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A surprisingly interesting integral
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- Опубликовано: 6 сен 2024
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Wow. The way it collapses to such a simple form is just amazing!
@6:40 (or thereabout) I went a different direction. Since 1/(k+3/4) = 4/(4*k+3) and 1/(k+5/4) = 4/(4*k+5), I got I = 4*SUM(k>=0; 4/(4*k+3) - 4/(4*k+5)) = 16*SUM(k>=0;1/(4*k+3) - 1/(4*k+5)). Then I made a power series out of the sum: S(x) = SUM(k>=0; x^(4*k+3)/(4*k+3) - x^(4*k+5)/(4*k+5)). Reindexing gave me S(x) = SUM(k>=1; (-1)^(k+1)*x^(2*k+1)/(2*k+1)), which is quite close to atan(x) = SUM(k>=0; (-1)^k*x^(2*k+1)/(2*k+1)), meaning S(x) = 1 - atan(x). Back substitution gave me I = 16*S(1) = 16*(1 - atan(1)) = 16*(1 - pi/4).
much simpler would be the subtituition t=xy and then s=sqrt(t)/y. It becomes a simple double integral
Double integral with two variables involving the Euler's constant and unique properties of the digamma function ultimately ending in a beautiful final result..... Nice 👍👍
A small error at 3:45 "fixed" at 5:23, and another at 8:53 that is pretty much ignored.
1:19 You used that abs(x) must be strictly less than 1, but the integral goes up to (including) 1. What am I missing here?
0 and 1 are the lower and upper LIMITS respectively....so x and y both approach 1 from the left. So the geometric series expansion is valid.
Very smart way to evaluate this double integral. Thank you.
really sad to see oily macaroni constant go away ;(
I know bro....I know😥
@@maths_505 Last time they appeared in one of your solutions they also cancelled out! The maths Gods toy with us.
They cancelled because use of the Digamma function was unnecessary in the first place, since the sum can be reduced to involve only the Leibniz formula for pi after some reindexing and adding and removing terms from the sum. I'm just saying that it's not a surprising coincidence that the Euler-Mascheroni constant happened to cancel out in evaluating this integral.
crazy was just looking into these double integrals for matters finance related
Hi math 505 would you miss solving this question? Prove that the gamma function is defined as : gamma(n) = Integral from negative infinity to positive infinity of {(x raised to the power of (2n-1)) times (e raised to the power of negative (x square)) dx
Thanks for your time
Take the gamma function with the integral defined w.r.t x and substitute x=root(u). That should do the trick.
This was chill. 😌👌
The way that beast collapsed to a simple product of integrals was amazing. The monster was chopped into pieces were the simple power rule of integration was to be applied. That was a smart trick to take it that way. 😂
I was thinking of solving two integrals separately by separating the addition, this way we'd have two "free variables" with respect to the sub I was thinking of, namely u = xy, in which we start with the variable that is not in the numerator, after the sub I think we'd get a reasonably simple integral.
Ah, your method is way more excellent G.
Thanks bro
Can you fix your playlists such that the most recent videos are at the end and the ones that have been done are at the top?
hey I have a question. When you converted to the geometric series, how to justify what happens at (x, y)=(1,1)? I have this small nagging question for a while and seeing your video reminds me to ask it
the ends of the two definite integrals denote open intervals, no? they're not included in the points that are plugged into the integrand and so it shouldn't matter
The point (1,1) is a set of measure zero and hence does not contribute to the integral.
Wild!
I was searching for this types 😢
Si staccano in 2 integrali uguali...I=2(8(-arctg1+1))=16(1-pi/4)
First!
SUIIIIIIIIIIIIIIIIII