This is almost identical to the question I got on my Complex Analysis exam last year, including using reverse triangle identity to use Jordan's lemma. Pretty sure I tackled it using residues though
I wanna just take a second to really appreciate this channel. I'm a college freshman, and it's 2 in the morning the Sunday before dead week. And I'm here. The math is over my head, but somehow you present this in a way I can follow, and it's better than any teacher or professor I've had. It's people like you and Grant from 3blue1brown and many other of these math channels that I can thank my love of math for. You just make it so interesting, so elegant, so... exciting, as dorky as that sounds. I can give channels like you a lot of credit to where I am now, a math major. I've always been good at math, but I've never seen it as something I could do for the rest of my life. Not till I found channels like yours. They have instilled a passion in me. I cannot thank you enough. Keep it up, you make some damn good videos
Thank you so much : I have been studying complex analysis on my own from books and for some reason the epsilon part was not clear : it is so much better with an example !! Great video (and channel !!)
First watched this video in 2018, in my highschool year, without any idea of what you are talking about. Now after learning some complex analysis, it feels so great to be able to fully understand it!
I tried this with complex analysis before watching the video. My closed curve was a little different, but I arrived at the same result. Instead of a half-circle through the complex plane around the first and second quadrant, I chose a quarter-circle, which resulted in a close curve around the 1st quadrant, minus the origin. My complex integrand was e^(iz)/z as well.
Integrals 2 and 4 on the contour are 2i * dirichlet integral (see this by u substitution u = -x) so we quickly get that the value we are looking for is -1/2i * ( integrals 1 and 3) and 1 has a bound with a factor of exp(-Rsin(x)) with sin(x) positive so quickly runs off to zero. The final integral 3 is 1/2 the residue with negative sign as it is clockwise around the origin
3:00 while harder to spot, perhaps (eⁱᶻ-1)/z would be a better choice of function (if real valued, its imaginary component is also sinx/x) For this function, 0 is a removable singularity rather than a pole, since eⁱᶻ shrinks faster than z, so we can simply ignore it, thus avoiding any circular arc to bypass the singularity. Proceeding this way, we would have to assume Jordan’s Lemma to eliminate one of the integrals as R → ∞, but if we are allowed to do that then this is a much “easier” function to deal with.
I remember in last year of high school learning integrals, that i asked myself out of boredum what would the integral of (sin x)/x be. By the methods we had learned then i was looping myself with substitutions. I actually did end up solving it a few weeks later after i read in an engineering book how to taylor expand functions. I expanded the sin(x) and i go in the end an infinite series which corresponded to pi/2 . Felt very nice for a 17 yr old kid.
That’s just a sine integral (Si) in disguise. At the start you can substitute u=e^x and you get the integral from 0 to infinity of sin(u)/(u)du. You can use Feynman method to integrate it i think. Imma do it and then edit when I’m done. Ye just say that I(a)= integral from 0 to infinity of (sin(u)/u)e^(-au)du and differentiate with respect to a. Now that I think about it it’s actually pretty trivial lol, anyways the integral evaluates to Pi/2
You should have clarified the Jordan lemma you utilized to drop the line integral on the Capital Lambda! This is worthy of another relevant presentation!
That's good, trouble is that the complex integral just came from thin air. Why for instance just the upper half plane? (I assume because we don't go to minus infinity?). Also no motivation for using that complex integral in the first place. It works really well but it's like as if you know the answer and work back to the beginning.
I'll just say that I spent my summer studying complex analysis since I saw this video back in early February and wanted to see if I would understand it later on. And well, no. I still don't understand jack-shit lmao. Good video papa, it's old but gold.
Good video! My teacher commented It equals the area of the main triangle (base = Pi, heigth =1). It seems pure black magic the cancellations that occur.
Its interesting to contrast the solution to this integral using the Feynman technique. You will not need to know Complex Analysis... A detailed Ytb video of how this is done, see FLM's nemesis BlackPenRedPen, entitled> _The main dish, integral of sin(x)/x from 0 to inf, via Feynman's Technique_
Yo Papa Flammy help me out here a bit. When you substituted x = - v and changed the limits, the negative sign is still present in the power of the exponential. When I re-substitute -v as x, the exponential has power is ix and not -ix. I did not understand why -iv became -x and not x, since -v is x. If anyone can help me out that would be extremely helpful. |e| love your videos boi. That part where we move around the singularity blew my mind. Keep up the good work of the mathematical gods.
