IBP failed me
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- Опубликовано: 19 сен 2024
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Beautiful! Loved the geometric series and the Laplace Transform🍻
Hi Doron. Thank you! 🙏
The ζ is just a Z with a little curl between the first two strokes, and a little hook at the end.
I know but I still can't draw it! 🤣🤣🤣 Honestly what has worked for me in the past was copying someone else's version of the symbol. Yeah math is hard. I wonder if people would be mad if i just draw a 'Z'
@@owl3mathi do a loop at the top like maths 505
I tried a different approach which seemed to work. First multiplied both the numerator and denominator by e^-x then chose 1-e^-x as u and then found that x^2 as (-ln(u+1))^2 which then I used IBP wherein I was going in circles thus I made use of the solution I and equated it with that like in the e^xsinx function integral. I don’t know if my approach is correct but that’s what I did. Also one thing to note is that I did it for the indefinite case since I don’t care for potentially transforming limits
Nice! 👍
I had fun with this. Started with e^-x/e^-x and then made a sub u = 1-e^-x to get ln^2(1-u)/u. I then went down the feynmann rabbit hole and it got messy fast. However, I'm going to keep playing. Currently at I'(a) = integral of -2 ln(1-au)/u(1-au) which I can partially decompose and get some Li(x) function.
sounds tricky!
funny enough i asked chatgpt for an integral a few days ago due to boredom and it gave me this same problem but with an x^3 in the numerator instead. Equally shocking is that it got the correct solution. We also encounter ζ(4) which has a known value of π^4 / 90 which is pretty nice in that integral.
Hi Josh. Nice! The general formula is zeta(n+1). Well not exactly but something similar 😀
My first thought was the gamma and zeta connection and the integral is ζ(3)Γ(3)
Yep exactly! And that’s a nice way to express it
I don't think that zeta(3) is any more or less unsatisfying than ln(3), sin(3), sqrt(3) or e^3. In every case it's a number that has to be approximated by some value if we want to use it to calculate a real world answer.
Hey adandap. Yep agree. No issues with zeta(3) 👍
When I stay up late I see that people really do comment at this time. I think because y’all are on UNSW time
@@owl3math Are you in the US?
bruh just remember zeta(s)=1/gamma(s)(integral 0 to+ ♾️ u^(s-1)/(e^u-1)
LETS GOOOO
Ha! Yeah!! 👍
Could have done all the same tricks with x on the top instead of x squared and had an actual solution
Yep pi^2/6 time. I think I did that one before. I suppose that is more satisfying then a decimal solution.
I=int[0,♾️](x^2•e^-x/(1-e^-x))dx
1/(1-e^-x)=sum[k=0,♾️](e^-kx)
I=int[0,♾️](x^2•sum[k=1,♾️](e^-kx))dx
I=sum[k=1,♾️](int[0,♾️](x^2•e^-kx)dx)
t=kx
dt=kdx
I=sum[k=1,♾️](k^-3•int[0,♾️](t^2•e^-t)dt)
I=sum[k=1,♾️](2!/k^3)
I=2•zeta(3)