# Finding if an improper integral is Convergent

• Yohan
In summary, the conversation discusses finding the values of p > 0 for which the integral ##\displaystyle{\int_0^\infty x^{p-1}e^{-x}\,dx}\;## converges. The speaker suggests breaking the integral into two parts and using the limit convergence test, but ultimately concludes that there are no values of p that make both parts converge. The concept of the Gamma function is mentioned as a possible approach to solving the problem. The speaker also mentions alternative methods such as substitution and using the Taylor series of the exponential function.
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..

Last edited:
What is ##\displaystyle{\int_0^\infty e^{-x}\,dx}\;?##

fresh_42 said:
What is ##\displaystyle{\int_0^\infty e^{-x}\,dx}\;?##
i think it is 1

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?

Last edited:
fresh_42 said:
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

Last edited:
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}##.

## What is an improper integral?

An improper integral is an integral that does not have both limits of integration as finite numbers, or where the function being integrated has a singularity within the limits of integration.

## How do you determine if an improper integral is convergent?

To determine if an improper integral is convergent, you must evaluate the integral by taking the limit as one or both of the limits of integration approaches infinity or a singularity. If the limit exists and is a finite number, then the integral is convergent. If the limit does not exist or is infinite, then the integral is divergent.

## What is the difference between a convergent and a divergent integral?

A convergent integral has a finite value, while a divergent integral either does not have a finite value or does not exist.

## Can an improper integral be both convergent and divergent?

No, an improper integral can only be either convergent or divergent. It cannot be both.

## What are some techniques for determining the convergence of an improper integral?

Some techniques for determining the convergence of an improper integral include using the comparison test, the limit comparison test, and the ratio test. These tests involve comparing the given integral to a known convergent or divergent integral, or taking the limit of the function being integrated and comparing it to a known function. Other techniques include using partial fractions, integration by parts, and trigonometric substitutions.

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