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  • hba
\[\int\limits _{0}^{\infty} \lfloor x \rfloor e^{-x} dx\]
  • hba
I am getting 1/(e-1)
absolute value of x right?

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  • hba
I get 1, but i'll check again
Hmm can't do the integral but the aproximation I do get to 1.
that initially seemed like floor value of x..... :P
trid partial integration?
i also get 1.
  • hba
Yeah one more which i was doing was Integral (x^2/x^2+1) and i got x- arctanx by trig sub.
thats correct....and can be done without substitution...
if u can use this integral as standard int 1/(1+x^2) dx = arctan x +c
btw forget what I said about partial integration... looks like geting no where.
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after IBP, i get the integral of: \[\large [-e^{-x}(x+1)]^\infty_0\] i made the leap that -e^(-x) tends to zero much faster than x+1 tends to infinity as x goes to infinity, but can't give a real justification right now
Looks just about right Mathmuse. Seems like I made a bad choise when I did IBP.
looks the same, wots 'sign(x)'?
The sign function?

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