If you like the videos and would like to support the channel: www.patreon.com/Maths505 You can follow me on Instagram for write ups that come in handy for my videos and DM me in case you need math help: instagram.com/maths.505?igshid=MzRlODBiNWFlZA== My LinkedIn: www.linkedin.com/in/kamaal-mirza-86b380252
Jokes aside, RUclips has no information about gender. The gender breakdown of viewership is based on a made-up gender it assigned based on stereotypes of what kind of videos folks watch by gender. Classic GIGO.
sorry mate a lil busy last two to three days so i missed out some videos i will quickly watch those tomorrow when u tried not to let intrinsive tought by not using gamma function i can't stop laughing because of the battle ur mind and heart finally u saved it for later video idea😂
It was nice to see you go through gamma withdrawal, and make it to the other side intact. What was even nicer is how it appears to have sparked a video idea?!
By looking at the end result being (3^2)*ζ(3), I wonder if the result is always of the form n^2 *ζ(n) ( or n^(n-1) * ζ(n) ) ,where n is the logarithm power in the starting integral.
Hello, there's this crazy integral and its answer that's just stated on Wiki without the derivation, and I was wondering if you'd be up to the challenge of solving it!
Why is it not valid to simply integrate by parts where u = ln( (1-x) / (1+x) )^3 and dv = dx? After some basic simplification, when you plug in the bounds you get two ln( 0 )'s which is undefined. Also when I plug it into a not super complex calculator it also says the integral is undefined but can be approximated as the result in the video.
When making geometric series -1/(1+u)^2, it should be appropriate to add 1 to the power of negative 1 in right side. So, it will be (-1)^(k). Anyhow, thank you for this interesting integral and innovative solution.
@@MrWael1970 No. The 0th term is 0 that's why he can disregard it and start at k=1. If he was re-indexing in k then both (-1)^k and u^(k-1) would have been affected but only (-1)^k was - to soak up the floating negative sign.
I have a feeling you know the half lady from the 3 and a half you mention. You could have used 3 and a quarter or 3 and three fourths, or really any fractional value lesser than 1. The very fact that you know its precisely 1/2 is a bit sus. Also nice integral.
@@maths_505my math ig page had a 60-40 gender split, which either means women are actually interested in integrals, or that there are guys lying about their gender… I think we both know which is really true
For x imaginary (ix) and x < 1 the integrand suggests the relation psi*/psi which might have a relation to quantum mechanics and relativity (which fail because |sigma1|+|sigma2| is not a group in SU(2) because of the existence operator +. I have provided links to a number of documents relating to the vector-free foundation of mathematical physics at the physicsdiscussionforum organization for which this perspective might be relevent. Drop in and give me a call... :)
My way is following Substitution y = (1-x)/(1+x) We will get integral 2\int\limits\_{0}^{1}\frac{ln\left(y ight)}{\left(y+1 ight)^2}dy Now integrate by parts with clever choice of integration constant \int\limits\_{0}^{1}\frac{\ln^3\left(y ight)}{\left(y+1 ight)^2}dy u = \ln^3\left(y ight) , du = \frac{3}{y}\cdot\ln^2\left(y ight)dy dv = \frac{1}{\left(y+1 ight)^2}dy , v = (-\frac{1}{y+1}+A) 0 - \lim_{x\to 0}\left((-\frac{1}{y+1}+A)\ln^3\left(y ight) ight)-3\int\limits_{0}^{1}(-\frac{1}{y+1}+A)\frac{\ln^2\left(y ight)}{y}dy Now if we choose A=1 we can evaluate limit and remaining integral will simplify We will get -6\int\limits_{0}^{1}\frac{\ln^2\left(y ight)}{1+y}dy Now I would use series expansion and then integration by parts
If you like the videos and would like to support the channel:
www.patreon.com/Maths505
You can follow me on Instagram for write ups that come in handy for my videos and DM me in case you need math help:
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I am a lady and I watch this channel. Happy to get the mention! All three and a half of us had better have commented.
2 and a half of you have commented so far 😂
I love watching these fun integrals while getting cozy before bed :)
LOOOOL well done for avoiding the gamma function today. must have been real tough.😅
Alright gentlemen and three and a half ladies,
Missed opportunity though: no irrational number of ladies ...
What next? A pi number of enbies?
@@boredd9410 I'm the 0.1416 enby lol
Jokes aside, RUclips has no information about gender. The gender breakdown of viewership is based on a made-up gender it assigned based on stereotypes of what kind of videos folks watch by gender. Classic GIGO.
That switching between integral and summation operators is as beautiful as that u=(1-x)/(1+x) substitution
It’s just due to uniform convergence of a function on the open interval of its power series expansion
sorry mate a lil busy last two to three days so i missed out some videos i will quickly watch those tomorrow
when u tried not to let intrinsive tought by not using gamma function i can't stop laughing because of the battle ur mind and heart
finally u saved it for later video idea😂
Hi,
"ok, cool" : 0:16 , 2:27 ,
"terribly sorry about that" : 1:36 , 2:23 .
It was nice to see you go through gamma withdrawal, and make it to the other side intact.
What was even nicer is how it appears to have sparked a video idea?!
It was one of the darkest moments of my life but I'm grateful for the lessons learned.
This can be also done by taking
ln²1-x/1+x as t,doing some subsitutions,
then integrating by parts and applying gamma function (at least i think so).
glad to be one of the 7/2 ladies here 😂
Amazing.
i'm the half lady LMAOOOOO
@@halfasleeptypist 💪💪🚀🚀🚀
Rearranging to ux + x + u = 1, you can see it's symmetrical, so goes the same both ways.
Nice observation
This would be really useful in long expressions like higher degree polynomials...
