MM87: contour integrals-branch points and branch cuts

[@prabalbaishya6179]

I started out learning complex analysis and your videos are really helpful. Also for the evaluation of the integral without complex analysis, I think there is way more less calculative procedure. But it requires the knowledge of the Beta function {β(m,n)} and Euler's mirror formula{ Γ(n)*Γ(1-n)= π/sin(nπ)for n between 0 and 1}.On substituting x= tanθ. We get the integral from 0 to π/2 √(tanθ)dθ. Which can be rewritten as integral from 0 to π/2 (sinθ)^[2*(3/4)-1](cosθ)^[2*(1/4)-1]dθ. Which is exactly the value of 1/2*β(3/4, 1/4). Which equals to 1/2*Γ(1/4)*Γ(3/4). And now using Euler's mirror equation we get this to be 1/2*π/sin(π/4). Which is indeed π/√2.

[@DR_VIV]

Very ingenious! Thanks for pointing this method out. I do have videos on beta and gamma functions, so this method is well within the scope of these lectures.

[@simrannahar8262]

this was an AMAZING video, so well explained, really helped me out, thank you so much!

[@shuewingtam6210]

On video 6:12 -i-i in denumerator of second fraction should be -2i .so you should change addition sign between two fraction into minus with the mentioned -2i switched to +2i. So total residue shoud be sqrt(2)*pi*i

[@DR_VIV]

You are correct... I made two small errors and they cancelled themselves out to give the right final answer. I will post a small correction video one of these days....

[@shuewingtam6210]

@@DR_VIV pls give me the link when it is ready.

[@DR_VIV]

@@shuewingtam6210 it is video 87(bis) posted about an hour back....

[@DR_VIV]

kzbin.info/www/bejne/hWbQhJupa5aHbcU

[@rohkofantti8673]

What fountain pens do you use?

[@DR_VIV]

I use many fountain pens…. I own hundreds, keep changing them according to my needs, moods, ink color…

[@Rondineli-dj4rg]

Please, would you solve this integral ( x^(1/3) )/( x² + 1 ) dx in interval [ 0 , ∞ ], I've tried both ways as presented by you in the video, but unfortunately I wasn't successful. Thank you 👍

[@DR_VIV]

kzbin.infoJqLlGL1Wnog?feature=share

[@DR_VIV]

See my short video showing a still photo of the work for this integral!

[@mnk4214]

Please, would you solve this integral ( x^(1/3) )/( x^4 + 1 )^2 dx in interval [ 0 , ∞ ], I've tried, but unfortunately I wasn't successful. Thank you 👍

[@DR_VIV]

Pi/(3 sqrt(3))… it is easier to do this with a simple u substitution and then partial fractions because the complex analysis will take too long. You have 4 poles of order 2 each and a branch cut at origin too!

[@mnk4214]

Thanks for the video, I figured out my mistake. I counted the residues not on [0 2pi], but on [-pi pi]

[@djgoswami360]

5:00 Why not √i = exp(i5π/5)? Please reply

[@DR_VIV]

Hello, the right hand side (exp i 5 pi/5) is equal to -1. So it cannot be equal to the left hand side.

[@djgoswami360]

@@DR_VIV i extremely sorry it was by mistake. I meant exp(i5π/4).

[@DR_VIV]

@@djgoswami360 sure you can also use that, it will merely pick up the root at the lower half plane. I stuck to the upper half plane for convenience.

[@djgoswami360]

@@DR_VIV but the thing is that The cetral purpose of contour integral with brach cut is to make the given multivalued function as singlevalued. And here the contour you choosen correspondents 0 to 2pi for argument. This thing is troubling me.

[@channelphys6887]

OK

[@hajsaifi3842]

Fonction bêta mieux peut être