Integral of (ln(cos x))^2
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- Опубліковано 7 вер 2024
- We calculate the definite integral of (ln(cos x))^2 over the interval 0 to pi/2 using Fourier series and some other tricks.
Playlist: • Interesting Integrals
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I feel like you just took a tremendously tricky problem and broke it into a few easy steps. Very interesting
That's exactly what we should always do with any kind of problem.
love it when integrals are evaluated exactly, pure math!
Wow! The way you proved the second bullet point blew my mind! A difficult integral with an awesome solution! Please keep putting out such quality content! Thank you 👍
I really like your tool layout and how you never use anything ‘new’ without either proving it or else it being obvious or famous.
Hey MP, I love your channel and it's freakin awesome! Just a small request- If you could make a video playlist / links to the previous videos, it would be very nice! Thanks and keep making awesome content!
A late comment, but this integral can also be solved by substituting cos(x) by u, then it will look like a beta function with a log^2 inside, which can be inferred by deriving the beta function twice on the first variable.
The derivative of the beta function involves gamma and digamma functions. Using pi/2 as parameter value as well as certain known values of the digamma function, we can solve this integral.
Truly amazing proof! At first I have no idea how those each bullet point works for the proof, while at the end I've got to say, "Aha!"
how many concepts, tricks, knowledge.
thanks for sharing!
This was super slick. Thank you so much for this elegant derivations. A related problem that can be solved in a similar way is $\int_0^{2\pi} ln(1+t * cos(x)) dx$ for 0
19:35 There is n*(m-n) in denominator for m=1 and n=1 we have division by 0.
hello from 2021 ! A nice problem I was thinking about is the integral of (lncosx)*(lnsinx) from 0 to pi/2. Thanks for your videos btw !
And how you calculated it
Have you tried the same approach as shown on video
It's a tough question.No wonder I don't solve it. I appreciate your lecture.
I love this videos so much, but could you put the playlists in the order they were uploaded so that its easier to watch in order?
Heavy integral,...but the beautiful solution of Penn. Good for you!
I managed to evaluate this my own way with some complex analysis. The biggest difficulty was evaluating the integral of log^2 (1 + z^2) / z from 1 to i using the normal series representation of log(1 - z). It took quite a while to work the whole problem out the first time, but I got the same final answer. I also got a bonus harmonic series result out of it, being the sum of H_(2k + 1) / (2k + 1)^2 for k from 0 to infinity = 21/16*zeta(3).
I take notes while your videos. Thank you!
It can be shown that the integral is equal to δ²/δx ⅛β(x+½,½) at x=0, and then use special values of the PolyGamma fxn to solve this as well, but your way is much prettier
Damn and i was just going to use this as my comeback video lol
My plan was to do (ln(cos x))^3 to "finish the trilogy". You can tackle it if you want -- I think the solution requires a sum identity involving harmonic numbers. This identity could be a "spin-off".
@@MichaelPennMath no I cant deny you the star wars medal now. Thatd just be cheating lol but seriously no you got it.
Very well done and interesting. Also, the solution is the same if you put a sin instead of a cos. This is easy once you have done the hard work. Solvable for the same prize (trivial however)
I suppose the integral of ln(cos) was not even needed. You see, when you have squared the Fourier-expansion, the middle term was only made up of cosines, whose integrals are all zeroes.
What was really done (as you have referenced it) is an orthogonal decomposition of ln(cos) in some sort of a "Fourier-ish" base. The only tweak I would like to introduce is adding the constant function to the base. Calculating an inner product (which the original integral definitely is) in such a case is quite easy. In a Fourier-ish case, this is called Parseval's theorem which is quite commonly used in Electrical Engineering.
That second bullet point must have taken some work to think of. That double sum could be done much simpler though by simply converting the square to a doublesum with two copies of the summand with two different dummy variables. I think he confuses it with squaring a taylor series, which does result in a convolution.
Nothing to do with integrals, but seeing an expression with (ln {something]^2 reminds me of my first calculus class. Not knowing better, I wrote it as "ln^2 [something[" and was told that only the trig functions are written that way.
Actually you frequently see ln^2 in math literature.
@@knivesoutcatchdamouse2137 That's awesome! I've told the story before that on a worksheet in the second grade I identified a woman as an actor, and was corrected ("actress"). 20-some years later, I was able to say I wasn't wrong; I was just ahead of the curve. This is a second time for me, I guess.
@@JayTemple Haha, keep on innovating, my man!
You are a mathematical beast
Muy interesante. Gracias por compartir el conocimiento.
How can this much mathematics fit inside a single mind?
Folds.
-Stephanie
MP Editors
17:55 how do you justify using the Cauchy product if the series is not absolutely convergent there?
Can somebody please give the link to the previous video of this. Since he said at the beginning "this is a continuation of a previous video". Thanks in advance!
That factorization e^2ix + e^(-2ix) + 2 can also be written as (e^ix +e^-ix)^2. Can we get to the conclusion from that?
Why not just take the direct product of the two infinite series? It (of course) produces the same result (using the orthogonality of the cosines), but it's much more clear than introducing the finite sum.
Dear Sir , hi , i would like to mention a correction the time is 6.39 ........it should be - ln 2 , thank you sir , its honor to having a chat with you may Allah give you hidaya and bless ........
I have two questions :
- how do you justify problem the step after writing cos 2x in term of e^2ix and e^-2ix, I mean you wrote ln of a complex number which is multivaluate actually
- you proved the formula for ln(1+u) with norm(u) < 1 I agree (you used absolute convergence) but why did you use it for u = e^2ix (norm(u) = 1)
Généralisation des séries alternées
You're not the only one to wonder about this. I'd like clarifications too.
Delightful, this!
How do you spell the name of the series that expands ln cos x?
Phoier, Phoiet, Phoie? Foier? Foyet? 😬
What happen if the upper bound is pi/4 instead of pi/2 ? 🤔🤔
¿Y la constante de integración de las primeras integrales?
Awesome !
Amazing
great vid
Cool!
AND THATS A GOOD PLACE TO STOP
There was no good place to stop here 🤣
@20:23 inversion of integral and infinite sum? Wow... 🙄
2.04662202447 exactly. Amazing
hello to those coming from today's video
形式计算，很多地方不严格
Chinese spotted
Sono arrivato al 14',poi ho tentato il suicidio
Int(ln(cos(x))^2,x=0..Pi/2)
u=cos(x)
du=-sin(x)dx
du=-sqrt(1-u^2)dx
dx=-du/sqrt(1-u^2)
Int(ln(u)^2*(-1)/(sqrt(1-u^2)),u=1..0)
Int(ln(u)^2/(sqrt(1-u^2)),u=0..1)
Int(sum((-1/2 choose n)(-1)^n*u^(2n)ln(u)^2,n=0..infinity),u=0..1)
sum((-1)^n(-1/2 choose n)Int(u^(2n)ln(u)^2,u=0..1),n=0..infinity)
Finally we should get this sum
2*sum((2*n)!/(2^n*n!)^2/(2*n+1)^3,n=0..infinity)
Following this way we need only single sum but it may not be easy to evaluate it