When in doubt, assume your integral equals zero. You’ll be right like half the time. I kid you not, one of my math TAs told the class this a few years ago 😂😂
the sum of the integrals is zero because the contour goes epsilon above the pole at z=0. but if we pick the contour that went epsilon below zero the sum of the integrals would be 2pi*residue with all the rest being the same?
@14:46 Why aren't able to just take the limit as epsilon goes to zero of integral bounds? In that case since range of the integral is 0, the integral would be 0. Why can't this be done? Obviously the value of the integral is different than what you got, but don't see how that is incorrect.
At 10:21 when you substitute the x back in, don't the limits of the integral also change and become negative again? Because you left the limits in terms of v when substituting it back in.
let's use one of the theorems of engineering: sin(x)=x. So now we have an integral from 0 to inf of x/x = 1, and constants are trivial to integrate so it's left as an excercise to readers
When you're so deep into math you don't even bother writing out the parametrization of your curve in the complex plane before applying jordan's lemma. Note to undergrads watching the video, you should definitely do it if you expect earning points on your complex analysis exam.
True it is a little disappointing seeing a math video not explaining every step. It kinda seems like watching someone rush through a homework problem. You can tell he wasn't too into this particular problem.
14:40 was expecting "like you play around with your girlfriend" 10:02 was expecting " you can call it your mother, i don't care"🤣 I guess it was to early in the past for you to make your usual jokes, the ones you make in later videos. One more thing is, i think it's easy to notice, that now you've become more relaxed, what i mean is, you use different words, your sentences are more fluent, which in leads to more spicy videos. 🔥
I am kinda confused by the -ε to ε part of the integral. You are going counterclockwise on that contour, shouldn't it somehow get an additional negative sign. I mean if you avoid the pole and go through the upper half then the original integral is 0. But what if you while closing the ε contour, you go through the lower plane? how does it work then?
let -v=x so -x=v and dv=-dx and dx=-dv then you declare that v=x (the same x) doesn't this imply by substitution that x=-x but: if kx = x then k=1 right? what is going on here? could x=kx for any k?
Hi! This might be a beginner question, but around 6 minutes you talk about this integrand being analytical, and how it would be possible to put it into a series form, because it's going around the singularity, but then a question rises to me: how about the tight corners in the -R, -epsilon, epsilon and R? Shouldn't differentiation be impossible in the corners because the derivative is undefined there? Or is it that breaking the integrals into parts and summing them up removes this problem?
When doing contour integrals on the complex plane, you mentioned cauchy’s integral therom about the closed contour being analytical therefore yielding a 0. Ultimately this just seems like an extension of fundamental thm of line integrals, but for the complex plane. This whole business seems a lot like green’s thm, but there’s obviously no vector field. What is the analog in this case, or is it just that one can view the integral in just the same way as if it were not a vector field? Hopeful my question is comprehensible, but ask if i'm leaving something out.
Wait so for the integral from Pi to 0, since the lower value should be down wouldn’t it be better to make it from Pi to 2Pi (Tao), which also allows for an infinite number of values if we keep adding 2Pi for each
curious as to where everyone learns this material. I'm learning complex analysis in a class called mathematical physics, because it's for the physics and applied physics majors, and the math majors take complex analysis in the math department. how common is it for non-math majors to take a class like this?
What happens if you take a contour integral and you include instead of exclude the z=0 residue? Will that make everything much more complicated? Probably yes but maybe it is fun to do?
Don’t -1=cos(x)+isin(x) and 1=cos(x)+isin(x) both have an infinite number of solutions because of their periodicity? Why did you choose the smallest one? Is it because epsilon needs to be as small as possible by definition??
He could've chose any pair of solutions and get the same result (as long as they differ by pi, which corresponds to parametrizing that semicircle by the angle phi)
13:40 what is this "chaltons lemme"? i understand gausses estimation lemme which works in most probelms but not this one, but have not come across this thm. can someone explain further, also i couldnt find anything online
good job! but you made a small mistake (which is more of notation really). The integral is on the small semi-circle is not from -ε to ε (thats an integral on the real line).
prof i have a question, i hope u reply as soon as possible cuz i have an exam next week how can i proof that an integrale from -pi to +pi of f(t)=dt/(1+sin^2(t)) equals to pi.√2 i switched sin^2(t) to [1-cos(2t)]/2, then cos(2t) to (z+z^-1)/2 ..etc and by residus theorem i found it equals pi/√2, i dont know what i did wrong, im dealing with (-pi) ti (+pi) like 0 to (2pi) but i see no difference sorry for my bad english, im arabian and study's language is french
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Your sense of humor is wonderful. Reminds me of myself some 55 years ago in grad school. Loved this presentation.
