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* ❤️
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.
I will be posting a solution addressing exactly this concern. Getting the right contour should be a straightforward mechanical process, not an art form.
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
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* ❤️
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
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?
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 RUclips would delete my comment otherwise. Sorry about that)
Double the contour integrals, double the amazing content!
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* ❤️
Very nice, this is the exact method I used when he posted that video.
Hello dear *QN³* ...
One integral with two methods?!
Complex one with You ( dear *QN³* ) and the other with Penn?!
*Super Duper Cool*
Yo wtf this is sick
Keep pumping it out maths god
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.
I will be posting a solution addressing exactly this concern. Getting the right contour should be a straightforward mechanical process, not an art form.
video posted on my channel.
31:56 Real analysis could never
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
The cursed bicycle.
Great content as always :P I hope you two could make a collab video sometime, I'm sure it would be awesome
what are branch points and branch cuts?
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* ❤️
All this mess just to make two pies square shaped, damn.
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
Four integrals for the price of one.
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?
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.
The Analytical Answer needs to use the incomplete gamma function.
@@vascomanteigas9433 what is the final answer?
How about solving
∞
∫ [ exp(-x) * ln (x) * ln (x) ] dx
0
Mirip sudirman wkwk
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 RUclips would delete my comment otherwise. Sorry about that)