A RIDICULOUSLY AWESOME INTEGRAL: int 0 to π/2 xsin(x)cos(x)ln(sin(x))ln(cos(x))
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- Опубликовано: 2 ноя 2023
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Complex analysis lectures:
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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
You make it look effortless haha
beautiful solution. keep rocking on integrals.
Very interesting solution. Smart and Simple solution. Thank you very much.
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!
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
Cool, loving your integral videos! Can you please make a video proving those gamma prime and digamma values on your shirts?
instagram.com/p/Cuak4YaNRy9/?igshid=MzRlODBiNWFlZA==
@@maths_505 Wow, you have a bunch of them proved on your IG! So cool, you just got a new follower! 🥳
That is, indeed, ridiculously awesome. 👏
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.
what happens to the x at the front
also what do you mean expand in powers of u
@@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.
Please solve putnam integrals 🙏
Can u make some videos on summ of series using calculus?
It's pretty interesting for me🎉😊
There are a few in my playlist. I'll add more along the way.
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 😢
ruclips.net/video/xM4z0ncARlw/видео.html
@@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 :)
@@HisMajesty99 tremendously appreciated your majesty
Can you make a video, if you haven't already, expaining the little gamma Euler-whatever constant? I haven't been able to find it online. Thanks!
In 2 or 3 videos
I was like "um how tf did I get π/16-π³/192" but then I saw the answer and went "oh"
Wow
Nice! Can you pkease solve the integral (lnx)^2/sqrt(4-x^2) from 0 to 1 the result is 7pi^3/216.
Sure
@@maths_505 thanks😃
Brasil!🎉
π/16-π^3/192=π/16(1-π^2/12)...dovrei ricontrollare....
🇧🇷💯