The first integral can be gotten easily by a substitution u = cos(x), the second can be calculated using integration by parts for Stieltjes integrals with integrators cos(x) and sin((x)ln(x) and the third can be deduced from the integrals of ln(tan(x))^2 and ln(sin(x))^2 from 0 to pi/2.
sin(x)ln(sin(x)) There is easy to find indefinite integral iven for calculus I or Calculus II students (I dont know which it is in US In Europe it is Calculus I) Integration by parts , Pythagorean identity , substitution for cosine ,partial fraction decomposition and we should get F(x)=cos(x)(1-ln(sin(x)))+ln((1-cos(x))/sin(x)) Integral Int(sin(x)^2ln(sin(x)),x=0..Pi/2) also can be calculated by parts (but without calculating antiderivative) Your solution is not as simple as possible at least for this two integrals 3 Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln^2(t)/(1+t^2),t=0..infinity) =Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(t)/(1+t^2),t=1..infinity) =Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(1/u)/(1+1/u^2)(-1/u^2),u=1..0) =Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(u)/(u^2+1),u=0..1) =Int(ln^2(u)/(1+u^2),u=0..1)+Int(ln^2(u)/(u^2+1),u=0..1) =2Int(ln^2(u)/(1+u^2),u=0..1) =2Int(sum((-1)^ku^{2k}ln^2(u),k=0..infinity),u=0..1) =2sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity) =2sum((-1)^k (2/(2k+1)u^(2k+1)ln^2(u)|_{0}^{1}-2/(2k+1)Int(u^(2k)ln(u),u=0..1)),k=0..infinity) =2sum((-1)^k (-2/(2k+1))Int(u^(2k)ln(u),u=0..1),k=0..infinity) =2sum((-1)^k (-2/(2k+1))(1/(2k+1)u^(2k+1)ln(u)|_{0}^{1}-1/(2k+1)Int(u^{2k},u=0..1)),k=0..infinity) =2sum((-1)^k (-2/(2k+1))(-1/(2k+1)Int(u^{2k},u=0..1)),k=0..infinity) =4sum((-1)^k/(2k+1)^3,k=0..infinity) Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln(sin(x)/cos(x))^2,x=0..Pi/2) =Int((ln(sin(x))-ln(cos(x))^2,x=0..Pi/2) =Int(ln^2(sin(x)),x=0..Pi/2)-2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)+Int(ln^2(cos(x))),x=0..Pi/2) Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln^2(sin(x)),x=0..Pi/2) - 2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2) 2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(sin(x)),x=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2) 2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(sin(Pi/2-t)),(-1)t=Pi/2..0) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2) 2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2) 2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = 2Int(ln^2(cos(t)),t=0..Pi/2) - 2sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity) Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) - sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity) Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) -2sum((-1)^k/(2k+1)^3,k=0..infinity) But Int(ln^2(cos(t)),t=0..Pi/2) can be problematic for Calculus II students also problematic may be calculating this sum I saw solution with double sum (Michael Penn's video)
I regconize that all 3 integrals end up with very neat result (doesn't imply digamma function or Euler's constant...) so is there anyway to apporach without using the digamma function or not
Yes. The first one for instance, just integrate by part, integrate sin(x) as 1-cos(x) and derive log(sin(x) as cos(x)/sin(x). The extra 1 term in 1-cos(x) is to avoid infinity of log(sin(x)) when x=0. You will end up with a rather simple integral of cos(x)^2-cos(x)/sin(x) = (1-sin(x)^2-cos(x))/sin(x) = 1/sin(x)-sin(x)-cot(x), which are easy to integrate. Second would I'd guess would be similar (I didn't try) by replacing sin(x)^2 by (1-cos(2x))/2 then similar approach as 1st one. Third one is a known integral. You can find many references to it.
@@maths_505 Bruh I learned about the digamma by inconvenience I was trying to understand analytic continuation, the absolute gigachad of a professor explained the concept flawlessly using gamma as the example.
I'm all for this turning into a cooking channel. I would still watch, religiously.
The first integral can be gotten easily by a substitution u = cos(x), the second can be calculated using integration by parts for Stieltjes integrals with integrators cos(x) and sin((x)ln(x) and the third can be deduced from the integrals of ln(tan(x))^2 and ln(sin(x))^2 from 0 to pi/2.
