A RIDICULOUSLY AWESOME INTEGRAL: int 0 to π/2 xsin(x)cos(x)ln(sin(x))ln(cos(x))

Рет қаралды 8,053

Күн бұрын

Пікірлер: 35

@venkatamarutiramtarigoppul2078 4 ай бұрын

After applying the kings property just multiply numerator & denominatior by 8 take sin^2 x = t sin2x dx = dt & cos^2 x =1-t. So the integral sorts out to pi/32 integral on 0 to 1 ln x ln (1-x )dx & just expand ln (1-x) obtain sum as sum of positive integers 1/n*1/(n+1)^2 which is a telescoping sum

@slavinojunepri7648 Ай бұрын

Smart substitution

@nicolastorres147 Жыл бұрын

Cool, loving your integral videos! Can you please make a video proving those gamma prime and digamma values on your shirts?

@maths_505 Жыл бұрын

instagram.com/p/Cuak4YaNRy9/?igshid=MzRlODBiNWFlZA==

@nicolastorres147 Жыл бұрын

@@maths_505 Wow, you have a bunch of them proved on your IG! So cool, you just got a new follower! 🥳

@jieyuenlee1758 8 ай бұрын

11:15 or plug specific values into the digamma function: f(z+1) =-A+Sum(k=[1,inf],(1/n-1/(n+1) where f is the digamma function; A is the euler masheroni constant

@Daniel-yc2ur Жыл бұрын

You make it look effortless haha

@michaelhoffman9172 Жыл бұрын

The beta and gamma functions aren't really necessary. Starting with the integral from 0 to Pi/2 of sin(x)*cos(x)*ln(sin(x))*ln(cos(x)), make the substitution u = sin(x) to reduce it to 1/4 the integral from 0 to 1 of u*log(u)*log(1-u^2). Expand in powers of u and integrate term-by-term to get 1/4 the sum over n \ge 1 of 1/n(n+1)^2. The sum can be written as \sum_{n\ge1} 1/n(n+1) (which telescopes to 1) minus \sum_{n\ge1} 1/(n+1) (which is Pi^2/6 - 1). The conclusion then follows.

@uggupuggu 9 ай бұрын

what happens to the x at the front

@uggupuggu 9 ай бұрын

also what do you mean expand in powers of u

@Sparky1_1 8 ай бұрын

@@uggupugguThe substitution is done after using the property of Definite integral”King Rule”. The x is removed by adding the I by both side then the LHS becomes 2I and we proceed further.

@slavinojunepri7648 Ай бұрын

Smart substitution

@trelosyiaellinika 6 ай бұрын

Indeed ridiculously awesome! The title is well-chosen! I have become an addict of your videos! You must take responsibility, there's no cure... 🤣 Jokes aside, thanks for your effort!

@MrWael1970 Жыл бұрын

Very interesting solution. Smart and Simple solution. Thank you very much.

@nicogehren6566 Жыл бұрын

beautiful solution. keep rocking on integrals.

@HisMajesty99 Жыл бұрын

My good sir, I have a special request for you! There is an infinite series which many years ago I ran into but was not able to solve. I would be absolutely delighted if you could attempt it and post a solution. It is a series that converges to a very interesting value, and though I can find the answer online, I cannot find a solution or explanation for how to obtain it. It is as follows: ∞ (n!)² Sum --- n=1 (2n)! The result is: (9+2√3π)/27 However, I was never able to figure out why 😢

@maths_505 Жыл бұрын

kzbin.info/www/bejne/rn6Xq2OkmKaFotk

@HisMajesty99 Жыл бұрын

@@maths_505 you are a gdm LEGEND mate! The absolute mad lad already has a video on it….wow, surreal. Much appreciated man, cheers. 🙌 And what a beautiful solution development :)

@maths_505 Жыл бұрын

@@HisMajesty99 tremendously appreciated your majesty