"Complex Variables" by John W. Dettman is a great read: the first part covers the geometry/topology of the complex space from a Mathematician's perspective, and the second part covers application of complex analysis to differential equations and integral transformations, etc. from a Physicist's perspective.
@@manstuckinabox3679 Here is a book completely dedicated to contour integration for real integrals “Complex Integration - Ron Gordon” Ron is a mathematician who solved some of the freakiest integrals on the Math Stack exchange.
Nice result and great explanation. However, some minor comments: 1- For the arc integrals, you didn't use Jordan lemma. Instead, you used ML estimate. 2- The ln function is not a function. Rather, a multivalued function. Therefore, you must define a branch cut. In your case, it worked out because your contour didn't go through it (you were lucky..) Note I'm not a mathematician, nor do I constrain myself to their church of rigorously writing down everything. But the steps you missed are essential and will cause harm in other examples if not taken into account.
"he wakes up amd becomes a real man, until he realizes contour integration is not defined on his feild." Oh, my lord I'm crying, this is the most beautiful string of words a youtuber ever said, I feel so involved in the scriptwriting process, the moment you started the video I knew you were going to reference the complex man, The Integral, the complex, the complex. the whole shebang, fills me with determination for my complex analysis exam (not a traditional one, but still challenging and "graded" non the less). Thank you so much for a beautiful explanation through an example of how this works technically. oh and at 6:35 it would have been nice if you explained a bit more why it is a simple pole, I hilariously have my faith in a little conjecture that I'm working to prove or disprove, that given a function of the form (F(z))/Product(Pvn(z))), where Pvn is a polynomial of the first order, the order of the pole depends on the power of Pvn (or by "how much it appears in the product ").
Thank you so much! What you've demonstrated in this video is a totally new concept for me. And you explained it so well!! I can finally have a slight understanding of how to use complex analysis of integrals thanks to you, so thank you!! I love your videos and with each video of yours my ability to deal with all sorts of integrals improves significantly. thank you so much!!
Bro... We need more of this! I seriously enjoyed this video, and considering the fact that I've never explicitly studied Contour Integration, this one felt super intuitive and to be honest, makes me wanna go continue my Complex Analysis course now!
@@daddy_myers Boi I just finished it, consider me your rival and show me your true form! (obvioisly your laurent expantion form WHICH I, FOR THE LIFE OF ME, CAN'T FIND THEM FOR THESE ANNOYING FREAKING RESIPROCALS OF POLYNOMIALS!)
I love residues and these integrals. While this is messy other integrals of this type demonstrate just how efficient mathematics can be (cos(x)/(x^2+1) for example). While this is complicated I'm always surprised why this isn't taught to capable high school students as its shows just how "simple" maths can be if the correct tool is chosen which is a powerful lesson.
Yeah but there is alot of pure mathematics that goes into this stuff so it's best to save it for later However, I do think that having a small survey of contour integration as part of Cal2 (I know it sounds ambitious but hey why not😂) would be pretty awesome
@@maths_505 showing the integral of 1/(1+x^2) from -inf to inf can be done 3 ways; arctan, partial fractions and complex analysis is beautiful. Telling someone to identify the poles (make them complex rather than on the number line), show how to calculate residue:f(z-pole)*f(z)|z=pole -> 2*pi*i*residue. I always found being shown different methods rather than feeling at school that there was only 1 way opened up the creativity that is so important for maths and means the "rigid" and "constrained" approach people often dislike goes. The Richard Feymann interview is the early 80s discussed this; arithmetic vs. algebra (czcams.com/video/VW6LYuli7VU/video.html). If you've not seen the whole interview....its a thing of beauty and the answer to the question: Dead or alive who would you have dinner with! Also, epsilon (E) limit just boils down to whether ln^2 argument becomes undefined faster than the exponential term goes to 0. As the denominator tends to 1 as E -> 0....of which I dropped the oscillatory to get leaving lim E -> 0 (ln^2 E)/E. Couple of l'Hopitals later and the limit = 0.
@@edcoad4930can also be done with trig sub => int 1/(x^2+1) dx x = tan x => int 1/(sec^2(x)) * sec^2(x) dx => int 1 dx => tan(x) + C It could also be done with Integration by parts => x *1/(x^2+1) - int x * -1x^2/(x^2+1)^2 = + int x^3/(x^2+1)^2 => here you could use polynomdivison
The integral with the log in first power can be shown to be zero easily by letting x=e^t which gives int_{t=-inf}^{+inf} t/(e^t+e^(-t)) which vanishes by symmetry as the integrand is an odd function.
Very valuable and concise presentation … but it would be highly appreciated if the steps to evaluate the integral on the little clockwise contour called “gamma” are spelled-out for completeness!
