@@maths_505 No, the original commenter @FatihKarakurt is correct. Either the sum needs to start at 1 instead of 0, or the summand needs to use 2k+1 instead of 2k-1. As written in the video, it includes a 1/(-1)² term, which it should not.
Something I noticed regarding the ln2 that keeps coming up in these gamma functions: ln(2) = ln(1 + 1) = ln(1 + e^0) = ln(1 + e^0) - 0 = ln(1 + e^0) - ln(1 + 0) = ln(1 + e^0) - ln(1 + e^-oo) = -1*ln(1 + e^(-oo)) -(-1)*ln(1 + e^(-0)) = Int(oo,0) e^-x/(1 + e^-x) * dx I am not sure what this means but this seems to relate to the logit function which is involved in categorical (n-ary) data inference in statistics, especially binary data (True/False). Maybe information entropy?
I think you might be on to something. We know there are deep connections between fields. Next to half-life, ln(2) appears in the formula for the entropy of a system with two equally likely microstates. On wiki there is a sentence on the gamma fn: " The fact that the integration is performed along the entire positive real line might signify that the gamma function describes the cumulation of a time-dependent process that continues indefinitely, or the value might be the total of a distribution in an infinite space." Which is pretty obvious, but can relate to entropy
Noice! .Michael penn solved this integral and also the integral (ln(cosx))^3 from 0 to pi/2,so now you can solve the integral (ln(cosx))^4 from 0 to pi/2 and complete the series..🙃😃
Here's my solution: given that the integral (ln(cosx))^2 from 0 to pi/2 equals the integral (ln(sinx))^2 from 0 to pi/2, which means the integral (ln(sinx)-ln(cosx))^2 from 0 to pi/2 plus the integral (ln(sinx)+ln(cosx))^2 from 0 to pi/2 equals 4 times the value of the integral we're trying to calculate, i.e. 4I. The integral (ln(sinx)+ln(cosx))^2 from 0 to pi/2 is quite easy to solve, because ln(sinx)+ln(cosx) = ln(sin(2x))-ln2, so it equals I + (ln2)^2*(pi/2) - ln4*the integral ln(sinx) from 0 to pi/2 = I + (ln2)^2*(3*pi/2), so the integral we're trying to calculate equals (ln2)^2*(pi/2) plus 1/3 of the integral (ln(sinx)-ln(cosx))^2 from 0 to pi/2. And this integral is also not that hard, because ln(sinx)-ln(cosx) = ln(tanx), and also the integral (ln(tanx))^2 from 0 to pi/4 equals the integral (ln(tanx))^2 from pi/4 to pi/2, if we let tanx = t, the integral (ln(tanx))^2 from 0 to pi/2 equals 2 times the value of the integral (lnt)^2/(1+t^2) from 0 to 1. And when 0
An easier and simple method is by evaluating the series for ln(cos x). Nevertheless this method by equating the partial differentiation of beta function in different ways is beautiful and interesting.
Let u=cosx,la lunzione integranda risulta (lnu)^2/sqrt (1-u^2) integrata da 0 a 1...utilizzo la I(a)=u^a/sqrt (1-u^2),percio risulta I=I''(0)..I(a)=1/2B((a+1)/2,1/2)...a questo punto devo derivare 2 volte la funzione beta...ma non lo so fare ..
Hola math ....😁 Estoy estudiando todas absolutamente todas las integrales del canal...para empezar a explicarlas pero en español..para que la otra parte del mundo se percate de lo increíble que son las matemáticas 😁😁😁🤑🤑🤑🐭🐭🐭👻🤓☝️😁
U cant? X! Is undefined outside N so u need 2 substitute gamma function. But this function is now in the under part of a fraction so cant solve that right?
How the hell did you manage to come up with this solution!? I tried the integral, ending up having to differentiate a Laplace transform of (exp(2t)-1)^(-1/2) twice and evaluate the result at zero... In other words, I reached a dead end. I was deeply saddened. To think of utilizing the Beta function from the get-go is just next-level.
I noticed we needed a log(trig) term so applying the Leibniz rule to the beta function seemed viable. I love using the gamma function so that's pretty much the first thing that comes to my mind.
You should include a link to the video you reference. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ikyVHEHmgP8.html (He also did the ln(cos) and ln(cos)^3 integrals. (While this one is easily found on his channel, a lot of his videos follow your tendency to provide meaningless titles that don't provide context and make your channel unsearchable.)
How is this Feynman's technique? He simply noticed the the integral was the second derivative of the beta function and evaluated that. There was no differentiation under the integral sign as far as I could tell?
I think that this integral is not really cool. It was a lot of non creative work to a not simple answer. Maybye no integral will be beautiful to me after that pi times lemniscate over 2 times sqrt(2) integral. 😢😊 I challenge you to find something even better.
