Finding if an improper integral is Convergent

  • Thread starter Yohan
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Homework Statement:

find out for what values of p > 0 this integral is convergent

Relevant Equations:

improper integrals
find out for what values of p > 0 this integral is convergent

##\displaystyle{\int_0^\infty x^{p-1}e^{-x}\,dx}\;##

so i broke them up to 2 integrals one from 0 to 1 and the other from 1 to ∞ and use the limit convergence test. but i found out that there are no vaules of p that makes both of parts of the integral to converge.. is it right?

it seems i posted this tread twice by mistake but i dont know how to delete one..
 
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Answers and Replies

  • #2
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What is ##\displaystyle{\int_0^\infty e^{-x}\,dx}\;?##
 
  • #3
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What is ##\displaystyle{\int_0^\infty e^{-x}\,dx}\;?##
i think it is 1
 
  • #4
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So the integral exists in case ##p=1##. The question is, does it for ##p=\varepsilon##, too?
Have you ever heard about the Gamma function?
 
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  • #5
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So the integral exists in case p=1p=1. The question is, does it for p=εp=ε, too?
i am not familiar with the gamma function

i understand how to find out if:
##\displaystyle{\int_0^\infty e^{-x}\,dx}\;##
is convergance. because i know how to calculate it's antiderivative. but for
##\displaystyle{\int_0^\infty x^{p-1}e^{-x}\,dx}\;##
when i try to use the limit test.

i defined ## f(x) = x^{p-1}e^{-x}## and ## g(x) = x^{p-1}## and then the ##\lim_{x \rightarrow \infty } \frac{f(x)}{g(x)}## is 0 so if the integral of g(x) from 0 to ∞ is convergent the the integral of f(x) is convergant. but then the integral of g(x) only is only convergante when p < 0 which doesnt make sense
 
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  • #6
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One possibility is to calculate ##\int_0^\infty e^{cx^\beta}\,dx## and substitute ##x^\beta = u##, or to consider the Taylor series of the exponential function. Another idea is to write ##x^{p-1}## as ##e^{cx}##.
 

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