For those who are unsatisfied with the explanation at around 15:20, here is an alternative way to evaluate the integral over Gamma, which requires further estimations. th-cam.com/video/yDiC_tZdEp8/w-d-xo.html
This is so cool! Thank you so much for this example; I was quite lost when my professor brought this kind of process up as a point tangential to the lecture.
Loved your series (no pun intended) of math videos. In your videos on the integral of sin(x)/x, I wondered whether there was a simple answer if the limits of integration went from a (>0) to infinity, rather than from 0 to infinity.
Thanks! Although such an value would exist, I'd presume that a closed form would be quite difficult to obtain without numerical methods for integration. Perhaps this could be a topic for investigation!
@@qncubed3 I know I'm very late, but suppose there was a closed form for from a>0 to infinity. This would mean you could express the integral from 0 to a in closed from as pi/2 - (integral from a to infinity) . But this would imply that the primitive for sinx/x can be expressed in closed form, which is a contradiction, so a closed form for the integral from a>0 to infinity cannot possibly be expressed in closed form , right?
lets say epsilon be r to reduce words at 10:35 integ [ exp(i r exp (i * t)) ] ~ 1/r (1 - exp(- r * constant)) when r goes to 0, this results in 0/0 and, this was the same as the jordan thing. or he could have done the same just to apply R going to infinity BEFORE the integration, but he didnt. but ok the limit goes to ~ r const / 1 = 0
What im most interested in is Issue 1 - How do you convert from a real function to a corresponding complex function? Here you changed sin(x) to e^iz and I do not see why. And you dropped the “2” in front of the real part when converting to complex Issue 2 - How do you select the best contour geometry for a given complex integral? You chose a semicircle although there are many other shapes
The integral was not equal to the contour, but rather the contour produces the integral when adding two integrals from the contour lying on the real axis. Although, due to Euler's formula, e^ix = cosx + isinx, so taking the imaginary part would equal sinx. Semicircles are convenient for contours, as e^ix produces values on the unit circle plotted on the real and imaginary axis. That's why he parameterized the contour integrals on the circular path with re^ix, where r is the radius of the circle. To select the "best" contour geometry, it's a good idea to use the simplest integrals that would create a closed contour while avoiding problems like singularities on the contour or branch cuts. For example, to evaluate the desired integral, you need a contour that produces integrals from -infinity to infinity. However, a singularity at zero means we can not use one integral that goes from -infinity to infinity. Instead, we can go from an arbitrarily large negative number to an arbitrarily small negative number, then form a path that goes around zero, to an integral from small to big positive numbers. A circular path of a small radius would be a good choice. To close the contour, we need to make a path that connects the ends of our current contour, where a circular path would also work.
@@snipergranola6359 We need to turn our contour into a closed loop in order to use Cauchy's integral theorem. To connect R and -R, using a semi circle on the upper half of the complex plane is usually a good approach. Hope that helps.
Hello Dear *QN^3* . I just love this playlist, that's it. Please be more active (actually I love your channel and your channel is one my favorite channel; you're in my top three). Thank you so much
Hello I’m so confused with this one concept. I’m sure if a singularity lies ON or is enclosed by a contour then the answer is the sum of the residues, so I don’t understand why we need to take a detour around the origin. Why can’t we let it lie on the semi circle?
I believe Cauchy's residue theorem only applies if all singularities lie within the interior of a contour. Generally, you would want to avoid integrating over these singularities anyways. Since the singularity at the origin is "obstructing" the direct path from -R to R, we must then take a detour. You can the detour above or below the origin, the final result will be the same. If you take the path below, you will need to use Cauchy's residue theorem. I chose to take the upper semicircle as it just saves you a step and you can use the fact that the contour will evaluate to zero instead. Hope that helped :)
qncubed3 I see now. The Cauchy Goursat theorem only applies if it analytic inside and ON C so I assumed residue theorems require singularities to also be inside and ON C. Which is not the case. Thank you!
If a curve is on an arc of a circle, we can parametrise the path using r*e^(i*t) as this will give us the set of complex numbers on a circle of radius r. We can choose the domain of t depending on where the curve starts and ends.
You could add that Theo absolut value of the integral is the Same Than the integral since its zero, also pi and zero are the values that you put into the antiderivative of the integrand so there are no Problems with that sin
thanks for your video ! I have a question : when you compute the integral over little gamma and big gamma, you say that the limit of the integral equal the integral of the limit of the function. But does this step require more justification ? I thought that these kind of equality had to be justified with the theorem of continuity of parameter dependant integral.
