Improper integral and the application of L'hopitals Rule

  • #1
in this problem im trying to find the mean of the probability density function of c*e^(-c*t) and by doing so i am multiplying the function stated previously by the variable t, which i know is correct. after taking the anti derivative and evaluating using the limit. I get an indeterminate form of infinity over infinity and I then take the derivative of the numerator and denominator of the expression -e^(-c*t)*t which = t/(e^(c*t))= and get 1/(e^(c*x) which tends to zero when i take the limit as x->∞. I'll show my attempt below. i know the answer i got is right, but am unsure if my application of l'hopitals rule is appropriate.


∫ t*c^(-c*t)dt after integrating by parts i get: lim x->∞ -e^(-c*x)*x-(e^(-c*x))/c
0

then when i evaluate from 0 to ∞: [lim x->∞ -(x/(e^(c*x))-((e^(-c*x))/c)]-[(-e^(0)*0)-1/c]
for the indeterminate part of this integral i use lhopitals rule and take derivative of numerator and denominator of just that part of the equation. is this correct? Can apply lhopitals to just part of an expression and not the rest. thanks
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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If [itex]lim_{x\to a} f(x)= A[itex] and [itex]\lim_{x\to a} g(x)= B[/itex] both exist then [itex]\lim_{x\to a} f(x)+ g(x)[/tex] also exists and is equal to A+ B. In your example, the first limit is 0 so you can use whatever method you want to determine the second.
 

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