The reverse triangular inequality is actually two inequalities bundled in one since |x| is always greater than both x and -x, so you can pick any one of |a-b| >= |a| - |b| or |a-b| >= |b| - |a|
@ Yeah, i don’t know if it’s somehow linked to the king’s property though. I haven’t had the time to play around with it. Thank you for your videos. Complex analysis is my favorite way of solving these types of integrals
im starting calc 2 next semester and i can’t wait. im self studying before the semester starts and ive gotten thru all the calc 2 integration techniques (parts, trig integrals/sub, partial fractions). i love it so much and now i want to make math content as well. gonna try and learn manim. keep up the great work bro
@@OscgrMaths in ration functions, sometimes there are two vertical asymptotes, but in between the outputs of the functions can cross the horizontal asymptote. can you expand on that by an chance?
@@leonardobarrera2816 Thats the same age I got interested in contour integration! There's lots of good resources online - qncubed3 is a great channel for helpful videos and the book 'Introduction to Complex Analysis' by Priestley has some good chapters on it too. Feel free to ask any questions about it, it's generally a second year topic in a pure maths degree.
It's because at the end of the problem we take the imaginary part of the answer the integral of (xe^ix)/(1+x^2) = the integral of (x(cosx+isinx))/(1+x^2) = the integral of (xcosx)/(1+x^2) + i the integral of (xsinx)/(1+x^2). This means if we take the Imaginary part of the integral of e^ix, we just get the integral of sinx. Hope this makes sense!
x is odd sin(x) is odd odd•odd=even 1+x^2 is even even/even=even I=2•int[0,♾️](xsin(x)/(1+x^2))dx sin(x)=sum[k=0,♾️]((-1)^k•x^(2k+1)/(2k+1)!) I=2•sum[k=0,♾️]((-1)^k/(2k+1)!•int[0,♾️](x^(2k+2)/(1+x^2))dx ß(u,v)=2•int[0,♾️](x^(2v-1)/(1+x^2)^(u+v))dt I=sum[k=0,♾️]((-1)^k•ß(k+3/2,1-(k+3/2))/(2k+1)!) ß(x,1-x)=pi/sin(pi•x) sin(x+3/2)=-cos(x) cos(kpi)=(-1)^k, k€Z I=-pi•sum[k=0,♾️](1/(2k+1)!) sinh(x)=sum[k=0,♾️](x^(2k+1)/(2k+1)!) I=-pi•sinh(1) I=pi/2e-epi/2 One of us is wrong, and for the life of me i cannot find a flaw.
Sorry I didn't see your whole video because something struck me immediately when I saw the integrand. It's an odd function. Shouldn't it's integral from some -a to +a better 0?
Hey! The function is actually even. If you substitute -x into the function you get (-x)sin(-x)/(1+(-x)^2) = xsinx/(1+x^2). Were the trig function cosine, it would be but since both x and sinx are odd, the resultant function is even. Hope this helps!
Yoo, the transition in 12:39 is flawless. Nice video!
Thanks so much! Really glad you enjoyed it.
happy new year bro love ur videos
happy new year to you too! thanks so much for the comment it means a lot. theres lots more coming in the next few days!
The reverse triangular inequality is actually two inequalities bundled in one since |x| is always greater than both x and -x, so you can pick any one of |a-b| >= |a| - |b| or |a-b| >= |b| - |a|
im absolutely loving this integral series you’ve been doing!! keep up the good work!!
Thanks I'm so glad!
Phenomenal teaching style!! Happy new year.
Thanks so much! Glad you enjoyed.
@@OscgrMaths All thanks to you bro, this video might have just somewhat taught me Contour integration, and I found this fun as well as helpful!
Interestingly enough, if you replace x*sin(x) by cos(x), you obtain the same answer
Wow that's very interesting thanks for sharing!
@ Yeah, i don’t know if it’s somehow linked to the king’s property though. I haven’t had the time to play around with it. Thank you for your videos. Complex analysis is my favorite way of solving these types of integrals
im starting calc 2 next semester and i can’t wait. im self studying before the semester starts and ive gotten thru all the calc 2 integration techniques (parts, trig integrals/sub, partial fractions). i love it so much and now i want to make math content as well. gonna try and learn manim. keep up the great work bro
Thanks so much! Glad you enjoyed it.