Yes, in general, you cannot exchange limits and integrals, and you would need to use the dominated convergence theorem (or other tests) to justify that.
The integrand is sin(x)/x . If we want to convert this to a complex function, we simply accept complex numbers as inputs. So we have sin(z)/z . However, it is generally nicer to work with the complex exponential function in place of sines and cosines (because we can use euler's formula to obtain the sine function again) This is how we get e^(iz)/z as our complex function.
at 5:53, it is still not logical enough. the limits of integral from -infinity to - epsilon where the changed angle pi which must be implimented, compared to the angle 0 from +eps to + inf.
Yeah. Do you know why he was able to just replace "u" with "x" beyond just it being a "dummy variable"? I don't seem to understand how he could equate u = -x to x, since it would be implying that x = -x, or am I missing something?
@@MrChicken1joe okay so a few months later I figured out what happened. Because a definite integral evaluates to just a number, a u-sub maintains that number, and an arbitrary change of variables (from u to x) doesn’t change that value. In sum, the introduction of “u” should make the integrand simpler, but in the end it is the same number so it doesn’t matter if it is in terms of u or x as they are both just dummy variables
at 4:38, can someone show me at least once, in stead of simply using x, use r and theta = pi from -R to -epsilon, because there was an angle change there. instead of using dx, one must use dR, once all is settled, then anyone change the notation from R to x. AND there is NO such as NEGATIVE R in complex, so using -R or -x is same as saying someone probably skipped the logic somewhere and became a hand waving method. yes -R is R with theta = pi in disguise. hope and need to see how mathematicians handle angle changes in a contour line
@@melvindebosscher826 whenever you see a trigonometric function you choose e^iz because its really easier to work with and since sin(z) is really big for im(z) big so it will cause some problem.
The contour I used is simply a semicircular contour of radius R in the complex plane traversed in the positive direction (counterclockwise), with the addition of another semicircular path of radius epsilon at the origin to avoid the singularity.
I assume you are talking about the integral over the upper semicircle. What I did was I found an upper bound for the absolute value of the integral using some inequalities. This gave me another integral which included the parameter "R". In the limit as R approaches infinity, the exponential term in the integral tends towards zero, hence evaluating the limit gives 0. Since the absolute value of the integral is less than or equal to zero, the integral must be zero itself in the limit (the absolute value can only be positive)
For those who are unsatisfied with the explanation at around 15:20, here is an alternative way to evaluate the integral over Gamma, which requires further estimations.
th-cam.com/video/yDiC_tZdEp8/w-d-xo.html
Every theorem is proved while solving question, very nice.
Thanks :)
No it isn’f. Its inefficient and cringeworthy
@@maalikserebryakovIt includes the proofs, you don't do all this when solving it. It's a helpful tecnique
@@maalikserebryakovNo it isnt, its pretty helpful
Thanks a ton! you're a blessing for high schoolers like me who want to learn contour integration
HIGH SCHOOLERS?????
@@yoylecake313 I'm freshman rn
This is so cool! Thank you so much for this example; I was quite lost when my professor brought this kind of process up as a point tangential to the lecture.
Thanks! Glad you found it helpful 👍
I *might* actually pass my qualifying exam because of your videos--thank you so much!
No worries, good luck!
Your videos are extremely helpful!❤
Loved your series (no pun intended) of math videos. In your videos on the integral of sin(x)/x, I wondered whether there was a simple answer if the limits of integration went from a (>0) to infinity, rather than from 0 to infinity.
Thanks! Although such an value would exist, I'd presume that a closed form would be quite difficult to obtain without numerical methods for integration. Perhaps this could be a topic for investigation!
@@qncubed3 I know I'm very late, but suppose there was a closed form for from a>0 to infinity. This would mean you could express the integral from 0 to a in closed from as pi/2 - (integral from a to infinity) . But this would imply that the primitive for sinx/x can be expressed in closed form, which is a contradiction, so a closed form for the integral from a>0 to infinity cannot possibly be expressed in closed form , right?
@@aadhavan7127never too late :) Yes, I agree with your logic 👌
Do you have to rewrite the interval from negative infinity to positive infinity as 2* from zero to infinity? I don’t see the point.
Very nice video, helped me a lot. Thank you so much.
You're welcome! Happy to help
Absolute brilliant video! Well explained and very clear. Thank you sir
Thanks! Glad you enjoyed it!
this was such a helpful vide, really really benefitted from it so thank you so much!
lets say epsilon be r to reduce words
at 10:35 integ [ exp(i r exp (i * t)) ] ~ 1/r (1 - exp(- r * constant)) when r goes to 0, this results in 0/0 and, this was the same as the jordan thing. or he could have done the same just to apply R going to infinity BEFORE the integration, but he didnt. but ok the limit goes to ~ r const / 1 = 0
What im most interested in is
Issue 1 - How do you convert from a real function to a corresponding complex function?
