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hba
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\[\int\limits _{0}^{\infty} \lfloor x \rfloor e^{-x} dx\]
hba
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I am getting 1/(e-1)
Frostbite
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absolute value of x right?
hba
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Right.
Mathmuse
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I get 1, but i'll check again
Frostbite
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Hmm can't do the integral but the aproximation I do get to 1.
hartnn
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that initially seemed like floor value of x..... :P
Frostbite
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trid partial integration?
hartnn
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i also get 1.
hba
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Yeah one more which i was doing was Integral (x^2/x^2+1) and i got x- arctanx by trig sub.
hartnn
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thats correct....and can be done without substitution...
hartnn
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if u can use this integral as standard
int 1/(1+x^2) dx = arctan x +c
Frostbite
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btw forget what I said about partial integration... looks like geting no where.
Mathmuse
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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
Frostbite
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Looks just about right Mathmuse. Seems like I made a bad choise when I did IBP.
Mathmuse
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looks the same, wots 'sign(x)'?
Frostbite
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The sign function?