again the method I mentioned in the other video is proven using contour integration. (and can be found on the VIIth chapter of Gamelin...) since it vanished here it was: we define I(s) = integral of log(x^2+1)x^s which is M{f(x)}(s+1) but M{f'(x)}(s) =2M{1/x^2+1}(s+1) (here is where contour integration is needed) which is 1/2*pi*csc(1/2pi(s+1)). then M{f(x)}(s) = -(s-1)1/2*pi*csc(1/2pi(s)) making I(s) = -s/2*pi*csc(1/2*pi*(s+1)). deriving and s = -2 should end it
@@maths_505 You might find my video useful. There I explain a method to find out the arg(z-i) and arg(z+i) for paths Psi1 and Psi2 ua-cam.com/video/y1UNUBiGemA/v-deo.html Thanks ;)
Thank you for including a detailed discussion of the branch cuts and the individual arguments of each branch. That stuff always trips me up when you have to deal with more than one cut.
Oh yeah me too....I was stuck on this problem for a few hours. Took a break, played football, assisted 2 goals in an hour's game and felt amazing. After a shower I came back to the problem and just thought "wait a minute.....let's break the contour down and rotate it here and there"....and it solved the integral quite nicely....
Apply Partial fraction become 1/4pi×(I1+I2) Where I1=Int(cosx)/(pi/2-x) at [0,inf) I2=lnt(cosx)/(pi/2+x) at[0,inf) For I1 Let y=pi/2-x I1=Int(siny/y) from(-inf,pi/2] For I2=Let y=pi/2+x I2=Int(siny/y) from[pi/2,inf) I1+I2=Int(siny/y) from(-inf,inf)=pi I=1/4pi×(I1+I2)=1/4pi×pi=1/4
I def missed out by doing engineering instead of math. My grad degree was almost exclusively calculus, but everything was in physical dimensions and I just used numerical techniques. Complex analysis blows my mind.
Correct me if I'm wrong, but wouldn't the angle phi range from -3*pi/2 - arcsin(delta/epsilon) to pi/2 - arcsin(delta/epsilon) regarding gamma_2 ? For the whole circle to be parametrized in the limit, one would then require lim delta/epsilon = 0
Thanks bud. That was a good piece. I put a link to a video I did for evaluating the arguments of z+/-i in the reply to one of your comments. Let me know how it's that when you get a chance.
Do you have a reference on how to measure angles in the complex plane? Or make a video with a thorough ecxplanation on the subject? Because, in this case, I could not understand your numbers. Regards.
Before watching the video, I used Feynman's technique to differentiate log(a*x^2+1) under the integral sign to simplify the branch cuts. But other than that I used the same method.
If you want very tough integrals just ask chat gpt 4 for tough integrals it gave me this (integral e^z×Ai(z)/z dz) Where Ai(z) is airy function ∫(ln(x)⋅Γ(x)⋅Ei(x)/x)dx When you ask it for more difficult it will give even more difficult questions on integration
Could someone please help me out? I tried to show that the integral over gamma2 vanishes for epsilon going to zero, but didn't succeed. I used the parametrization shown at 7:55. But then for epsilon --> 0, we have that log(z) goes to log(i), log(1+z²) diverges and z² goes to -1, so the whole fraction f(z) does _not_ vanish at all. What am I doing wrong?
You also have an epsilon in the numerator (because of the differential element) being multiplied by the log term. In the limit as epsilon approaches zero, epsilon times log of the function of epsilon goes to zero
Need??? Bro you think I make these videos to show techniques "needed" to solve particular integrals😂 Nah man I just solve em for fun....I found the application of contour integration to these integrals quite beautiful especially because both of them could be solved in one go ......the contour was pretty exotic too and lots of people learning contour integration struggle with branch cuts along the vertical axis so this was a nice example for explanation purposes.
again the method I mentioned in the other video is proven using contour integration. (and can be found on the VIIth chapter of Gamelin...)
since it vanished here it was: we define I(s) = integral of log(x^2+1)x^s which is M{f(x)}(s+1) but M{f'(x)}(s) =2M{1/x^2+1}(s+1) (here is where contour integration is needed) which is 1/2*pi*csc(1/2pi(s+1)). then M{f(x)}(s) = -(s-1)1/2*pi*csc(1/2pi(s)) making I(s) = -s/2*pi*csc(1/2*pi*(s+1)). deriving and s = -2 should end it
🔥🔥🔥
It didn't vanish
I pinned it bro😂
@@maths_505 oh LOL could'nt find it in the previous video, so I thought I must have forgotten to press enter or somethin, thanks for the pin my G.
