Convergence for improper integral $int_0^infty x^re^{-x} dx$












3












$begingroup$


I'm trying to find for which values $r$ the following improper integral converges.
$$int_0^infty x^re^{-x} dx$$
What I have so far is that $x^r < e^{frac{1}{2}x}$ for $x geq a$, which splits the integral into
$$int_0^a x^re^{-x} dx + int_a^infty e^{-frac{1}{2}x}$$
We know the latter interval converges, but I don't know what to do with the first one. (For reference, graphing the functions reveals the answer to be $x > -1$.)



Edit: I would like a proof without the gamma function. Preferably one that uses the comparison test to compare limits.






calculus integration improper-integrals







share|cite|improve this question











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  • 6




    $begingroup$
    This is actually one of the more interesting integrals in mathematics. try plugging in $ r = 1, 2, 3, 4...$ etc. What do you notice?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 3:41






  • 1




    $begingroup$
    Hint:$$intlimits_0^{infty}mathrm dz, x^n e^{-x}=n!=Gamma(n+1)$$
    $endgroup$
    – Frank W.
    Jan 18 at 3:42






  • 3




    $begingroup$
    Did anybody even read the OP's question or do they just want to use the post to mention the gamma function?
    $endgroup$
    – Dionel Jaime
    Jan 18 at 3:49








  • 1




    $begingroup$
    @AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges.
    $endgroup$
    – Chase K
    Jan 18 at 3:53










  • $begingroup$
    @DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards.
    $endgroup$
    – John Omielan
    Jan 18 at 3:58
















3












$begingroup$


I'm trying to find for which values $r$ the following improper integral converges.
$$int_0^infty x^re^{-x} dx$$
What I have so far is that $x^r < e^{frac{1}{2}x}$ for $x geq a$, which splits the integral into
$$int_0^a x^re^{-x} dx + int_a^infty e^{-frac{1}{2}x}$$
We know the latter interval converges, but I don't know what to do with the first one. (For reference, graphing the functions reveals the answer to be $x > -1$.)



Edit: I would like a proof without the gamma function. Preferably one that uses the comparison test to compare limits.






calculus integration improper-integrals







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    This is actually one of the more interesting integrals in mathematics. try plugging in $ r = 1, 2, 3, 4...$ etc. What do you notice?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 3:41






  • 1




    $begingroup$
    Hint:$$intlimits_0^{infty}mathrm dz, x^n e^{-x}=n!=Gamma(n+1)$$
    $endgroup$
    – Frank W.
    Jan 18 at 3:42






  • 3




    $begingroup$
    Did anybody even read the OP's question or do they just want to use the post to mention the gamma function?
    $endgroup$
    – Dionel Jaime
    Jan 18 at 3:49








  • 1




    $begingroup$
    @AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges.
    $endgroup$
    – Chase K
    Jan 18 at 3:53










  • $begingroup$
    @DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards.
    $endgroup$
    – John Omielan
    Jan 18 at 3:58














3












3








3





$begingroup$


I'm trying to find for which values $r$ the following improper integral converges.
$$int_0^infty x^re^{-x} dx$$
What I have so far is that $x^r < e^{frac{1}{2}x}$ for $x geq a$, which splits the integral into
$$int_0^a x^re^{-x} dx + int_a^infty e^{-frac{1}{2}x}$$
We know the latter interval converges, but I don't know what to do with the first one. (For reference, graphing the functions reveals the answer to be $x > -1$.)



Edit: I would like a proof without the gamma function. Preferably one that uses the comparison test to compare limits.






calculus integration improper-integrals







share|cite|improve this question











$endgroup$




I'm trying to find for which values $r$ the following improper integral converges.
$$int_0^infty x^re^{-x} dx$$
What I have so far is that $x^r < e^{frac{1}{2}x}$ for $x geq a$, which splits the integral into
$$int_0^a x^re^{-x} dx + int_a^infty e^{-frac{1}{2}x}$$
We know the latter interval converges, but I don't know what to do with the first one. (For reference, graphing the functions reveals the answer to be $x > -1$.)



Edit: I would like a proof without the gamma function. Preferably one that uses the comparison test to compare limits.






calculus integration improper-integrals




calculus integration improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 3:55







Chase K

















asked Jan 18 at 3:38









Chase KChase K

675




675








  • 6




    $begingroup$
    This is actually one of the more interesting integrals in mathematics. try plugging in $ r = 1, 2, 3, 4...$ etc. What do you notice?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 3:41






  • 1




    $begingroup$
    Hint:$$intlimits_0^{infty}mathrm dz, x^n e^{-x}=n!=Gamma(n+1)$$
    $endgroup$
    – Frank W.
    Jan 18 at 3:42






  • 3




    $begingroup$
    Did anybody even read the OP's question or do they just want to use the post to mention the gamma function?
    $endgroup$
    – Dionel Jaime
    Jan 18 at 3:49








  • 1




    $begingroup$
    @AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges.
    $endgroup$
    – Chase K
    Jan 18 at 3:53










  • $begingroup$
    @DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards.
    $endgroup$
    – John Omielan
    Jan 18 at 3:58