There are people who claim that calculus is a complex science, while others say that it is a simple and easy science. Is this related to its uses? For example, in physics, it is only arithmetic, but in mathematics, it is complicated and requires high analysis skills. Lately, this science has seemed interesting to me, and my curiosity to learn more about it has begun to increase
The integral of ln x/(x^2+1) can be evaluated in a very elementary way: split the integral up in two parts : from 0 to 1 and from 1 to inf.From 0 to 1 use the new variable exp (-t ) = x and in the interval 1 to inf choose exp t= x . Then after some elementary steps you find that the two integrals are equal up to the sign , i.e. the sum is zero.
In fact integral on the left hand side can be splitted into interval 0..1 and 1..infinity Then after integrating by parts twice we have to calculate the sum (sum of alternating reciprocals of odd cubes)
RHS integral is easy with x = 1/u substitution LHS integral is easy if you can calculate sum It is possible to calculate these integrals without contour integration
Question: When deciding on a contour, doesn’t it make more sense to mark down all the branch points, cuts, forbidden points etc, FIRST and THEN choose a contour appropriately ?
Thanks so much for this elegant solution and the comprehensive explanations. Just one question: why do you circumvent the point z=0 in the upper half plane and not in the lower half plane? My tentative answer is: z=0 is actually a branch point of log(z), so we have to draw the cut starting at z=0 in an appropriate manner. If we chose it to lie in the lower half plane and if, in addition, we would like our path close to the real axis to avoid the cut, we need to circumvent the branch point in the upper half plane. Is this reasoning correct?
Yes indeed Our branch cut to ensure the log function is single valued is [0,-i\infty). So we draw a small semi circle around z=0. We avoid the singularity and ensure everything's nice and meromorphic
Although this isnt really the best explanation I've given on using contour integration so far....it is the first one though. For much better explanations, check out the 2 most recent videos in the contour integration playlist. I explained things more formally there.
Does the shape of the contour matter? If yes, how do we choose it? For example why don't you choose a full circle so that both residues would be considered?
Not sure I agree with the handwaving that, because the modulus was zero and the integral was less than or equal to the modulus, the integral itself was zero. Isn't the integral still allowed to be negative potentially-- LESS THAN or equal to? What did I miss there?
11:52 : Is this right? IMO the sign of the i*pi term needs to be switched when squaring, which means that there is no 2*pi*i/x^2+1 integral after expanding the square. In the end of the video you just set it zero, which seems only valid had you proved that the integral cannot yield a complex value. That doesn’t seem to be obvious at all… Rather, the term doesn’t contribute because it doesn’t exist in the first place;-) Could you pls check and comment? Seems to be a major issue
Switch the sign? There is no sign mistake there and as far as the second integral is concerned ofcourse it exists and it's zero. Also it's plainly obvious that you won't get a complex value for it since you're integrating a real valued function on the interval of integration. Thankfully there are no major issues.
I wonder whether the last steps are valid - suppose I-2 yielded a complex number, then pi*i*I-2 had been real. I guess one would have to mention that I-2 is real, no? Also, for the parametrization of z in both Gamma-integrals, i never really understood why R can be treated as a constant in those situations. We are integrating over what’s inside the contour, not just on its border. Shouldn’t R vary from 0 to infinity? Would be great if you could enlighten me…
oh finally ... sometimes google surprises me ... i have a question can you solve the first integral using the Faynman trick .... using the variable ln(tx) ?? because for some reason does not give me the right answer
"Complex Variables" by John W. Dettman is a great read: the first part covers the geometry/topology of the complex space from a Mathematician's perspective, and the second part covers application of complex analysis to differential equations and integral transformations, etc. from a Physicist's perspective.
Does it have exercises?
@@manstuckinabox3679 Yes.
Schaum's Outline on Complex Variables is a good compliment to Dettman.
@@douglasstrother6584 Ah thanks! I needed a good book to supplement my course!
@@manstuckinabox3679
Here is a book completely dedicated to contour integration for real integrals
“Complex Integration - Ron Gordon”
Ron is a mathematician who solved some of the freakiest integrals on the Math Stack exchange.
Nice introduction and entertainment.
Nice result and great explanation. However, some minor comments:
1- For the arc integrals, you didn't use Jordan lemma. Instead, you used ML estimate.
2- The ln function is not a function. Rather, a multivalued function. Therefore, you must define a branch cut. In your case, it worked out because your contour didn't go through it (you were lucky..)
Note I'm not a mathematician, nor do I constrain myself to their church of rigorously writing down everything. But the steps you missed are essential and will cause harm in other examples if not taken into account.
