More complicated than it looks???

[@bigjazbo9217]

When you say you can use EITHER integration by parts OR Laplace transform, isn't that really the same thing? The tables of Laplace Teansforms for most functions were derived using IBP. So, invoking LT is really IBP but skipping several steps and going straight to a table.

[@owlsmath]

That sounds pretty fair to me. IBP is the most common way to derive most of the Laplace formulas so yea

[@erfanmohagheghian707]

That's not true! For many of the Laplace transform formulae, contour integration and residue theorem is required.

[@MikeMagTech]

Nice job! There was definitely some work in that one!

[@owlsmath]

Thanks! Yeah a lot going on in this one today. 👍

[@adandap]

Cool. An alternative another start is to write the integral as x . e^(-x)/(1-e^(-x)) and do parts. You get ln(e^2 - 1) - integral[ln(1 - e^(-x)) dx] then expand the ln out etc.

[@owlsmath]

I already recorded that alternative method video about it 🤣 But yes you’re right and it gives 2 different perspectives I think. Thanks 🙏

[@doronezri1043]

Excellent video 👏👏👏 Logarithm, dilogarithm and Basel in one solution (and there was a Laplace windowing opportunity there...) 🍻

[@owlsmath]

That's a lot! And I did notice the Laplace windowing situation but I already recorded another video about that. I'm somewhat obsessed with that idea now. 🤣🤣 thanks Doron. 🍻

[@maxvangulik1988]

I=int[0,2](xe^-x/(1-e^-x))dx
e^-x/(1-e^-x)=sum[k=1,♾️](e^-kx), x>0
I=sum[k=1,♾️](int[0,2](xe^-kx)dx)
u=x
dv=e^-kx•dx
du=dx
v=-1/k•e^-kx
I=sum[k=1,♾️](-2/k•e^-2k+1/k•int[0,2](e^-kx)dx)
I=sum[k=1,♾️](-2/k•e^-2k+(1-e^-2k)/k^2)
I=pi^2/6+2ln(1-e^-2)-Li_2(e^-2)
Li_2(x)-Li_2(1-x)=pi^2/6-ln(x)ln(1-x)
Li_2(e^-2)-Li_2(1-e^-2)=pi^2/6+2ln(1-e^-2)
I=-Li_2(1-e^-2)

[@maxvangulik1988]

I=sum[k=1,♾️](L{(1-U(x-2))x}|k)
I=sum[k=1,♾️](1/k^2-L{U(x-2)x}|k)
L{U(x-a)f(x)}=e^-as•L{f(x+a)}
I=pi^2/6-sum[k=1,♾️](e^-2k•(1/k^2+2/k)
I=(pi^2/6-ln(e^-2)ln(1-e^-2))-Li_2(e^-2)
I=-Li_2(1-e^-2)

[@owlsmath]

thanks Max :)