  • 6




    $begingroup$
    This is actually one of the more interesting integrals in mathematics. try plugging in $ r = 1, 2, 3, 4...$ etc. What do you notice?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 18 at 3:41






  • 1




    $begingroup$
    Hint:$$intlimits_0^{infty}mathrm dz, x^n e^{-x}=n!=Gamma(n+1)$$
    $endgroup$
    – Frank W.
    Jan 18 at 3:42






  • 3




    $begingroup$
    Did anybody even read the OP's question or do they just want to use the post to mention the gamma function?
    $endgroup$
    – Dionel Jaime
    Jan 18 at 3:49








  • 1




    $begingroup$
    @AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges.
    $endgroup$
    – Chase K
    Jan 18 at 3:53










  • $begingroup$
    @DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards.
    $endgroup$
    – John Omielan
    Jan 18 at 3:58








6




6




$begingroup$
This is actually one of the more interesting integrals in mathematics. try plugging in $ r = 1, 2, 3, 4...$ etc. What do you notice?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 3:41




$begingroup$
This is actually one of the more interesting integrals in mathematics. try plugging in $ r = 1, 2, 3, 4...$ etc. What do you notice?
$endgroup$
– Aniruddh Venkatesan
Jan 18 at 3:41




1




1




$begingroup$
Hint:$$intlimits_0^{infty}mathrm dz, x^n e^{-x}=n!=Gamma(n+1)$$
$endgroup$
– Frank W.
Jan 18 at 3:42




$begingroup$
Hint:$$intlimits_0^{infty}mathrm dz, x^n e^{-x}=n!=Gamma(n+1)$$
$endgroup$
– Frank W.
Jan 18 at 3:42




3




3




$begingroup$
Did anybody even read the OP's question or do they just want to use the post to mention the gamma function?
$endgroup$
– Dionel Jaime
Jan 18 at 3:49






$begingroup$
Did anybody even read the OP's question or do they just want to use the post to mention the gamma function?
$endgroup$
– Dionel Jaime
Jan 18 at 3:49






1




1




$begingroup$
@AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges.
$endgroup$
– Chase K
Jan 18 at 3:53




$begingroup$
@AniruddhVenkatesan I know it equals $n!$ for natural numbers, but for this question I'm mainly interested in for which real values of $r$ the interval converges.
$endgroup$
– Chase K
Jan 18 at 3:53












$begingroup$
@DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards.
$endgroup$
– John Omielan
Jan 18 at 3:58




$begingroup$
@DionelJaime As I recall, the original question didn't mention not using the gamma function, with this being added afterwards.
$endgroup$
– John Omielan
Jan 18 at 3:58










1 Answer
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Note that for all $r in mathbb{R}$,



$$lim_{x to infty} frac{x^r e^{-x}}{x^{-2}} = lim_{x to infty} frac{x^{r+2}}{e^x} = 0 $$



since the exponential function tends to infinity faster than any polynomial. Hence, by the limit comparison test the integral over $[1, infty)$ converges since $int_1^infty x^{-2}, dx = 1$.



See if you can finish by finding the condition on $r$ such that the integral over $[0,1]$ converges.






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    active

    oldest

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    6












    $begingroup$

    Note that for all $r in mathbb{R}$,



    $$lim_{x to infty} frac{x^r e^{-x}}{x^{-2}} = lim_{x to infty} frac{x^{r+2}}{e^x} = 0 $$



    since the exponential function tends to infinity faster than any polynomial. Hence, by the limit comparison test the integral over $[1, infty)$ converges since $int_1^infty x^{-2}, dx = 1$.



    See if you can finish by finding the condition on $r$ such that the integral over $[0,1]$ converges.






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      Note that for all $r in mathbb{R}$,



      $$lim_{x to infty} frac{x^r e^{-x}}{x^{-2}} = lim_{x to infty} frac{x^{r+2}}{e^x} = 0 $$



      since the exponential function tends to infinity faster than any polynomial. Hence, by the limit comparison test the integral over $[1, infty)$ converges since $int_1^infty x^{-2}, dx = 1$.



      See if you can finish by finding the condition on $r$ such that the integral over $[0,1]$ converges.






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        Note that for all $r in mathbb{R}$,



        $$lim_{x to infty} frac{x^r e^{-x}}{x^{-2}} = lim_{x to infty} frac{x^{r+2}}{e^x} = 0 $$



        since the exponential function tends to infinity faster than any polynomial. Hence, by the limit comparison test the integral over $[1, infty)$ converges since $int_1^infty x^{-2}, dx = 1$.



        See if you can finish by finding the condition on $r$ such that the integral over $[0,1]$ converges.






        share|cite|improve this answer









        $endgroup$



        Note that for all $r in mathbb{R}$,



        $$lim_{x to infty} frac{x^r e^{-x}}{x^{-2}} = lim_{x to infty} frac{x^{r+2}}{e^x} = 0 $$



        since the exponential function tends to infinity faster than any polynomial. Hence, by the limit comparison test the integral over $[1, infty)$ converges since $int_1^infty x^{-2}, dx = 1$.



        See if you can finish by finding the condition on $r$ such that the integral over $[0,1]$ converges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 4:33









        RRLRRL

        51.1k42573




        51.1k42573






























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