It feels pretty amazing to finally see someone actually _calculate_ these things. This is something that I normally never see - I always see math teachers bring up transform tables. I guess this is because the actual calculations take a very long time, but it is still really cool to see what is going on "behind the scenes" there.
"Applied Complex Variables" by John W. Dettman is a great read: the first part covers the geometry/topology of the complex space from a Mathematicans' perspective, and the second part covers application of complex analysis to differential equations and integral transformations, etc. from a Physicists' perspective. It is a detailed compliment to QN^3 videos.
Maybe I didn't understand it very well, why the Psi 1 and Psi 2 don't they cancel here? In the residue theorem proof something like that used to be cancelled
i dont get why does the integral over omega vanish. Just because it encloses a branch point? I'm rusty in this. But I feel like correct argument should appreciate that the integrand is unbounded, but only because of a factor |ε|^(-1/2) and that is taken care off by multiplying by ε (radius of omega). Is it always guaranteed to work out like that if we have a branch point?
In a video about sinx/x integral, the small omega part maybe horizontal he computed was -πi or something but here it vanishes, I'm totally confused and also bad at it I can't make sure where it vanishes or not
Wouldn't it be easier to complete the contour as a semicircle-like curve to the right, a kind of D shape, to avoid so many integrals? Thanks for the video!
In a curve to the right, we have e^st (t>0 as per the domain of functions which we take the laplace transform of) going to infinity exponentially as s is in the right half plane and Re(s)>>0 on the semicircle. This means the semicircle integral doesn’t go to zero, and is very difficult to calculate.
In Bromwich Contour lemma it is required that the degree of numerator exceeds that from the denominator by at least one. Such doesn't happen for the inverse square root. Am I missing any other lemma?
By analysing your video about the Bromwich contour lemma I came to the conclusion that in fact it is just needed that the power of the denominator be greater than that of the denominator, which is verified for the inverse square root!
After the substitution t = (1-x)/(1+x), that integral changes into a form which has already been evaluated on this channel. Namely, the integral from a to b of (x-a)^p*(b-x)^(1-p)*dx. Well, actually, only if you change (x-1)^3 to (1-x)^3 in your integrand, which is not real-valued over the domain of integration.
It feels pretty amazing to finally see someone actually _calculate_ these things.
This is something that I normally never see - I always see math teachers bring up transform tables. I guess this is because the actual calculations take a very long time, but it is still really cool to see what is going on "behind the scenes" there.
"Applied Complex Variables" by John W. Dettman is a great read: the first part covers the geometry/topology of the complex space from a Mathematicans' perspective, and the second part covers application of complex analysis to differential equations and integral transformations, etc. from a Physicists' perspective. It is a detailed compliment to QN^3 videos.
Does this make 1/sqrt(t) an eigenfunction of the Laplace transform, or does it not count because the input space is different?
When they forgot to give you a laplace table on the exam
every video you seem to drag the way to say greetings.
grrrrrrrrreeeeeeeeeeeeetttings
Yeah! Love your lectures. Can you make lectures on the method of steepest descent and also contour integral for bessel or hankel function
Maybe I didn't understand it very well, why the Psi 1 and Psi 2 don't they cancel here? In the residue theorem proof something like that used to be cancelled
i dont get why does the integral over omega vanish. Just because it encloses a branch point? I'm rusty in this. But I feel like correct argument should appreciate that the integrand is unbounded, but only because of a factor |ε|^(-1/2) and that is taken care off by multiplying by ε (radius of omega). Is it always guaranteed to work out like that if we have a branch point?
In a video about sinx/x integral, the small omega part maybe horizontal he computed was -πi or something but here it vanishes, I'm totally confused and also bad at it
I can't make sure where it vanishes or not
Wouldn't it be easier to complete the contour as a semicircle-like curve to the right, a kind of D shape, to avoid so many integrals? Thanks for the video!
In a curve to the right, we have e^st (t>0 as per the domain of functions which we take the laplace transform of) going to infinity exponentially as s is in the right half plane and Re(s)>>0 on the semicircle. This means the semicircle integral doesn’t go to zero, and is very difficult to calculate.
In which book can I find the theory and approach used in the video?
You can find it in most complex analysis textbooks
Yeaaa! Finally!
In Bromwich Contour lemma it is required that the degree of numerator exceeds that from the denominator by at least one. Such doesn't happen for the inverse square root. Am I missing any other lemma?
By analysing your video about the Bromwich contour lemma I came to the conclusion that in fact it is just needed that the power of the denominator be greater than that of the denominator, which is verified for the inverse square root!
Could you please solve Integral
x^m/(a^n+x^n), from 0 to infinity
Solution will be like 2pi/[na^(n-1)]* sin[(m+1)pi/n]
I've already done something similar on this channel
Thanks man I checked them those were very enjoyful.
May you resolve this integral, Integral of ((x-1)³x)^(1/4) / (x+1)³ from 0 to 1 ? this integral is equal to 3(2⁴)PI/64
A similar integral has been done on my channel - the dogbone contour examples.
After the substitution t = (1-x)/(1+x), that integral changes into a form which has already been evaluated on this channel. Namely, the integral from a to b of (x-a)^p*(b-x)^(1-p)*dx. Well, actually, only if you change (x-1)^3 to (1-x)^3 in your integrand, which is not real-valued over the domain of integration.
Fantástico
Perfect🥰🥰🥰