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Complex Analysis: An Integral from  

qncubed3
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Today, we evaluate the integral from 0 to infinity of log^2(x)/(sqrt(x)(1-x)^2).
Michael Penn's video: • I really like this int...

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25 июл 2022

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Комментарии : 25   
@Decrupt
@Decrupt 2 года назад
Double the contour integrals, double the amazing content!
@user-wu8yq1rb9t
@user-wu8yq1rb9t 2 года назад
Min: 22:42 I think you need *Hagoromo Chalk* , because the *Hagoromo Chalk* are antigravity!! Min 25:00 .... 27:00 thanks for details. Just Great (one of the best of yours) 🤓 Thank you so much dear *QN³* ❤️
@The1RandomFool
@The1RandomFool 2 года назад
Very nice, this is the exact method I used when he posted that video.
@Alex_Deam
@Alex_Deam 2 года назад
31:56 Real analysis could never
@user-wu8yq1rb9t
@user-wu8yq1rb9t 2 года назад
Hello dear *QN³* ... One integral with two methods?! Complex one with You ( dear *QN³* ) and the other with Penn?! *Super Duper Cool*
@jkid1134
@jkid1134 Год назад
Yo wtf this is sick
@rahulnadig9873
@rahulnadig9873 2 года назад
Keep pumping it out maths god
@kummer45
@kummer45 2 года назад
Contour integral is the real deal. I always hate when people on some books come up with the paths without explaining how they where constructed. The contour path should be selected localizing the poles in the situation. This should be explained further. The rest is easy to follow and the classical exercises of the theorem. The deal here is how the practitioner comes up with the adequate contour and the identification of the relevant poles.
@djangogeek
@djangogeek Год назад
I will be posting a solution addressing exactly this concern. Getting the right contour should be a straightforward mechanical process, not an art form.
@djangogeek
@djangogeek Год назад
video posted on my channel.
@ameliaC-xg8rg
@ameliaC-xg8rg 2 года назад
Great content as always :P I hope you two could make a collab video sometime, I'm sure it would be awesome
@benjaminjohnson-martin1248
@benjaminjohnson-martin1248 2 года назад
The cursed bicycle.
@holyshit922
@holyshit922 Год назад
Firsty I substituted y^2=x to get rid of square root then I split interval into [0,1] and [1,infinity] Int(8ln^2(y)/(1-y^2)^2,u=0..1)+Int(8ln^2(y)/(1-y^2)^2,u=1..infinity) then substituted y = 1/u in the integral Int(8ln^2(y)/(1-y^2)^2,u=1..infinity) to get 8Int((1+u^2)ln^2(u)/(1-u^2)^2,u=0..1) Then I integrated indefinite integral by parts 8Int((1+u^2)ln^2(u)/(1-u^2)^2,u) (differentiating log) and then calculate the limits Finally I got -16Int(ln(u)/(1-u^2),u=0..1) From here I expanded 1/(1-u^2) as geometric series and calculated integral Int(u^(2n)ln(u),u=0..1) by parts and got 16sum(1/(2n+1)^2,n=0..infinity) 16(sum(1/n^2,n=1..infinity) - sum(1/(2n)^2,n=1..infinity)) 16(sum(1/n^2,n=1..infinity) - 1/4sum(1/n^2,n=1..infinity)) 16*3/4sum(1/n^2,n=1..infinity) 12sum(1/n^2,n=1..infinity) 2π^2
@rishetmehra4364
@rishetmehra4364 3 месяца назад
what are branch points and branch cuts?
@user-wu8yq1rb9t
@user-wu8yq1rb9t 2 года назад
Hello dear *QN³* Probably you watched the last video of Dr. Penn (looks crazy) already .... You think its answer is right?! Honestly, I have some problem with it. I'm not sure, but I think we have some problem on it (at least in details)! I hope you'll have a plan for it. Dear *QN³* ❤️
@soup1649
@soup1649 Год назад
hello can you do a video on the integral from 0 to infinity of cosx/(x^3+x^2+x+1) dx please! i've tried to evaluate it by contour integration but i'm stuck
@stephyjose9533
@stephyjose9533 Год назад
Can you help compute the integral $\int_ {- \infty} ^ {\infty} dk\frac {e^ {-a\sqrt {s+\iota k}} e^ {-a\sqrt {s-\iota k}}} {(s+\iota k) (s-\iota k)} $ using complex analysis?
@Leomenxd
@Leomenxd 2 года назад
All this mess just to make two pies square shaped, damn.
@burpleson
@burpleson 2 года назад
Four integrals for the price of one.
@GreenMeansGOF
@GreenMeansGOF Год назад
Is it possible do solve integral of sin^2(x)/(1+x^2) from 1 to infinity? Wolframalpha only gives a decimal answer so I assume the answer is no.
@vascomanteigas9433
@vascomanteigas9433 Год назад
The Analytical Answer needs to use the incomplete gamma function.
@GreenMeansGOF
@GreenMeansGOF Год назад
@@vascomanteigas9433 what is the final answer?
@nasim09021975
@nasim09021975 Год назад
How about solving ∞ ∫ [ exp(-x) * ln (x) * ln (x) ] dx 0
@nurhaudi112
@nurhaudi112 4 месяца назад
Mirip sudirman wkwk
@KStarGamer_
@KStarGamer_ 2 года назад
There are some alternative methods on MSE for anyone interested: questions/4427604/evaluate-int-0-infty-frac-lnx2-sqrtx1-x2dx (I had to split up the URL because RU-vid would delete my comment otherwise. Sorry about that)
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