Finding if an improper integral is Convergent

[Yohan]

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 don't know how to delete one..

 

[fresh_42]

What is $\displaystyle{\int_0^\infty e^{-x}\,dx}\;?$

 

[Yohan]

What is $\displaystyle{\int_0^\infty e^{-x}\,dx}\;?$

i think it is 1

 

[fresh_42]

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?

 

[Yohan]

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 doesn't make sense

 

[fresh_42]

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 $e^{cx}$.