This integral is actually one of your favorite constants

[@maths_505]

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

[@0oAstro]

Alright gentlemen and three and a half ladies,

Missed opportunity though: no irrational number of ladies ...

[@boredd9410]

What next? A pi number of enbies?

[@ribozyme2899]

​@@boredd9410 I'm the 0.1416 enby lol

[@daliasprints9798]

Jokes aside, KZbin 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.

[@emma5068]

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.

[@maths_505]

2 and a half of you have commented so far 😂

[@jennisaurus715]

I love watching these fun integrals while getting cozy before bed :)

[@jneen]

same >_>

[@keyaanmatin4804]

LOOOOL well done for avoiding the gamma function today. must have been real tough.😅

[@satyam-isical]

That switching between integral and summation operators is as beautiful as that u=(1-x)/(1+x) substitution

[@Harmonicaoscillator]

It’s just due to uniform convergence of a function on the open interval of its power series expansion

[@CM63_France]

Hi, "ok, cool" : 0:16 , 2:27 , "terribly sorry about that" : 1:36 , 2:23 .

[@renerpho]

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?!

[@maths_505]

It was one of the darkest moments of my life but I'm grateful for the lessons learned.

[@aravindakannank.s.]

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😂

[@boltez6507]

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).

[@nizogos]

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.

[@sergiokorochinsky49]

Looking at 10:07... How about this? n! (2-2^(2-n)) Zeta(n)

[@mikeoffthebox]

Rearranging to ux + x + u = 1, you can see it's symmetrical, so goes the same both ways.

[@satyam-isical]

Nice observation This would be really useful in long expressions like higher degree polynomials...

[@pluieuwu]

glad to be one of the 7/2 ladies here 😂

[@isavenewspapers8890]

Amazing.

[@halfasleeptypist]

i'm the half lady LMAOOOOO

[@pluieuwu]

@@halfasleeptypist 💪💪🚀🚀🚀

[@SnowWolfie11]

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.

[@maths_505]

That's cuz I haven't told the super complex calculator how to solve it yet😎😎

[@wqltr1822]

I think the integral could also be written as a multiple of the integral of (artanhx)^3, which is pretty cool

[@johnmich3459]

Hi, thanks for the fantastic video, just curious what software are you using for handwriting.

[@jhonnyrock]

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!

[@maths_505]

DM me the integral on Instagram.

[@MrWael1970]

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.

[@mjpledger]

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.

[@MrWael1970]

@@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.

[@mjpledger]

@@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.

[@Unidentifying]

request for some nonlinear PDEs

[@aymanalgeria7302]

Pure nostalgie ❤

[@shivamdahake452]

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_505]

😂😂😂😂

[@merouan3922]

Thanks can you publish about PDE problems

[@sandyjr5225]

Why do you say 3.5 ladies in the beginning? I'm really curious..

[@shebo96]

it's a joke because most followers of this channel are guys

[@rafiihsanalfathin9479]

Probably 3.5%

[@maths_505]

Yeah it's just a joke😂

[@abdulllllahhh]

@@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

[@Unidentifying]

shoutout to the 1/2 females out there, they exist

[@renerpho]

Challenge completed 🙂 Thank you!

[@yoav613]

Nice! My favorite sub u=(1-x)/(1+x)😊

[@brickie9816]

i feel dumb cuz idk how you would continue after invoking the gamma function. gotta watch more maths 505 videos!

[@slowf2l263]

how would you do it with gamma function tho

[@tsa_gamer007]

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

[@Listenpure]

Yes even I tried to find it out but he used to say it so fast I never knew what he said

[@Listenpure]

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

[@ambiguousheadline8263]

I believe you mean the lemniscate constant, denoted with an omega and bar on top

[@bennetdiesperger4080]

I think you are refering to the Lemniscate constant which is denoted by a ϖ (\varpi)

[@4wdsquirrel48]

I think it's the lemniscate constant, it's something like gamma^2(1/4) / 2sqrt(2pi)

[@thomasblackwell9507]

Thanks for avoiding the Gamma and Beta functions. There use to me seems like cheating.

[@KarlSnyder-jh9ic]

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.

[@Jalina69]

I was never taught about the gamma function in uni. feeling so robbed.

[@maths_505]

Pretty much every other video here makes use of the gamma function so you've come to the right place.

[@felipealonsoobandolopez9957]

incredibleeeeee kjakjkj

[@bennokrickl8135]

I really like the video style and pace of the video, but say "oook, cool" one more time and I might break something 😅

[@maths_505]

You'll get used to it eventually 😂

[@BuleriaChk]

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... :)

[@JosBergervoet]

His name was Roger Apéry, not Avery!

[@surajbhatt4713]

Challenge completed ❤

[@sarahakkak408]

Nice video

[@maths_505]

Thanks

[@calculus8399]

Nice

[@holyshit922]

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

[@dalibormaksimovic6399]

Aperony constant

[@edmundwoolliams1240]

Never would have guessed that initial substitution! Seems too obvious 😂

[@anestismoutafidis4575]

∫ 0->1 ln^3(0)•dx -[ln^3(1)•dx] ∫ [0] 0->1->0 [ 0]

[@DD-ce4nd]

Ooooookaay cool :-)

[@agrimmittal]

three and a half ladies lol

[@garvittiwari6266]

I guess it should be gentleman and imaginary ladies

[@fahadibrar379]

First'

[@maths_505]

Second

[@felipealonsoobandolopez9957]

@@maths_505 kjkjkj i was the first? kjkjkj😁