Here you changed sin(x) to e^iz
and I do not see why.
And you dropped the “2” in front of the real part when converting to complex
Issue 2 - How do you select the best contour geometry for a given complex integral?
You chose a semicircle although there are many other shapes
The integral was not equal to the contour, but rather the contour produces the integral when adding two integrals from the contour lying on the real axis. Although, due to Euler's formula, e^ix = cosx + isinx, so taking the imaginary part would equal sinx.
Semicircles are convenient for contours, as e^ix produces values on the unit circle plotted on the real and imaginary axis. That's why he parameterized the contour integrals on the circular path with re^ix, where r is the radius of the circle.
To select the "best" contour geometry, it's a good idea to use the simplest integrals that would create a closed contour while avoiding problems like singularities on the contour or branch cuts. For example, to evaluate the desired integral, you need a contour that produces integrals from -infinity to infinity. However, a singularity at zero means we can not use one integral that goes from -infinity to infinity. Instead, we can go from an arbitrarily large negative number to an arbitrarily small negative number, then form a path that goes around zero, to an integral from small to big positive numbers. A circular path of a small radius would be a good choice. To close the contour, we need to make a path that connects the ends of our current contour, where a circular path would also work.
Plz upload more videos and concept u just saving our time ,question and theorem at the same time,killed 2 birds with one stone
Sure! I have more complex analysis videos coming up if that's what you're after. Glad you found it useful!
@@qncubed3 can give me a PDF of your book
@@snipergranola6359 I don't use/own any books. I learn everything from TH-cam and online :)
@@qncubed3 ok tell me why we take upper semi circle only to evaluate this integral
@@snipergranola6359 We need to turn our contour into a closed loop in order to use Cauchy's integral theorem. To connect R and -R, using a semi circle on the upper half of the complex plane is usually a good approach. Hope that helps.
Thank you I was frustrated while doing maths these videos are restoring my faith in my maths
Hello Dear *QN^3* .
I just love this playlist, that's it.
Please be more active (actually I love your channel and your channel is one my favorite channel; you're in my top three).
Thank you so much
:D
A nice video and logic, but I have some difficulties to recognize your handwrinting of x, u, and n, quite often thery all look the same.
I think it's because he writes too small. Simple fix is to simply write larger scripts and not squash too much writing in small spaces.
Min: 1:50 :
The best semi circle ever!
That was very helpful, thanks for sharing!
Really helpful master!
Hello I’m so confused with this one concept. I’m sure if a singularity lies ON or is enclosed by a contour then the answer is the sum of the residues, so I don’t understand why we need to take a detour around the origin. Why can’t we let it lie on the semi circle?
I believe Cauchy's residue theorem only applies if all singularities lie within the interior of a contour. Generally, you would want to avoid integrating over these singularities anyways. Since the singularity at the origin is "obstructing" the direct path from -R to R, we must then take a detour.
You can the detour above or below the origin, the final result will be the same. If you take the path below, you will need to use Cauchy's residue theorem. I chose to take the upper semicircle as it just saves you a step and you can use the fact that the contour will evaluate to zero instead. Hope that helped :)
qncubed3 I see now. The Cauchy Goursat theorem only applies if it analytic inside and ON C so I assumed residue theorems require singularities to also be inside and ON C. Which is not the case. Thank you!
@@ZainAGhani That's true! You're welcome :)
Great video. Someday I'm going to teach a class in how to properly write the letter "x." Let me know if you'd like to attend.
This is beautiful. Can you explain how you picked the correct parameterizations for the integrals?
If a curve is on an arc of a circle, we can parametrise the path using r*e^(i*t) as this will give us the set of complex numbers on a circle of radius r. We can choose the domain of t depending on where the curve starts and ends.
You could add that Theo absolut value of the integral is the Same Than the integral since its zero, also pi and zero are the values that you put into the antiderivative of the integrand so there are no Problems with that sin
thanks for your video ! I have a question : when you compute the integral over little gamma and big gamma, you say that the limit of the integral equal the integral of the limit of the function. But does this step require more justification ? I thought that these kind of equality had to be justified with the theorem of continuity of parameter dependant integral.
Yes, in general, you cannot exchange limits and integrals, and you would need to use the dominated convergence theorem (or other tests) to justify that.
