Complex analysis is such an amazing thing! I’m watching all your videos to get exposure to the main ideas, so I can link the theory to practice. Thank you for doing this, I’m feeling better about this stuff now, less intimidated by it
Hi, I don't understand why do you replace sin(z) by e^iz in the begining ? Because sin(z)=(e^iz-e^-iz)/2i ??? I know it's correct but I don't understand :)
How do we explain after all why can we guarantee the existance of the integral (and so its equality to Principal Value and limR->oo of the integral in [-R,R])? Usually, we proove that f is bounded by a riemann integrable function (which is more obvious to show) and so we have the existance of f integral ... but I can't seem to find a riemann integrable bound for this one, any ideas????
Hello sir, and thank you for this video. I wanted to try solving this by writing sin(z) as (e^iz - e^-iz)/2i and then calculating the residue of this new f(z) at z=i which turns out to be Res(f(z), z=i) = (i/4)*(e - 1/e). Then I = 2πi*Res(f) = 2πi*(i/4)*(e - 1/e) = -(π/2)*(e - 1/e) which is not the same as your answer! Could you tell me where the mistake is?
Hello, thanks for this video. I just have one question. If we evaluate the limit of an integral, we can only interchange the order of limit and integration if the integrand is a monotonically increasing function and measurable according to monotone convergence theorem, right? The integrand e^-Rsint is measurable but it is a decreasing function as R becomes larger. Do you think you can apply monotone convergence theorem here? Another theorem I think that can be invoked is Lebesgue Dominated Convergence Theorem if MCT won't work.
It's easier when calculating the residues and integral over Gamma as you only need to deal with one exponential, rather than 2 if you use the complex definition of sine.
How do you generally choose your contour in a given integral? If I choose a different contour, will I get another answer? What's the thought process? It's the hardest part
you can choose a different contour, but it may be harder to parameterise (the semi circle is generally the easiest, but you could choose, for example, a box contour). but if your newly chosen contour encloses different residues or more poles of your f(z) you will get a different answer
because the real part of the function f(x) disappears when we evaluate, because it is now an odd integrand (even function/odd function = odd function) we dont have to worry about it
Complex analysis is such an amazing thing! I’m watching all your videos to get exposure to the main ideas, so I can link the theory to practice. Thank you for doing this, I’m feeling better about this stuff now, less intimidated by it
Such a complex thing explained simply. Thanks mate
Thanks a lot dear *QN³*
Now I'm so interesting about the Laplace one.
Great videos. Didn’t you drop the I you brought out front around 10 minutes?
Very good explanation.This helped a lot. Thank you!
Hi, I don't understand why do you replace sin(z) by e^iz in the begining ? Because sin(z)=(e^iz-e^-iz)/2i ??? I know it's correct but I don't understand :)
Ahh okey I just realized that you build a function which is different about the real fonction, but if I use my method why it's not good ?
Sorry for my english, it isn't my native language
Because if I use my method, I will have a expression with sinh(1) and his coefficient because sin(iz)=isinh(z) when I pass the limit by z=i
sin(z) doesn't vanish in the limit in the upper half plane while e^iz does. We want this property so the integral over Gamma goes to 0.
@@qncubed3 okey thanks you :)
How do we explain after all why can we guarantee the existance of the integral (and so its equality to Principal Value and limR->oo of the integral in [-R,R])? Usually, we proove that f is bounded by a riemann integrable function (which is more obvious to show) and so we have the existance of f integral ... but I can't seem to find a riemann integrable bound for this one, any ideas????
Hello sir, and thank you for this video.
I wanted to try solving this by writing sin(z) as (e^iz - e^-iz)/2i and then calculating the residue of this new f(z) at z=i which turns out to be Res(f(z), z=i) = (i/4)*(e - 1/e).
Then I = 2πi*Res(f) = 2πi*(i/4)*(e - 1/e) = -(π/2)*(e - 1/e) which is not the same as your answer!
Could you tell me where the mistake is?
I really have the same curiosity and don't know why D:
That function doesn't decay for large modulus of z, hence the integral over the upper arc will not vanish
@@qncubed3Thank you!
Pourquoi on ne peut pas trouver des vidéos comme ça en Francais 😢
Merci beaucoup tu es incroyable
Hello, thanks for this video. I just have one question. If we evaluate the limit of an integral, we can only interchange the order of limit and integration if the integrand is a monotonically increasing function and measurable according to monotone convergence theorem, right? The integrand e^-Rsint is measurable but it is a decreasing function as R becomes larger. Do you think you can apply monotone convergence theorem here? Another theorem I think that can be invoked is Lebesgue Dominated Convergence Theorem if MCT won't work.
keep up the good work!!!
sir, is this integral twice that of integral done from 0 to infinity?
Yes, the integrand is an even function
That is really helpful!!!!! Thank you!
You're welcome!
why do you need to do e^iz and take imaginary instead of doing sin(z) directly
It's easier when calculating the residues and integral over Gamma as you only need to deal with one exponential, rather than 2 if you use the complex definition of sine.
How do you generally choose your contour in a given integral? If I choose a different contour, will I get another answer? What's the thought process? It's the hardest part
A lot of practice and examples!
@@qncubed3thoughtless answer.
@@maalikserebryakov then give an answer with thoughts dude.
@@maalikserebryakov he aint wrong though
you can choose a different contour, but it may be harder to parameterise (the semi circle is generally the easiest, but you could choose, for example, a box contour). but if your newly chosen contour encloses different residues or more poles of your f(z) you will get a different answer
I like your vedio thank you ..
Explain sin(x)=e^(iz) ?how
They are not equal, however when we construct our function we use e^(iz) as it is easier to deal with.
@@qncubed3 👍👍
Same…i am also confused with it😩
@@kunalburdak8964converting a real integrand into a complex integrand is a complicated on its own
because the real part of the function f(x) disappears when we evaluate, because it is now an odd integrand (even function/odd function = odd function) we dont have to worry about it
so we replaced sin(x) by e^iz. Can I also replace cos(x) by e^iz too ?
i believe so, just make sure you solve for the real part when you evaluate the integral
Sir what is answer if the denominator is(x^2-r^2).
Can you help me sir. If r is real
In general the integral will diverge as you will get isolated singularities at x = +/- r.
Thankyou love you😊🥰
thank you sir
good