Hi, thanks for the fantastic video, just curious what software are you using for handwriting.
Nice! My favorite sub u=(1-x)/(1+x)😊
I think the integral could also be written as a multiple of the integral of (artanhx)^3, which is pretty cool
By looking at the end result being (3^2)*ζ(3), I wonder if the result is always of the form n^2 *ζ(n) ( or n^(n-1) * ζ(n) ) ,where n is the logarithm power in the starting integral.
Looking at 10:07... How about this?
n! (2-2^(2-n)) Zeta(n)
Challenge completed 🙂 Thank you!
Hello, there's this crazy integral and its answer that's just stated on Wiki without the derivation, and I was wondering if you'd be up to the challenge of solving it!
DM me the integral on Instagram.
Pure nostalgie ❤
Why is it not valid to simply integrate by parts where u = ln( (1-x) / (1+x) )^3 and dv = dx? After some basic simplification, when you plug in the bounds you get two ln( 0 )'s which is undefined. Also when I plug it into a not super complex calculator it also says the integral is undefined but can be approximated as the result in the video.
That's cuz I haven't told the super complex calculator how to solve it yet😎😎
When making geometric series -1/(1+u)^2, it should be appropriate to add 1 to the power of negative 1 in right side. So, it will be (-1)^(k). Anyhow, thank you for this interesting integral and innovative solution.
He did do that but he decreased by 1 in the exponent rather than increase by 1, going from (-1)^k to (-1)^(k-1) to include the extra -1.
@@mjpledger (-1)^(k-1) this is due to index of the summation to start with 1 instead of 0, but this is not the issue. Anyway Thanks.
@@MrWael1970 No. The 0th term is 0 that's why he can disregard it and start at k=1. If he was re-indexing in k then both (-1)^k and u^(k-1) would have been affected but only (-1)^k was - to soak up the floating negative sign.
Thanks can you publish about PDE problems
i feel dumb cuz idk how you would continue after invoking the gamma function. gotta watch more maths 505 videos!
request for some nonlinear PDEs
how would you do it with gamma function tho
Sir (math 505) could you please tell me the name of the constant which somewhat related to gamma^2(1/4)
I believe you denote it with omega symbol
Yes even I tried to find it out but he used to say it so fast I never knew what he said
I also want to know the name of that constant which is denoted by L which once he told to a result of a summation
I believe you mean the lemniscate constant, denoted with an omega and bar on top
I think you are refering to the Lemniscate constant which is denoted by a ϖ (\varpi)
I think it's the lemniscate constant, it's something like gamma^2(1/4) / 2sqrt(2pi)
I have a feeling you know the half lady from the 3 and a half you mention.
You could have used 3 and a quarter or 3 and three fourths, or really any fractional value lesser than 1. The very fact that you know its precisely 1/2 is a bit sus.
Also nice integral.
😂😂😂😂
A virtuoso performance, eh? I am humbled. Which is good -- that's exactly where I should be. Maybe I'll revisit this in a year or two. Or three.
Why do you say 3.5 ladies in the beginning? I'm really curious..
it's a joke because most followers of this channel are guys
Probably 3.5%
Yeah it's just a joke😂
@@maths_505my math ig page had a 60-40 gender split, which either means women are actually interested in integrals, or that there are guys lying about their gender… I think we both know which is really true
shoutout to the 1/2 females out there, they exist
Thanks for avoiding the Gamma and Beta functions. There use to me seems like cheating.
His name was Roger Apéry, not Avery!
Challenge completed ❤
Nice
I was never taught about the gamma function in uni. feeling so robbed.
Pretty much every other video here makes use of the gamma function so you've come to the right place.
For x imaginary (ix) and x < 1 the integrand suggests the relation psi*/psi which might have a relation to quantum mechanics and relativity (which fail because |sigma1|+|sigma2| is not a group in SU(2) because of the existence operator +. I have provided links to a number of documents relating to the vector-free foundation of mathematical physics at the physicsdiscussionforum organization for which this perspective might be relevent. Drop in and give me a call... :)
I really like the video style and pace of the video, but say "oook, cool" one more time and I might break something 😅
You'll get used to it eventually 😂
Nice video
Thanks
Aperony constant
Never would have guessed that initial substitution! Seems too obvious 😂
∫ 0->1 ln^3(0)•dx -[ln^3(1)•dx]
∫ [0] 0->1->0 [ 0]
incredibleeeeee kjakjkj
My way is following
Substitution y = (1-x)/(1+x)
We will get integral
2\int\limits\_{0}^{1}\frac{ln\left(y
ight)}{\left(y+1
ight)^2}dy
Now integrate by parts with clever choice of integration constant
\int\limits\_{0}^{1}\frac{\ln^3\left(y
ight)}{\left(y+1
ight)^2}dy
u = \ln^3\left(y
ight) , du = \frac{3}{y}\cdot\ln^2\left(y
ight)dy
dv = \frac{1}{\left(y+1
ight)^2}dy , v = (-\frac{1}{y+1}+A)
0 - \lim_{x\to 0}\left((-\frac{1}{y+1}+A)\ln^3\left(y
ight)
ight)-3\int\limits_{0}^{1}(-\frac{1}{y+1}+A)\frac{\ln^2\left(y
ight)}{y}dy
Now if we choose A=1 we can evaluate limit and remaining integral will simplify
We will get -6\int\limits_{0}^{1}\frac{\ln^2\left(y
ight)}{1+y}dy
Now I would use series expansion and then integration by parts
Ooooookaay cool :-)
three and a half ladies lol
I guess it should be gentleman and imaginary ladies
First'
Second
@@maths_505 kjkjkj i was the first? kjkjkj😁