16:08
gamma(3/2)gamma(1/2)
÷4gamma2
=1/2gamma²(1/2)÷4
=pi/8
sin(x)ln(sin(x))
There is easy to find indefinite integral iven for calculus I or Calculus II students
(I dont know which it is in US In Europe it is Calculus I)
Integration by parts , Pythagorean identity , substitution for cosine ,partial fraction decomposition and we should get
F(x)=cos(x)(1-ln(sin(x)))+ln((1-cos(x))/sin(x))
Integral Int(sin(x)^2ln(sin(x)),x=0..Pi/2)
also can be calculated by parts
(but without calculating antiderivative)
Your solution is not as simple as possible at least for this two integrals
3
Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln^2(t)/(1+t^2),t=0..infinity)
=Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(t)/(1+t^2),t=1..infinity)
=Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(1/u)/(1+1/u^2)(-1/u^2),u=1..0)
=Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(u)/(u^2+1),u=0..1)
=Int(ln^2(u)/(1+u^2),u=0..1)+Int(ln^2(u)/(u^2+1),u=0..1)
=2Int(ln^2(u)/(1+u^2),u=0..1)
=2Int(sum((-1)^ku^{2k}ln^2(u),k=0..infinity),u=0..1)
=2sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity)
=2sum((-1)^k (2/(2k+1)u^(2k+1)ln^2(u)|_{0}^{1}-2/(2k+1)Int(u^(2k)ln(u),u=0..1)),k=0..infinity)
=2sum((-1)^k (-2/(2k+1))Int(u^(2k)ln(u),u=0..1),k=0..infinity)
=2sum((-1)^k (-2/(2k+1))(1/(2k+1)u^(2k+1)ln(u)|_{0}^{1}-1/(2k+1)Int(u^{2k},u=0..1)),k=0..infinity)
=2sum((-1)^k (-2/(2k+1))(-1/(2k+1)Int(u^{2k},u=0..1)),k=0..infinity)
=4sum((-1)^k/(2k+1)^3,k=0..infinity)
Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln(sin(x)/cos(x))^2,x=0..Pi/2)
=Int((ln(sin(x))-ln(cos(x))^2,x=0..Pi/2)
=Int(ln^2(sin(x)),x=0..Pi/2)-2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)+Int(ln^2(cos(x))),x=0..Pi/2)
Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln^2(sin(x)),x=0..Pi/2) - 2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(sin(x)),x=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(sin(Pi/2-t)),(-1)t=Pi/2..0) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = 2Int(ln^2(cos(t)),t=0..Pi/2) - 2sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity)
Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) - sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity)
Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) -2sum((-1)^k/(2k+1)^3,k=0..infinity)
But Int(ln^2(cos(t)),t=0..Pi/2) can be problematic for Calculus II students
also problematic may be calculating this sum
I saw solution with double sum (Michael Penn's video)
Congrats for getting a brilliant sponsor! So glad to see your channel is growing :)
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Nice trick! the first one can also be done by u=sinx and integration by part
Yes guy! You are right! The indefinite integral of the first one has answer! This make the 5:00am morning more easy and beautiful! 😆
I use u=cosx
This is a great video, and you are very personable as well as clear in explaining the math :) keep it up!
Your integral solving skill is best 👍
You can write a book on integrals
7:42 Vinculum, yes. It is Latin for "bond".
So I suppose it's a nice name for a spy, too :)
Vinculum Jacobus?
Amazing video as usual! How did you get to exploiting the beta function? Did you remember that expression by heart? Thanks!
Yup
Thank you very much. I find out the Feynman's technique thanks to you.
Congrats to the sponsor ship!
very nice approachs
Today I literally came to see your Brilliant ad 😂
I regconize that all 3 integrals end up with very neat result (doesn't imply digamma function or Euler's constant...)
so is there anyway to apporach without using the digamma function or not
Yes. The first one for instance, just integrate by part, integrate sin(x) as 1-cos(x) and derive log(sin(x) as cos(x)/sin(x). The extra 1 term in 1-cos(x) is to avoid infinity of log(sin(x)) when x=0. You will end up with a rather simple integral of cos(x)^2-cos(x)/sin(x) = (1-sin(x)^2-cos(x))/sin(x) = 1/sin(x)-sin(x)-cot(x), which are easy to integrate.
Second would I'd guess would be similar (I didn't try) by replacing sin(x)^2 by (1-cos(2x))/2 then similar approach as 1st one.
Third one is a known integral. You can find many references to it.
@@riadsouissi thank you so much. I appreciate it.
First two are relatively simple because d/dx (ln(sinx)) is cotx but I’m excited for the last one
Please Please please integrate the probability density function of the student's t distribution from -∞ to ∞, it's such a satisfying result
Wow, you can't even cook. I mean Papa Flammy bakes, that's what I expect out of math UA-camrs. 😊
srsly though, I've been watching this channel because I need my fix of integration under the integral sign and contour integrals
What is the name of program you use for writing please
Really Amazing and i enjoyed with this brilliant solution.
Amazing masterpieve
Il n1, integrazione per parti.. ln2-1.. Il n2, pi/8-ln2(pi/4)..il 3 boh...mi deve ancora venire l'idea giusta
Nice to have u in crescent
Really awesome :)
More feynmann?? surely you must be joking mister Kamaal.
Not just feynman.....we got them digamma bois too
@@maths_505 Bruh I learned about the digamma by inconvenience I was trying to understand analytic continuation, the absolute gigachad of a professor explained the concept flawlessly using gamma as the example.
@@maths_505 dId YoU kNoW yOu CaN eAsiLy ProVe Bessel uSinG cOmPleX AnAlySis?
BOOOO for saying 'zee'.