:))
When your integral quickly devolves into a Dirichlet integral
Every time I see an integral done with contours I am amazed how it works without knowing why :D Love your videos!
This is almost identical to the question I got on my Complex Analysis exam last year, including using reverse triangle identity to use Jordan's lemma. Pretty sure I tackled it using residues though
Estimation lemma is quick too
but what countor to choose??
I wanna just take a second to really appreciate this channel. I'm a college freshman, and it's 2 in the morning the Sunday before dead week. And I'm here. The math is over my head, but somehow you present this in a way I can follow, and it's better than any teacher or professor I've had. It's people like you and Grant from 3blue1brown and many other of these math channels that I can thank my love of math for. You just make it so interesting, so elegant, so... exciting, as dorky as that sounds. I can give channels like you a lot of credit to where I am now, a math major. I've always been good at math, but I've never seen it as something I could do for the rest of my life. Not till I found channels like yours. They have instilled a passion in me. I cannot thank you enough. Keep it up, you make some damn good videos
4:26 "Mathematicians hate him! Avoid integrating over singularities using this one weird trick"
This video is more than one year old... Why is the german boi still checking the comments?
Thank you so much : I have been studying complex analysis on my own from books and for some reason the epsilon part was not clear : it is so much better with an example !! Great video (and channel !!)
16:10 "So what is a -1" now that's my kind of question
Real shit
@@ranjitsarkar3126 no Complex Shit
First watched this video in 2018, in my highschool year, without any idea of what you are talking about. Now after learning some complex analysis, it feels so great to be able to fully understand it!
i times phi from 0 to pi
Yes I do
amazing!
lol
I tried this with complex analysis before watching the video. My closed curve was a little different, but I arrived at the same result. Instead of a half-circle through the complex plane around the first and second quadrant, I chose a quarter-circle, which resulted in a close curve around the 1st quadrant, minus the origin. My complex integrand was e^(iz)/z as well.
Not only excellent explanation and clear layout, but excellent engagement. I'll recommend you to my fellow students
Integrals 2 and 4 on the contour are 2i * dirichlet integral (see this by u substitution u = -x) so we quickly get that the value we are looking for is -1/2i * ( integrals 1 and 3) and 1 has a bound with a factor of exp(-Rsin(x)) with sin(x) positive so quickly runs off to zero. The final integral 3 is 1/2 the residue with negative sign as it is clockwise around the origin
At 5:42 The integral of -€ to +€, should be the integral over semicircle of radius €!
negative euro to positive euro
This video is an absolute jolly for me as I am new in complex analysis. It helped me clear a lot of concepts
You have no idea how helpful your videos are to me. I'm a second year physics student.
Lovely approach. Well done.
3:00 while harder to spot, perhaps (eⁱᶻ-1)/z would be a better choice of function (if real valued, its imaginary component is also sinx/x)
For this function, 0 is a removable singularity rather than a pole, since eⁱᶻ shrinks faster than z, so we can simply ignore it, thus avoiding any circular arc to bypass the singularity.
Proceeding this way, we would have to assume Jordan’s Lemma to eliminate one of the integrals as R → ∞, but if we are allowed to do that then this is a much “easier” function to deal with.
2:20 MAH BOIS
thats the cue for us to strap in for the ride!
I've just found your channel today and it is the best thing that could happen on this Sunday morning.
Yeah awesome like that first substitution dx=du/u. I'm probably going to laugh all day.
MAH BOIS
Not Me I just heard that line when also read this, :v
I just started taking Complex Variables so I really appreciate this video, thanks.
I remember in last year of high school learning integrals, that i asked myself out of boredum what would the integral of (sin x)/x be. By the methods we had learned then i was looping myself with substitutions. I actually did end up solving it a few weeks later after i read in an engineering book how to taylor expand functions. I expanded the sin(x) and i go in the end an infinite series which corresponded to pi/2 . Felt very nice for a 17 yr old kid.
man that thing looks simple but it is hard af
yo blammable baths this is a really nice video of yours
makes me wish you had a complex analysis playlist wink wink
the residue theorem is over 9000 !!!
I bet you secretly chose phi so you’d have an arsenal of rhymes to use.
dat name
That’s just a sine integral (Si) in disguise.
At the start you can substitute u=e^x and you get the integral from 0 to infinity of sin(u)/(u)du.
You can use Feynman method to integrate it i think.
Imma do it and then edit when I’m done.
Ye just say that I(a)= integral from 0 to infinity of (sin(u)/u)e^(-au)du and differentiate with respect to a.
