"But what if my improper integral diverged? Does that mean I should start crying, drop out of school, and ask my student loans to be forgiven? Of course not! Because student loans can't be forgiven! But staying on topic..."
Thankfully? Those tremendous costs have just been shifted to the people who actually DID pay off their loans. There is no “forgiven” - just idiocy in government and unaccountable former students.
I really appreciate the fast pace of your videos. This way anyone can adjust the tempo using essential YT tools. I used to hate Khan Academy videos for their slow explaining, but this is perfect!
Under certain conditions, yes you can. Those conditions happen to apply for our integrals given the conditions we stipulated at 5:58. You can look here for more details: math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral
The residue theorem resolution in case of this example is basically not needed, because d/dx (arctan x) = 1/(x^2+1) 😉 =>. Integral(-inf,inf) {1/(X^2+1) } = arctan(inf)-arctan(-inf) = pi/2 + pi/2 = pi. For cos(x)/(x^2+1) the residue theorem does a very good jjob!
Of course. I'd love complicated examples especially the ones which are complicated to do in real numbers. I had a question though: What if we write 1/(x^2+1) as (1/2i)[1/(x-i) - 1/(x+i)] and then integrate to get (1/2i)log([x-i]/[x+i]) Would that be correct and would I get a physical relation when I integrate with limits like definite integral in real gives area under the curve or something like that?
I'll work on the complicated examples then; thanks for the request! As for your method, I'm not sure it's going to make things simpler if we evaluate the natural log of an imaginary number. That opens up another can of worms.
Hello! Have been watching your videos in preparation for next year and I absolutely love them! What hardware and software do you use to make the videos? The quality is superb! :)
Hi Jeff, thanks for the questions! Here are my responses: - Bachelors in Physiology and Physics + Masters in Chem Eng. - For recording/editing, I'm using Camtasia. - I'm using a stylus.
Thank you for your prompt responses! I read on wikipedia that Residue theorem generalizes Stoke's theorem (the sum of the interior is equal to the line integral of the exterior, particularly, the curl of a vector field is 0 iff it is conservative). However, Stoke's theorem applies to R^n whereas Residue theorem applies to the complex plane. As I understand it, C does not equal R^2 so how do these theorems relate?
No problem! You're correct that C does not equal R^2, but in some of my videos, I've mentioned how complex numbers are like 2-dimensional versions of real numbers (others have also talked about this analogy, e.g. www.math.brown.edu/~banchoff/Beyond3d/chapter8/section07.html). Since a complex number z can be expressed as x + yi, we can plot z on a complex plane: x units along the Real axis and y units along the Imaginary axis (just like how we can plot an ordered pair of real numbers (x,y) on a Cartesian plane). Since the complex plane and R^2 are so similar in how they're visualized, it isn't that surprising that Stokes' Theorem, which generally applies to R^2 (or beyond), can be extended to the complex plane in the Residue Theorem, even though the complex plane is usually utilized in a different context. Hope that helps!
"But what if my improper integral diverged? Does that mean I should start crying, drop out of school, and ask my student loans to be forgiven? Of course not! Because student loans can't be forgiven! But staying on topic..."
Thankfully that last part didn't age too well
@@asapvarg I was just gonna say that
Thankfully? Those tremendous costs have just been shifted to the people who actually DID pay off their loans. There is no “forgiven” - just idiocy in government and unaccountable former students.
I really appreciate the fast pace of your videos. This way anyone can adjust the tempo using essential YT tools. I used to hate Khan Academy videos for their slow explaining, but this is perfect!
THANK YOU SO MUCH. Beautifully organized and clear! Life saver dude.
This saved me! I barely understood anything in class because the instructor was going so fast, but now I do.
My loans don't have to be forgiven anymore. Thank you
You deserve more than you got sir.
Learning it this week. This explanation is much better than the one on our textbook.
Excellent teaching method sir...Respect from Pakistan
Thank you sir , you have helped me soo much for my exams !!! Great series
8:20 does that mean you can just move the limit of R->inf inside the integral?
Under certain conditions, yes you can. Those conditions happen to apply for our integrals given the conditions we stipulated at 5:58. You can look here for more details: math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral
Great Organized Video. Thank you so much
Amazing videos as usual
Thank you! Good to hear your kind feedback!
very helpful, thank you for the great work
10:10 how do you know it has at least one zero above the real axis?