@@qncubed3 ok thanks for your answer
tkt sa marche
noice All this just to prove inverse fourier transform of Fourier transform is the function itself .
Can you help me with determining the function. Why did we called f(z)= e^iz/z
The integrand is sin(x)/x . If we want to convert this to a complex function, we simply accept complex numbers as inputs. So we have sin(z)/z . However, it is generally nicer to work with the complex exponential function in place of sines and cosines (because we can use euler's formula to obtain the sine function again) This is how we get e^(iz)/z as our complex function.
@@qncubed3 its a little bit late but thank you anyways i figured it out after i watched couple of videos. Thanks again
why is there a singularity since it is removable with L'Hopital?
sin(z)/z has a removable singularity but e^(iz)/z doesnt.
for tht integr for £ we can use residu simplary -ipi
What residues are you referring to?
Best video ever
Thanks 😊
@@qncubed3 can you make a video on the dominant convergence theorem?
Very good explanation 😍😍😍😍
If only I could remember this, I had to watch this like 20 times. I need to know this for an exam 😳😭
at 5:53, it is still not logical enough. the limits of integral from -infinity to - epsilon where the changed angle pi which must be implimented, compared to the angle 0 from +eps to + inf.
Yeah. Do you know why he was able to just replace "u" with "x" beyond just it being a "dummy variable"? I don't seem to understand how he could equate u = -x to x, since it would be implying that x = -x, or am I missing something?
@@varusername nah didnt get that either
@@MrChicken1joe okay so a few months later I figured out what happened. Because a definite integral evaluates to just a number, a u-sub maintains that number, and an arbitrary change of variables (from u to x) doesn’t change that value. In sum, the introduction of “u” should make the integrand simpler, but in the end it is the same number so it doesn’t matter if it is in terms of u or x as they are both just dummy variables
at 4:38, can someone show me at least once, in stead of simply using x, use r and theta = pi from -R to -epsilon, because there was an angle change there. instead of using dx, one must use dR, once all is settled, then anyone change the notation from R to x. AND there is NO such as NEGATIVE R in complex, so using -R or -x is same as saying someone probably skipped the logic somewhere and became a hand waving method. yes -R is R with theta = pi in disguise. hope and need to see how mathematicians handle angle changes in a contour line
Why can we replace sin(x) by exp(iz)
I get it, I had to watch the whole video. Thanks, great help.
@@melvindebosscher826 whenever you see a trigonometric function you choose e^iz because its really easier to work with and since sin(z) is really big for im(z) big so it will cause some problem.
why did you take the condition R limit to inf and epsilon towards 0 at 7:06?
We want to recover our original integral
Why do we need to cut out singularity point?
You cannot integrate over a singularity
super helpful thank you!
Thank you!
You're welcome!
Why did you use e^{iz} and not \frac{e^{iz}-cos\left(z
ight)}{i} in accordance with Euler's formula?
e^iz is easier to work with when contour integrating.
What is \frac{...... supposed to be?
@@azzteke It is LaTeX and represents a fraction. Equivalently that would be read as:
(e^(iz)-cos(z))/i
@@qncubed3 woudnt it be wrong ... plzz answer😭😭😭😭 if you take e^iz as sin z
brilliant
this is how to show a solution, you showed every step and why. it infuriates me to follow a proof and they skip several steps
Where are you from sir
Nice video! But please do not write = in the two equations from 17:02 on. They are not equal to terms on the line before!
Oops... That was probably meant to be an implication arrow.
sir - i am msc physics - i do not under stand how to draw contour i.e contour of present problem -please explain -thank u sir
The contour I used is simply a semicircular contour of radius R in the complex plane traversed in the positive direction (counterclockwise), with the addition of another semicircular path of radius epsilon at the origin to avoid the singularity.
@@qncubed3and why did you choose a semi circle
Disliked and unsubbed byere
Thanks
Should there be an Im operator around the whole integral after making the exp(ix)/x substitution? It will all fall away at the end.
How the second step become zero? Can you explain one more time
I assume you are talking about the integral over the upper semicircle.
What I did was I found an upper bound for the absolute value of the integral using some inequalities. This gave me another integral which included the parameter "R". In the limit as R approaches infinity, the exponential term in the integral tends towards zero, hence evaluating the limit gives 0. Since the absolute value of the integral is less than or equal to zero, the integral must be zero itself in the limit (the absolute value can only be positive)
beautiful
BANGER!
Gracias!
Quite long but gud
That x is a perfect n
Laplace simplifie est peut être mieux
Laplace mieux
ALLAHU AKBAR