MM87: contour integrals—branch points and branch cuts

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MM87: contour integrals—branch 0:21:39

MM87: contour integrals—branch points and branch cuts

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

simrannahar
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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 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.

prabalbaishya
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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

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

djgoswami
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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 👍

mnk
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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 👍

Rondineli-djrg