Because the denominator x^2+1 = 0 when x = +/- i. The x = +i zero counts as above the real axis.
The residue theorem resolution in case of this example is basically not needed, because d/dx (arctan x) = 1/(x^2+1) 😉 =>. Integral(-inf,inf) {1/(X^2+1) } = arctan(inf)-arctan(-inf) = pi/2 + pi/2 = pi.
For cos(x)/(x^2+1) the residue theorem does a very good jjob!
Thanks a lot. This really helped me!
last two minutes make me chuckle, all the laborious multi-step work with residual theorem can be replaced with a simple trig integral
Thank you for the great work.
great video, thank you so much
i agree with you
Hello!
Can you recommend any books related to this topic?
Thank You
Thanks for this video!
nice explanation sir
Wohoo, residue theorem!
Very helpful!
Of course. I'd love complicated examples especially the ones which are complicated to do in real numbers. I had a question though:
What if we write 1/(x^2+1) as (1/2i)[1/(x-i) - 1/(x+i)] and then integrate to get (1/2i)log([x-i]/[x+i])
Would that be correct and would I get a physical relation when I integrate with limits like definite integral in real gives area under the curve or something like that?
I'll work on the complicated examples then; thanks for the request! As for your method, I'm not sure it's going to make things simpler if we evaluate the natural log of an imaginary number. That opens up another can of worms.
Faculty of Khan Thanks. Please let me know when you work on that as I have nobody I can ask my doubts too. Please help :)
The integral of 1 / (z^2 + 1) is arctan(z)
Hello! Have been watching your videos in preparation for next year and I absolutely love them! What hardware and software do you use to make the videos? The quality is superb! :)
Thanks you help a lot😭
Do you have some recommendations for references/books regarding complex analysis or mathematical methods in general?
By the way, how to deal when the zeros (in fact the two zeros) are purely real ? Is there a way ?
amazing!
when you say poles above the real axis, does that count for z=0 also?(origin)
No, there is different technique for the case when there are poles on the real axis
Do you have something to do with Khan academy? Or do you just have the same last name?
Wonderful
Do not forget one has the option to slow down (or speed up) the rate of the lecture....I have done this with other YOU TUBE videos
What is your degree in? What software are you using? Mouse or stylus?
Hi Jeff, thanks for the questions! Here are my responses:
- Bachelors in Physiology and Physics + Masters in Chem Eng.
- For recording/editing, I'm using Camtasia.
- I'm using a stylus.
Thank you for your prompt responses!
I read on wikipedia that Residue theorem generalizes Stoke's theorem (the sum of the interior is equal to the line integral of the exterior, particularly, the curl of a vector field is 0 iff it is conservative). However, Stoke's theorem applies to R^n whereas Residue theorem applies to the complex plane. As I understand it, C does not equal R^2 so how do these theorems relate?
No problem!
You're correct that C does not equal R^2, but in some of my videos, I've mentioned how complex numbers are like 2-dimensional versions of real numbers (others have also talked about this analogy, e.g. www.math.brown.edu/~banchoff/Beyond3d/chapter8/section07.html). Since a complex number z can be expressed as x + yi, we can plot z on a complex plane: x units along the Real axis and y units along the Imaginary axis (just like how we can plot an ordered pair of real numbers (x,y) on a Cartesian plane).
Since the complex plane and R^2 are so similar in how they're visualized, it isn't that surprising that Stokes' Theorem, which generally applies to R^2 (or beyond), can be extended to the complex plane in the Residue Theorem, even though the complex plane is usually utilized in a different context. Hope that helps!
What if the zeroes of q(x) are all real?
Exemple: cos(x/a)/(x^2-a^2) , a is real
(integrate from -inf to +inf)
Niko dandam ra babu
Thanks!!
Why degree of q is at least 2 greater than degree of q
loved it
lol @ the joke at 1:10
Sir please take some more example.....
. . . which is why one should get their education in Europe, where it's free - even for immigrants . . .
Not in Germany, definitely not in the UK.
1:10 😂
Videos are awesome.
But plz slow down Ur Speed of explaination!!
Thank you for the feedback!
yeah , it is kind too fast, I have to pause video frequently to figure out what's going on. Great content though
Faculty of Khan sir pls make video on Jordon theorem
Great explanation. But man does ur voice sound monotonous!