A surprisingly fascinating integral
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- Опубликовано: 6 сен 2024
- This is one of those integration results that you'd least expect given the structure of the integral and that makes the solution development all the more satisfying.
Series expansion used in the video for ln(2sin(x)):
• My take on this on won...
What i found really cool with this integral is that once it is reduced to -6 time the integral from 0 to pi/2 of xln(sin(x)) if we rewrite it as -6 time the integral from 0 to pi/2 of x(ln(e^ix)+ ln(1-e^-2ix)-ln(2)) using complex exposant form for sin and proprieties of the log then we can split this into three integrals the first and third one being obvious and using the series expansion for ln(1-e^-2ix) witch is - the sum as n goes from 1 to infinity of e^-2in/n and integrating by part we end up with an expression of zeta of 3 for the real part and pi time zeta of 2 for imaginary part however since we are integrating a real function on a real interval the imaginary part must be zero so we can conclude that zeta of 2 is pi^2/6 and we end up with the same result as you. Sorry for my bad english and keep up the good work !
That's cool and all, but how do y'all know the series expansions of all these random ass functions?
Might just be cuz I haven't had complex anal yet. Can't wait for that!
The reduction of powers of x
make me recall about the nitric acids:the higher the concentration of nitric acid, the higher the oxidation number of nitrogen in reactants😮😮😊
The result is aesthetically pleasant indeed. Thanks for sharing.
Indeed, it's not an exaggeration when you say "a really cool integral" and "nice to evaluate". Really beautiful. Thanks! Shouldn't we also consider (3/4)(π^2)ln2 as one of the solutions for even values of k? Why did we discard it altogether?
I appreciate you using ln instead of log, as for some reason, most mathtubers I've watched use log :)
The correct operator is in fact log
Nice. Btw today michael penn showed today the integral identity 2/(npi) ((-1)^n-cos(nx))ln(2sin(x/2)) from 0 to pi=1/n^2. He proved it in tricky way but now thanks to you , i see that with the series of ln(2sin(x/2)) we can prove it easily😃💯
Yeah it's not really a tricky series....all we need is some basic complex analysis
I've actually solved the integral of x^n * log(sin(x)) from 0 to pi/2 where n is a natural number. It's a complicated finite sum in terms of the Riemann zeta function and Dirichlet eta function. So if that is equal to L(n), then the integral of x^n / sin(x)^2 from 0 to pi/2 is -n*(n-1)*L(n-2).
Thank you for this amazing effort.
Note for later let x be arcsin x => Leibnitz => u = x^2 => u = tan x should be doable from there
Very nice.for the series of ln(2sinx): ln(2sinx)=ln(-ie^(ix)(1-e^-(2ix))=-0.5pi i+ix -sum e^-2nix/n from 1 to inf.now if we take the real part we find that sum -cos(2nx)/n from 1 to inf is ln(2sinx),and for abonus if we take the im part we find that the sum sin(2nx)/n from 1 to inf is 0.5pi-x (for 0
beautiful. can you do more on abreys constant when it comes to these integrals?
Apery
I trted integration by parts twice and everything would be ok if we have x^2 in the numerator
Did you really think I'd post something that trivial 😂😂😂
@@maths_505
After integrating by parts twice i would substitute t=Pi/2-x
to get cosine inside ln
then i would have to integrals
one is trivial
and in second one i would try to use double angle identity for cosine and
expand ln(1-u) into power series
I tried to play with substitution t=Pi-x
but there was a problem with interval
(subsititution t=Pi-x can get rid x but changes this interval of integration)
This means that i started the same way as you but the problem for me was this additional x which left after integration by parts
moreover interval of integration did not allow me to easily eliminate this remaining x
@@maths_505 I like your solution, it is well that you linked video with derivation of this series expansion
Thanks mate....always a pleasure seeing your solution developments in the comments
I think michael penn or dr peyam made a video on integral xln(sinx) and solved it without any series expansions. But that took em a long time as far as I remember.
Interesting
Nice
Great!!👏👏👏
Can you integrate from 0 to pi/3? (if you have time)🤪
Sure😂
We surely can, by abusing the polygamma function.
With bermoulli numners and integration by parts..my result Is I=3pi^2/8*S...S(series)=(-1)^k pi^(2k)B(2k)/(2k)!(k+1).... Perhaps is correct
Can anyone suggest me an youtube channel which deals with integration but on an easier lever than this😅
Blackpenredpen
eww ln(4)