There is a very short and elementary way to calculate this integral I. First write sin[x] as 2*sin[x / 2]*sin[x / 2+pi/2] Then log[sin[x]= log[2]+log[sin[x] /2] +log[sin[x/2+pi/2] . The substitutions t= x/2 and t=x/2+pi/2] immediately lead to the equation I+ log[2]*pi = 2*I ,i.e. I = - pi*log[2] .
6:10 it depends on the branch cut of the logarithm, I'll pick it to be the negative real axis, then sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(x), the imaginary part has to be equal to 0, and the real part has to be less than 0, so sin(x)
30:24 use the taylor expansion of e^x, then you have x ln( x(2ie^itheta+(2ie^itheta)^2 x/2! +...) = xlnx+xln(2ie^itheta+....), the limit as x goes to 0 is then 0.
If you decompose log(sin(x))=iz-log(2)-i*Pi/2+log(1-exp(-2i*x)) and integrate from 0 to Pi, then Int_0^Pi log(sin(x)) dx = -Pi*log(2) + Int_0^Pi log(1-exp(-2i*x)) dx. You can then use complex analysis to show that the last integral is 0. Substitute z=exp(-2i*x), then this integral becomes Int_{|z|=1} log(1-z) dz/(-2i*z) where the contour is going clockwise. log(1-z)/z has a branch-cut on (1,infinity) and is holomorphic inside the disk of radius 1 about z=0. Hence this integral is zero by Cauchy. You may want to check the integral over the semi-circle of radius epsilon about z=1, but it is not difficult to see that it is zero as epsilon -> 0. Btw: What do you mean by standard solution? I once saw a solution using integration by parts twice (using the symmetry about Pi/2) and you end up with Int_0^{Pi/2} (x/sin(x))^2 dx and then there was a trick, but I don't remember.
I know the solution you're thinking of. It uses integration by parts once and then Feynman's trick after substituting y = tan(x). So you were right about there being a trick! It's a nice alternative to the normal solution that exploits symmetry within the integrand.
This is amazing! Made me sub as primitive functions of logarithms of trigonometric functions are currently in my scope. As much as I enjoy a purely real proof... :)
The Chaotic Good to flammy's Chaotic Evil
¡Gracias!
There is a very short and elementary way to calculate this integral I. First write sin[x] as 2*sin[x / 2]*sin[x / 2+pi/2] Then log[sin[x]= log[2]+log[sin[x] /2]
+log[sin[x/2+pi/2] . The substitutions t= x/2 and t=x/2+pi/2] immediately lead to the equation I+ log[2]*pi = 2*I ,i.e. I = - pi*log[2] .
What a great solution
Holy smokes. That is brilliant. I just worked through it. Absolutely brilliant. Thank you.
discovered your channel today and I got to say that I've fallen in love with the complex analysis series, keep it up!
6:10 it depends on the branch cut of the logarithm, I'll pick it to be the negative real axis, then sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(x), the imaginary part has to be equal to 0, and the real part has to be less than 0, so sin(x)
30:24 use the taylor expansion of e^x, then you have x ln( x(2ie^itheta+(2ie^itheta)^2 x/2! +...) = xlnx+xln(2ie^itheta+....), the limit as x goes to 0 is then 0.
If you decompose log(sin(x))=iz-log(2)-i*Pi/2+log(1-exp(-2i*x)) and integrate from 0 to Pi, then Int_0^Pi log(sin(x)) dx = -Pi*log(2) + Int_0^Pi log(1-exp(-2i*x)) dx. You can then use complex analysis to show that the last integral is 0. Substitute z=exp(-2i*x), then this integral becomes Int_{|z|=1} log(1-z) dz/(-2i*z) where the contour is going clockwise. log(1-z)/z has a branch-cut on (1,infinity) and is holomorphic inside the disk of radius 1 about z=0. Hence this integral is zero by Cauchy. You may want to check the integral over the semi-circle of radius epsilon about z=1, but it is not difficult to see that it is zero as epsilon -> 0.
Btw: What do you mean by standard solution? I once saw a solution using integration by parts twice (using the symmetry about Pi/2) and you end up with Int_0^{Pi/2} (x/sin(x))^2 dx and then there was a trick, but I don't remember.
Not exactly the same integral, but maybe related to this? ua-cam.com/video/7wiybMkEfbc/v-deo.html
@@Alex_Deam That's a neat method too, thanks.
I know the solution you're thinking of. It uses integration by parts once and then Feynman's trick after substituting y = tan(x). So you were right about there being a trick! It's a nice alternative to the normal solution that exploits symmetry within the integrand.
This is amazing! Made me sub as primitive functions of logarithms of trigonometric functions are currently in my scope.
As much as I enjoy a purely real proof... :)
Can you make videos on some of jee advanced tough complex number questions?
i dont eevn know complex analysis. I just love your handwriting
That is very kind of you
Thank you for the integral and your way to explain it
I need a help if you can sur
I want integrat
[0,inf] x^2*exp(-x)/(x^2+pi^2)
solved this using derivative of the beta function.
what an elegant problem!
Can you suggest book for Complex analysis? Thanks
rly nice explanation bruh subbed
Do we really need to let n-> infinite? As long as n> ε, keep n as a constant should work well.
The e^(-2n) in the logarithm won't vanish if n is finite
@@qncubed3 Thank you!
the opening with the chair is interesting though
Thankyou
12:21 🚁👍
Thanks professor
I am merely a baby undergrad
@@qncubed3 I dream of being like you someday
So change the x to z. Then just show everyone that you know how to do it.
Proof: Let x = z. This reduces to a trivial problem.
Grade 12
NCERT Problem :)
Example 36
very fav question of cbse
he is really cute.
That proposition is false
大学题