@@OscgrMaths did you find real analysis difficult? im not a math major but i want to take it for fun when i get to uni
been here for a bit. youve grown a lot, happy new year dawg, im still learning ration functions but this is cool lmao! congrats on 5k!
its great to have you and i'm really glad you found it cool! let me know if you come across any other ideas you'd want me to cover at some point.
@@OscgrMaths in ration functions, sometimes there are two vertical asymptotes, but in between the outputs of the functions can cross the horizontal asymptote. can you expand on that by an chance?
Proof by common sense of the mod integral theorem🙏 Good stuff again keep it up for this coming year!
Love the complex analysis videos. Congrats on 5k!
So glad you enjoyed! Thanks so much for the congratulations, it means a lot.
Happy new year Dr Oscar
Thank you! Happy new year to you too.
Nice explain
Thank you!
Happy new year mate and a very well made and informative video
Thanks so much! Happy new year to you too.
Happy new year😊
You too ☺️!
@@OscgrMathsthank you
And, at what year does it normal people learn this
(I am learning from the 16 years unluckly
@@leonardobarrera2816 Thats the same age I got interested in contour integration! There's lots of good resources online - qncubed3 is a great channel for helpful videos and the book 'Introduction to Complex Analysis' by Priestley has some good chapters on it too. Feel free to ask any questions about it, it's generally a second year topic in a pure maths degree.
2:13 you've written sin(x) as e^iz.
But shouldn't it be (e^(iz) - e^(-iz))/(2i)
It's because at the end of the problem we take the imaginary part of the answer
the integral of (xe^ix)/(1+x^2) = the integral of (x(cosx+isinx))/(1+x^2) = the integral of (xcosx)/(1+x^2) + i the integral of (xsinx)/(1+x^2).
This means if we take the Imaginary part of the integral of e^ix, we just get the integral of sinx. Hope this makes sense!
Thanks for clarifying
Happy new yar and congrats on 5k!! :)
Thanks for being here :)! Happy new year.
Nice result at the end there, happy new year 🥳
you too 🥳🥳
Happy new year! Neat video, and congrats on 5k 🎉
Thanks so much! Happy new year to you too.
Happy new year bruh✨ your videos are just amazing keep it up man💗
happy new year to you too and thanks for making it such a great one with all the support!
Happy New Year, and massive congratulations on the 5,000.
Thanks so much for being a part of it! Happy new year to you too.
mr Oscar have you ran out of video ideas?
Of course not! There's another video coming along later today and three more to follow. Did you have any ideas you wanted me to cover?
interestingly not Lebesgue integrable
x is odd
sin(x) is odd
odd•odd=even
1+x^2 is even
even/even=even
I=2•int[0,♾️](xsin(x)/(1+x^2))dx
sin(x)=sum[k=0,♾️]((-1)^k•x^(2k+1)/(2k+1)!)
I=2•sum[k=0,♾️]((-1)^k/(2k+1)!•int[0,♾️](x^(2k+2)/(1+x^2))dx
ß(u,v)=2•int[0,♾️](x^(2v-1)/(1+x^2)^(u+v))dt
I=sum[k=0,♾️]((-1)^k•ß(k+3/2,1-(k+3/2))/(2k+1)!)
ß(x,1-x)=pi/sin(pi•x)
sin(x+3/2)=-cos(x)
cos(kpi)=(-1)^k, k€Z
I=-pi•sum[k=0,♾️](1/(2k+1)!)
sinh(x)=sum[k=0,♾️](x^(2k+1)/(2k+1)!)
I=-pi•sinh(1)
I=pi/2e-epi/2
One of us is wrong, and for the life of me i cannot find a flaw.
hmm...nice. txs
Sorry I didn't see your whole video because something struck me immediately when I saw the integrand.
It's an odd function. Shouldn't it's integral from some -a to +a better 0?
Hey! The function is actually even. If you substitute -x into the function you get (-x)sin(-x)/(1+(-x)^2)
= xsinx/(1+x^2).
Were the trig function cosine, it would be but since both x and sinx are odd, the resultant function is even. Hope this helps!
@@OscgrMaths Am sorry. My bad. Blush.
@@bobbybannerjee5156 No problem!