How do you integrate #xe^x# from #-oo# to 0?

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Answer 1 · 2016-07-01T08:12:46

#= - 1 #

#int _(-oo)^0 dx qquad x e^x#

using IBP ie #int u v' = uv - int u' v# where, here:

#u = x, u' = 1#
#v' = e^x, v = e^x#

So we have

#[x e^x]_(-oo)^0 - int dx qquad e^x#

#= [ e^x (x - 1)]_(-oo)^0#

#= [ e^0 (0 - 1)] - [ \color{blue}{e^(-oo) (-oo) } - 1)] #

#= - 1 #

NB for the bit in blue, #lim_{x to - oo} x e^x = lim_{x to - oo} x/e^(-x) # which is indeterminate #oo/oo#so we use L'Hopital

#= lim_{x to - oo} 1/(-e^(-x)) #

#=- lim_{x to - oo} e^(x)#

#=- exp( lim_{x to - oo} x)#

#= -e^(- oo) = 0#