series such as cos(z), etc, are generally analytic over some domain. And then yes, the contour integral can be "shrunk" to a tiny one round any singularity(ies), which gives us Cauchy's theorem.
But, in this case, the real part of the whole contour integral is pi/e. Both the real and imaginary parts should have been zero, then. I was wondering what is broken here.
No it is not. It is zero if and only if function is analytic everywhere inside the contour. However there's an i point which is a simple pole.
@@juandesalgado It's the singularity. The zero result only applies if the function is holomorphic within the contour, which is a fancy way of saying there aren't any singularities.
E.g. f(z) = 1/z
f(z) has a simple pole (a first order pole) at z=0
Integrate round a circle of radius r, centred on the origin
Put z=exp(i theta), dz = i z dtheta
I = integ (theta=0 to 2 pi) iz / z dtheta = 2 pi i
This illustrates the Cauchy residue theorem. The residue is the limit of z f(z) as z->0. In general if a is a simple pole the limit is (z-a)f(z) as z->a.
Nice animation
A ridiculous instructive video at introducing and using the residue theorem to an audience that is assumed to not have ever been introduced to it before. In that case, the viewer will be totally confused with what you just did. Why not just spend 10 minutes introducing the residue theorem and then an application of it using this example.
It’s on his other channel, MathMajor.
I agree. There seems to be a fixation on technique, but knowing a technique is not the same as understanding it. I did find this quite helpful, as I'd forgotten the details of residues, but it doesn't offer much to newbies.
In the first 20 seconds of the video he mentions he will be using methods from complex analysis and links to a channel with a 20+ part series, the title of one of which is "the residue theorem". How much hand-holding do you want?
You know, you are right. Why don’t you replace that downloaded Nancy Sinatra video on your channel with a 10 minute video introducing the residue theorem?
@@m44lshannon I shouldn't labour this, but you kind of made my point. Knowledge doesn't just lie there in some book or video course. It is something the student has inside them. Generally acquired by practice, working on problems and proofs. The buzz-phrase is "active learning".
Michael is great, he knows the material, his presence and delivery are superb, but he can actually be a bit too convincing. Sometimes. Seems to me