Now that I think about it it’s actually pretty trivial lol, anyways the integral evaluates to Pi/2
I hadn’t seen that move at 11:30 before, great stuff
Good stuff man! Recommended you again today :D
Superb presentation ...! Keep up the good works!
The one that states the closed contour integral of an analytic function is 0 is Cauchy's integral theorem. Cauchy's integral formula is different.
Pakai Maclaurin Series: Sin x= x- x^3/ 3!+ x^5/ 5!- .....etc.
(Sin x)/ x= 1- x^2 /3!+ x^4 /5!- ....etc. Intergral (Sin x)/x= Integral [1- x^2/ 3!+ x^4/ 5!- ....etc) dx
Well yes. I actually ended up here trting to solve Sinx/x with substitution . thank you!
You should have clarified the Jordan lemma you utilized to drop the line integral on the Capital Lambda! This is worthy of another relevant presentation!
That's good, trouble is that the complex integral just came from thin air. Why for instance just the upper half plane? (I assume because we don't go to minus infinity?). Also no motivation for using that complex integral in the first place. It works really well but it's like as if you know the answer and work back to the beginning.
lim sin(u)/u as u - - > 0 is 1 and not 0, am i right?
I'll just say that I spent my summer studying complex analysis since I saw this video back in early February and wanted to see if I would understand it later on. And well, no. I still don't understand jack-shit lmao. Good video papa, it's old but gold.
Man your videos are just wonderful!!!!! Thnx
Good video! My teacher commented It equals the area of the main triangle (base = Pi, heigth =1). It seems pure black magic the cancellations that occur.
10:39 thanks for the explanation!
I thoroughly enjoyed this video. Good work.
This 2018 video lasts 20:18
dem boiiii
My favorite German mathematician. Master race lol!!!!
at 3:03, i need few more words why you need to use exp(iz)/z
Its interesting to contrast the solution to this integral using the Feynman technique. You will not need to know Complex Analysis...
A detailed Ytb video of how this is done, see FLM's nemesis BlackPenRedPen, entitled>
_The main dish, integral of sin(x)/x from 0 to inf, via Feynman's Technique_
I also did a video using Leibniz^^
Yo Papa Flammy help me out here a bit. When you substituted x = - v and changed the limits, the negative sign is still present in the power of the exponential. When I re-substitute -v as x, the exponential has power is ix and not -ix. I did not understand why -iv became -x and not x, since -v is x. If anyone can help me out that would be extremely helpful.
|e| love your videos boi. That part where we move around the singularity blew my mind. Keep up the good work of the mathematical gods.
When in doubt, assume your integral equals zero. You’ll be right like half the time. I kid you not, one of my math TAs told the class this a few years ago 😂😂
So here you assume sin(u) = e^it
-Also thanks you have really inspired me over my time here and taught me a lot
my personal favorite is using laplace transform to evaluate the integral.
That was brilliant!
a cute boy explaining math: this is great
the sum of the integrals is zero because the contour goes epsilon above the pole at z=0. but if we pick the contour that went epsilon below zero the sum of the integrals would be 2pi*residue with all the rest being the same?
I love you for this
Excellent video, awesome
to generalize it it's Pi from -inf to inf
for cos(x)/x it's a 0
@14:46 Why aren't able to just take the limit as epsilon goes to zero of integral bounds? In that case since range of the integral is 0, the integral would be 0. Why can't this be done? Obviously the value of the integral is different than what you got, but don't see how that is incorrect.
my boys !
At 10:21 when you substitute the x back in, don't the limits of the integral also change and become negative again? Because you left the limits in terms of v when substituting it back in.
He didn't substitute x back in. He just changed the name of the variable.
That was pretty, might haft to watch that few more times and polish up on lemmas.
That was awesome !
19:06 "Our *final solution* is..." in a German accent!
let's use one of the theorems of engineering: sin(x)=x. So now we have an integral from 0 to inf of x/x = 1, and constants are trivial to integrate so it's left as an excercise to readers
Brilliant. Thanks for posting
Great video keep up the good work
Is the video made deliberately of the length 20:18???
How did you know that you had to integrate( e^iz)/z at 3:01? Was there a theorem or something that I missed?
4:33
We can find da wey?
The only "wae" I know is the way of the Lord
u do not no da wae!
Dirischlett-Integral? that guy was French :D
edit: actually he was german, but still his name was in French
If you hate pronouncing stuff in the wrong way then why don't you pronounce phi as a Greek person would do? (fee)
Gábor Králik
Here in the US, that is often pronounced as phy or phee.
Flammable Maths
No offense, but the German accent is not helping pronunciation.
😁
Lucas M His accent is flammy, of course it helps boi
Lucas M Yes, and pronouncing it Phy is incorrect.