"he wakes up amd becomes a real man, until he realizes contour integration is not defined on his feild."
Oh, my lord I'm crying, this is the most beautiful string of words a youtuber ever said, I feel so involved in the scriptwriting process, the moment you started the video I knew you were going to reference the complex man, The Integral, the complex, the complex. the whole shebang, fills me with determination for my complex analysis exam (not a traditional one, but still challenging and "graded" non the less). Thank you so much for a beautiful explanation through an example of how this works technically.
oh and at 6:35 it would have been nice if you explained a bit more why it is a simple pole, I hilariously have my faith in a little conjecture that I'm working to prove or disprove, that given a function of the form (F(z))/Product(Pvn(z))), where Pvn is a polynomial of the first order, the order of the pole depends on the power of Pvn (or by "how much it appears in the product ").
Thank you so much!
What you've demonstrated in this video is a totally new concept for me. And you explained it so well!!
I can finally have a slight understanding of how to use complex analysis of integrals thanks to you, so thank you!!
I love your videos and with each video of yours my ability to deal with all sorts of integrals improves significantly. thank you so much!!
Brilliant! Nothing more to add, this was a wonderful integral and a great proof!
Thanks mate
Great channel...! challenges define the man behind the channel!
I feel like a little kid on Christmas day :,)
Bro... We need more of this!
I seriously enjoyed this video, and considering the fact that I've never explicitly studied Contour Integration, this one felt super intuitive and to be honest, makes me wanna go continue my Complex Analysis course now!
I'll upload more videos like cool and unusual contours and more complicated integrals.
@@daddy_myers Boi I just finished it, consider me your rival and show me your true form! (obvioisly your laurent expantion form WHICH I, FOR THE LIFE OF ME, CAN'T FIND THEM FOR THESE ANNOYING FREAKING RESIPROCALS OF POLYNOMIALS!)
one of the best teachers ever
Thanks mate
the changement of variable u=1/x allows us to do the integral on right
listening to the opening like eye of the tiger before my exam tmrw. great stuff
😂😂😂
Good luck bro
I love residues and these integrals. While this is messy other integrals of this type demonstrate just how efficient mathematics can be (cos(x)/(x^2+1) for example). While this is complicated I'm always surprised why this isn't taught to capable high school students as its shows just how "simple" maths can be if the correct tool is chosen which is a powerful lesson.
Yeah but there is alot of pure mathematics that goes into this stuff so it's best to save it for later
However, I do think that having a small survey of contour integration as part of Cal2 (I know it sounds ambitious but hey why not😂) would be pretty awesome
@@maths_505 showing the integral of 1/(1+x^2) from -inf to inf can be done 3 ways; arctan, partial fractions and complex analysis is beautiful. Telling someone to identify the poles (make them complex rather than on the number line), show how to calculate residue:f(z-pole)*f(z)|z=pole -> 2*pi*i*residue.
I always found being shown different methods rather than feeling at school that there was only 1 way opened up the creativity that is so important for maths and means the "rigid" and "constrained" approach people often dislike goes. The Richard Feymann interview is the early 80s discussed this; arithmetic vs. algebra (czcams.com/video/VW6LYuli7VU/video.html). If you've not seen the whole interview....its a thing of beauty and the answer to the question: Dead or alive who would you have dinner with!
Also, epsilon (E) limit just boils down to whether ln^2 argument becomes undefined faster than the exponential term goes to 0. As the denominator tends to 1 as E -> 0....of which I dropped the oscillatory to get leaving lim E -> 0 (ln^2 E)/E. Couple of l'Hopitals later and the limit = 0.
@@edcoad4930can also be done with trig sub => int 1/(x^2+1) dx x = tan x => int 1/(sec^2(x)) * sec^2(x) dx => int 1 dx => tan(x) + C
It could also be done with Integration by parts => x *1/(x^2+1) - int x * -1x^2/(x^2+1)^2 = + int x^3/(x^2+1)^2 => here you could use polynomdivison
finally we’re doing the good techniques
Actually made me chuckle at a couple points
The integral with the log in first power can be shown to be zero easily by letting x=e^t which gives int_{t=-inf}^{+inf} t/(e^t+e^(-t)) which vanishes by symmetry as the integrand is an odd function.
The intro was on point
Great speech in the beginning. Goku is good, good complex man.
Thanks 😂
Very valuable and concise presentation … but it would be highly appreciated if the steps to evaluate the integral on the little clockwise contour called “gamma” are spelled-out for completeness!
There are people who claim that calculus is a complex science, while others say that it is a simple and easy science. Is this related to its uses? For example, in physics, it is only arithmetic, but in mathematics, it is complicated and requires high analysis skills.
