Noob Integral Question, Indeterminate form of ?

[ken.]

Hi

I need INTegral from 0 to ∞ of : x*e^(-x) dx

I use IBP: u = x, du = dx
dv = e^(-x) dx, v = -e^(-x)

uv - ∫v du = -x*e^(-x) - ∫ -e^(-x) dx

I am trying to evaluate the uv term : -x*e^(-x) from 0 to ∞
for the ∞ term of -x*e^(-x) =
-inf * 1/e^(inf) = -inf * 1/inf = 1
OR
-inf * 1/e^(inf) = -inf * 1/inf = -inf * 0 = 0
OR
it is indeterminate form, use l hospital...
which path is correct one ?

for the 0 term of -x*e^(-x) = 0

I am really sorry for the noob calculus question.
This is part of larger question regarding Gauss Quadrature. but..
Kindly pls remind me of the correct one.

Thank you for your help

 

[HallsofIvy]

Hi

I need INTegral from 0 to ∞ of : x*e^(-x) dx

I use IBP: u = x, du = dx
dv = e^(-x) dx, v = -e^(-x)

uv - ∫v du = -x*e^(-x) - ∫ -e^(-x) dx

I am trying to evaluate the uv term : -x*e^(-x) from 0 to ∞
for the ∞ term of -x*e^(-x) =
-inf * 1/e^(inf) = -inf * 1/inf = 1
OR
-inf * 1/e^(inf) = -inf * 1/inf = -inf * 0 = 0

Neither of those is correct. You cannot treat "infinity" as if it were a number.

OR
it is indeterminate form, use l hospital...

Yes, this if of the form "$\infty/\infty$".

which path is correct one ?

for the 0 term of -x*e^(-x) = 0

I am really sorry for the noob calculus question.
This is part of larger question regarding Gauss Quadrature. but..
Kindly pls remind me of the correct one.

Thank you for your help

 

[ken.]

Thanks. u re right. This really shows how indeterminate it can be..

But, what about the zero term
do I also need to apply the L'Hospital there ?
because without L'Hospital, the term evaluates to 0
and with L'Hospital, the term evaluates to -1

I think it is 0. Because the L'hospital applies to individual limits and not to the whole term before evaluated 0 to inf
But.. just making sure.

Thank you.

 

[DrewD]

An important point about L'Hopital is that it only works for indeterminate forms. If the limit is defined, not only do you not need L'Hopital, but you should not use it. So
$\lim_{x\rightarrow0} xe^{-x}=0\cdot1=0$
is already defined.
Only use L'Hopital if you have already shown that the limit is an indeterminate limit.