When you're so deep into math you don't even bother writing out the parametrization of your curve in the complex plane before applying jordan's lemma.
Note to undergrads watching the video, you should definitely do it if you expect earning points on your complex analysis exam.
True it is a little disappointing seeing a math video not explaining every step. It kinda seems like watching someone rush through a homework problem. You can tell he wasn't too into this particular problem.
14:40 was expecting "like you play around with your girlfriend"
10:02 was expecting " you can call it your mother, i don't care"🤣
I guess it was to early in the past for you to make your usual jokes, the ones you make in later videos. One more thing is, i think it's easy to notice, that now you've become more relaxed, what i mean is, you use different words, your sentences are more fluent, which in leads to more spicy videos. 🔥
I am kinda confused by the -ε to ε part of the integral. You are going counterclockwise on that contour, shouldn't it somehow get an additional negative sign.
I mean if you avoid the pole and go through the upper half then the original integral is 0. But what if you while closing the ε contour, you go through the lower plane? how does it work then?
let -v=x
so -x=v
and dv=-dx and dx=-dv
then you declare that v=x (the same x)
doesn't this imply by substitution that x=-x
but:
if kx = x
then k=1
right?
what is going on here? could x=kx for any k?
Awesome!!!!!
That moment your an future engineer in and in cal 1 and just watching it for fun
Great video!
integral of (e^xdx)-1. can we find this?
Hi! This might be a beginner question, but around 6 minutes you talk about this integrand being analytical, and how it would be possible to put it into a series form, because it's going around the singularity, but then a question rises to me: how about the tight corners in the -R, -epsilon, epsilon and R? Shouldn't differentiation be impossible in the corners because the derivative is undefined there? Or is it that breaking the integrals into parts and summing them up removes this problem?
Is no one gonna talk about how the length of the video is the same as the year it was uploaded in?
Amazing intro
I defy.
/r/Anarchy
This guy needs to see a head doctor ASAP.
When doing contour integrals on the complex plane, you mentioned cauchy’s integral therom about the closed contour being analytical therefore yielding a 0. Ultimately this just seems like an extension of fundamental thm of line integrals, but for the complex plane. This whole business seems a lot like green’s thm, but there’s obviously no vector field. What is the analog in this case, or is it just that one can view the integral in just the same way as if it were not a vector field? Hopeful my question is comprehensible, but ask if i'm leaving something out.
Wait so for the integral from Pi to 0, since the lower value should be down wouldn’t it be better to make it from Pi to 2Pi (Tao), which also allows for an infinite number of values if we keep adding 2Pi for each
Thank you so much
Very Enjoyable:-)
"i times phi from 0 to pi" Somebody's speaking my language 😂❤️🎊
Hey Flammi boi, i don't understand when do you have to include the residue at infinity and when you do not! Can you enlighten me?
curious as to where everyone learns this material. I'm learning complex analysis in a class called mathematical physics, because it's for the physics and applied physics majors, and the math majors take complex analysis in the math department. how common is it for non-math majors to take a class like this?
I'm learning this as an electrical engineering student
Memes and math, now we're talking!
What happens if you take a contour integral and you include instead of exclude the z=0 residue? Will that make everything much more complicated? Probably yes but maybe it is fun to do?
Don’t -1=cos(x)+isin(x) and 1=cos(x)+isin(x) both have an infinite number of solutions because of their periodicity? Why did you choose the smallest one? Is it because epsilon needs to be as small as possible by definition??
He could've chose any pair of solutions and get the same result (as long as they differ by pi, which corresponds to parametrizing that semicircle by the angle phi)
13:40 what is this "chaltons lemme"? i understand gausses estimation lemme which works in most probelms but not this one, but have not come across this thm. can someone explain further, also i couldnt find anything online
Lemma
So nice
Could You make some more videos about that method of solving integrals? Best regards
It can become too easy by Laplace's transformation
good job! but you made a small mistake (which is more of notation really). The integral is on the small semi-circle is not from -ε to ε (thats an integral on the real line).
prof i have a question, i hope u reply as soon as possible cuz i have an exam next week
how can i proof that an integrale from -pi to +pi of f(t)=dt/(1+sin^2(t)) equals to pi.√2
i switched sin^2(t) to [1-cos(2t)]/2, then cos(2t) to (z+z^-1)/2 ..etc and by residus theorem i found it equals pi/√2, i dont know what i did wrong, im dealing with (-pi) ti (+pi) like 0 to (2pi) but i see no difference
sorry for my bad english, im arabian and study's language is french
D E F Y
Awesome!