Lately, this science has seemed interesting to me, and my curiosity to learn more about it has begun to increase
integral calculus is a complex science without a doubt
@@MaalikJohannes nice name
The integral of ln x/(x^2+1) can be evaluated in a very elementary way: split the integral up in two parts : from 0 to 1 and from 1 to inf.From 0 to 1 use the new variable exp (-t ) = x and in the interval 1 to inf choose exp t= x . Then after some elementary steps you find that the two integrals are equal up to the sign , i.e. the sum is zero.
Or by simple substitution x=1/u
In fact integral on the left hand side can be splitted into interval 0..1 and 1..infinity
Then after integrating by parts twice we have to calculate the sum (sum of alternating reciprocals of odd cubes)
RHS integral is easy with x = 1/u substitution
LHS integral is easy if you can calculate sum
It is possible to calculate these integrals without contour integration
Question:
When deciding on a contour, doesn’t it make more sense to mark down all the branch points, cuts, forbidden points etc, FIRST and THEN choose a contour appropriately ?
What a introduction….MAN to REAL MAN to COMPLEX MAN …. Whoaaaa…
Thanks so much for this elegant solution and the comprehensive explanations.
Just one question: why do you circumvent the point z=0 in the upper half plane and not in the lower half plane?
My tentative answer is: z=0 is actually a branch point of log(z), so we have to draw the cut starting at z=0 in an appropriate manner. If we chose it to lie in the lower half plane and if, in addition, we would like our path close to the real axis to avoid the cut, we need to circumvent the branch point in the upper half plane.
Is this reasoning correct?
Yes indeed
Our branch cut to ensure the log function is single valued is [0,-i\infty).
So we draw a small semi circle around z=0. We avoid the singularity and ensure everything's nice and meromorphic
Although this isnt really the best explanation I've given on using contour integration so far....it is the first one though.
For much better explanations, check out the 2 most recent videos in the contour integration playlist. I explained things more formally there.
We need more simplicity and more exploration for these kinds of hard integrals
Il secondo integrale, pongo lnx=t, risulta l'integrale da - inf a +inf di t/2cht che è una funzione dispari, perciò I=0
Does the shape of the contour matter? If yes, how do we choose it? For example why don't you choose a full circle so that both residues would be considered?
Not sure I agree with the handwaving that, because the modulus was zero and the integral was less than or equal to the modulus, the integral itself was zero. Isn't the integral still allowed to be negative potentially-- LESS THAN or equal to? What did I miss there?
No, because:
1. Suppose | f(z) |
Where is the branch cut ?
Yes; getting into complex analysis is like puberty for mathematicians.
"Today I am a man."
At 21:42 wouldn't their be a e^(2i phi) instead of just e^(i phi) , it doesn't matter later tho ik
11:52 : Is this right? IMO the sign of the i*pi term needs to be switched when squaring, which means that there is no 2*pi*i/x^2+1 integral after expanding the square. In the end of the video you just set it zero, which seems only valid had you proved that the integral cannot yield a complex value. That doesn’t seem to be obvious at all… Rather, the term doesn’t contribute because it doesn’t exist in the first place;-) Could you pls check and comment? Seems to be a major issue
Switch the sign?
There is no sign mistake there and as far as the second integral is concerned ofcourse it exists and it's zero. Also it's plainly obvious that you won't get a complex value for it since you're integrating a real valued function on the interval of integration. Thankfully there are no major issues.
Cauchy| Cow - Chi
Please do teach us contour integration in your new channel
Jackson Larry Jones Michael Harris Paul
I wonder whether the last steps are valid - suppose I-2 yielded a complex number, then pi*i*I-2 had been real. I guess one would have to mention that I-2 is real, no?
Also, for the parametrization of z in both Gamma-integrals, i never really understood why R can be treated as a constant in those situations. We are integrating over what’s inside the contour, not just on its border. Shouldn’t R vary from 0 to infinity? Would be great if you could enlighten me…
Integration on the contour not inside it
oh finally ... sometimes google surprises me ... i have a question can you solve the first integral using the Faynman trick .... using the variable ln(tx) ?? because for some reason does not give me the right answer
You don't need Feynman's trick for the integral. Here's a solution development using purely real methods:
czcams.com/video/XuZCzlMlKfw/video.html
Il n1 4B(3)...B f.beta di Dirichlet
0:56 😂😂😂
HEN HAO! ("very good" in Chinese)
Time for some more manliness: "Men, Men, Men!" ~ Martin Mull
czcams.com/video/2lCaIZE7Wuo/video.html
It's testosterone time!
@@daddy_myers "Yeah, Baby!", Austin Powers
Man ! Are you drunk ?
😂😂😂