A RIDICULOUSLY AWESOME INTEGRAL! (yes another one)
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- Опубликовано: 30 янв 2025
- A few transformations here and there, some special functions and Feynman's trick works every time.
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What a shame that the euler mascaroni constant got cancelled in the end :(
Lil gammas are going extinct in this cruel calculus world😭
Indeed cuz there was a simpler way to evaluate the integral 🥲 2:46 here if u notice that he takes t² = u then u = sin theta , the integral is simply 1/4* 0 to pi/2 of Ln(sinx) dx
9:18 bro managed to plug his merch in the middle of solving an integral, math RUclips (mathtube? methtube?) is on a whole new level
He's cooking
@@ericthegreat7805meth
Hi,
"terribly sorry about that" : 1:18 , 3:52 , 8:16 ,
"ok, cool" : 2:57 , 4:38 , 6:45 , 8:56 .
Always look for this comment
2:06 there is a simpler approach by letting t^2 = u then integration by parts afterwards you would be left with integral of arcsin(x) / x by letting x = sin(t) you will find the value easily, nice video
There is an alternative solution. If we use substitution t=sqrt(cos(phi)) at 1:58, we get a well known integral from 0 to pi/2 of ln(cos(x)).
you can also do t=sqrt(sin(phi)) which is easier because you dont have to switch the limits of integration due to the negative created by cosine's derivative
pov: u just learnt about complex functions and have decided that they are the BEST way to solve any integral (which is true, and they are incredibly beautiful)
I think another possible method is to let sin(x) = sqrt(y), it will reduce the integral to a simpler form. Anyhow great work as usual.
why does euler macaroni constant always cancel out??? i am going to cry.
macaroni☠☠
@@lakshay3745💀💀💀💀
Awesome result. Thank you for your innovative videos.
2:09 you can let t^2=u and then after u=sin(theta), you have 1/4 euler's log trig integral
Substituting t=sin(x) then cos(y)=t/sqrt(2) gives 1/4*integral(log(cos(y),0,pi/2). Almost straightforward result.
Edit: another substitution is required, z=2y.
Awesome video!!!
It's just made my morning 🎉
after some substitutions i found it was equal to a quarter of (0 to π/2) ∫ ln(cos(x)) dx which can be solved using the kings symmetry property
Wonderful end result
After the integral at 3:28, I used the substitution p=1-t and symmetry to get that I=(1/32)*\int_0^1 ln(t(1-t))/sqrt(t(1-t)) dt. Wonder if this approach leads anywhere...
love this channel/
Nice flag
This is art!
Love ur vids❤
what a genius.
"Talking is a lot harder than math"
That cracked me out.
Awesome! Thanks for this. I was really getting thirsty for some Beta, Gamma and Digamma...
"Talking is harder than doing math" - Truth
I would have used logarithmic differentiation to go directly to ψ
the plug went crazy
What software do you use???
Samsung notes
Yayyy
again first like and comment
Good job Ronaldo
My urge to make things compact are telling me that you should make the π/8 an exponent inside the logarithm 😀
I think the result is close enough that we can just call it -e/10
You always use the beta function, why not the alpha function? That's not very sigma of you
The person solving it is the alpha 😎🔥🤣
@@maths_505touche