But this integral can be calculated easier and faster using integration by parts
@@maths_505 You might find my video useful. There I explain a method to find out the arg(z-i) and arg(z+i) for paths Psi1 and Psi2
ua-cam.com/video/y1UNUBiGemA/v-deo.html
Thanks ;)
Thank you for including a detailed discussion of the branch cuts and the individual arguments of each branch. That stuff always trips me up when you have to deal with more than one cut.
Oh yeah me too....I was stuck on this problem for a few hours. Took a break, played football, assisted 2 goals in an hour's game and felt amazing. After a shower I came back to the problem and just thought "wait a minute.....let's break the contour down and rotate it here and there"....and it solved the integral quite nicely....
Pure gold. Didn't know you were that complex, man!
Feynman trick be looking fine to try on this integral🔥 but the contour integration is very satifsying
Yes indeed
A subscriber pointed that out on the last video. But I wanted to evaluate both of em so I decided on contour integration.
Apply Partial fraction become
1/4pi×(I1+I2)
Where
I1=Int(cosx)/(pi/2-x) at [0,inf)
I2=lnt(cosx)/(pi/2+x) at[0,inf)
For I1 Let y=pi/2-x
I1=Int(siny/y) from(-inf,pi/2]
For I2=Let y=pi/2+x
I2=Int(siny/y) from[pi/2,inf)
I1+I2=Int(siny/y) from(-inf,inf)=pi
I=1/4pi×(I1+I2)=1/4pi×pi=1/4
I def missed out by doing engineering instead of math. My grad degree was almost exclusively calculus, but everything was in physical dimensions and I just used numerical techniques. Complex analysis blows my mind.
You have proven that you can be an illustrator for Playboy besides being a mathematician.
😂😂😂
@@maths_505 I would not have noticed it unless you mentioned it though. I am definitely getting too old if I missed something like that!
Correct me if I'm wrong, but wouldn't the angle phi range from -3*pi/2 - arcsin(delta/epsilon) to pi/2 - arcsin(delta/epsilon) regarding gamma_2 ? For the whole circle to be parametrized in the limit, one would then require lim delta/epsilon = 0
Continuity allows us to apply the limits and simplify the writing
I did it but by using Feynman technique. It's cool
Thanks bud. That was a good piece. I put a link to a video I did for evaluating the arguments of z+/-i in the reply to one of your comments. Let me know how it's that when you get a chance.
Do you have a reference on how to measure angles in the complex plane? Or make a video with a thorough ecxplanation on the subject? Because, in this case, I could not understand your numbers.
Regards.
Before watching the video, I used Feynman's technique to differentiate log(a*x^2+1) under the integral sign to simplify the branch cuts. But other than that I used the same method.
If you want very tough integrals just ask chat gpt 4 for tough integrals
it gave me this (integral e^z×Ai(z)/z dz)
Where Ai(z) is airy function
∫(ln(x)⋅Γ(x)⋅Ei(x)/x)dx
When you ask it for more difficult it will give even more difficult questions on integration
5:40 All the poles are outside tge contour so there is no reisdue
Could someone please help me out? I tried to show that the integral over gamma2 vanishes for epsilon going to zero, but didn't succeed. I used the parametrization shown at 7:55. But then for epsilon --> 0, we have that log(z) goes to log(i), log(1+z²) diverges and z² goes to -1, so the whole fraction f(z) does _not_ vanish at all. What am I doing wrong?
You also have an epsilon in the numerator (because of the differential element) being multiplied by the log term. In the limit as epsilon approaches zero, epsilon times log of the function of epsilon goes to zero
@@maths_505 Thanks! I really totally forgot about the differential... :/
No problem. I'm just happy you asked an actual question instead of coming up with some weird math theory 😂😂😂
You lost me after the calling the integral I part. There is much homework generated by the branch cuts stuff.
5:10 thanks for the heads-up. My parents walked in on me while I was watching this and I switched to porn as that was easier to explain
No need for contour integration for either of them
Need???
Bro you think I make these videos to show techniques "needed" to solve particular integrals😂
Nah man I just solve em for fun....I found the application of contour integration to these integrals quite beautiful especially because both of them could be solved in one go ......the contour was pretty exotic too and lots of people learning contour integration struggle with branch cuts along the vertical axis so this was a nice example for explanation purposes.
DIOS LOS MALDIGA
echese a dormir
@@yonalejandrogutierrezpaez6159 yo vere
@@PINILLAALARCONWILLIAMFELIPE que atrevido