Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?












1












$begingroup$


Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function







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$endgroup$








  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47
















1












$begingroup$


Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47














1












1








1


1



$begingroup$


Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function







share|cite|improve this question









$endgroup$




Let $a_n = intlimits_{0}^{n} e^{-x^4} dx$. Does ${ a_n }_{n rightarrow infty}$ converge?



${ a_n } ={ intlimits_{0}^{1} e^{-x^4} dx, intlimits_{0}^{2} e^{-x^4} dx, ..., intlimits_{0}^{infty} e^{-x^4} dx }$



So we need need only to check that the definite integral



$intlimits_{0}^{infty} e^{-x^4} dx$ converges



By using Wolfram Alpha,



$intlimits_{0}^{infty} e^{-x^4} dx } = Gamma left( frac54 right) approx 0.906402$



where $Gamma$ is the Gamma function



Therefore ${ a_n }$ converges, to $Gamma left( frac54 right)$.



But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.



Thanks in advance!






calculus integration sequences-and-series gamma-function




calculus integration sequences-and-series gamma-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 31 '13 at 7:33









PandaManPandaMan

1,19911333




1,19911333








  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47














  • 1




    $begingroup$
    Exponential are 'always' quite strong.
    $endgroup$
    – Felix Marin
    Oct 31 '13 at 7:47






  • 1




    $begingroup$
    Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
    $endgroup$
    – Lucian
    Oct 31 '13 at 7:47








1




1




$begingroup$
Exponential are 'always' quite strong.
$endgroup$
– Felix Marin
Oct 31 '13 at 7:47




$begingroup$
Exponential are 'always' quite strong.
$endgroup$
– Felix Marin
Oct 31 '13 at 7:47




1




1




$begingroup$
Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
$endgroup$
– Lucian
Oct 31 '13 at 7:47




$begingroup$
Generally speaking, $$n!=Gleft(frac1nright)qquad where qquad G(n)=int_0^infty e^{-x^n}dxqquadforall ngeqslant0$$
$endgroup$
– Lucian
Oct 31 '13 at 7:47










2 Answers
2






active

oldest

votes


















7












$begingroup$

We do not need an explicit expression to show that an improper integral converges.



The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



Any increasing sequence which is bounded above converges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You are welcome.
    $endgroup$
    – André Nicolas
    Nov 19 '13 at 3:07



















0












$begingroup$

Recall the Taylor series for $e^x$ at $x = 0$ is given by:



begin{equation}
e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
end{equation}



Thus,



begin{equation}
e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
end{equation}



Thus,



begin{align}
int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
&= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
end{align}



We now apply the Ratio Test:



begin{align}
R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
&= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
end{align}



Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






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    2 Answers
    2






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    2 Answers
    2






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    active

    oldest

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    7












    $begingroup$

    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07
















    7












    $begingroup$

    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07














    7












    7








    7





    $begingroup$

    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.






    share|cite|improve this answer











    $endgroup$



    We do not need an explicit expression to show that an improper integral converges.



    The sequence $(a_n)$ is obviously increasing. It is bounded above by $int_0^1 e^{-x^4},dx+int_1^infty e^{-x},dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.



    Any increasing sequence which is bounded above converges.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 31 '13 at 7:58

























    answered Oct 31 '13 at 7:40









    André NicolasAndré Nicolas

    455k36432819




    455k36432819












    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07


















    • $begingroup$
      You are welcome.
      $endgroup$
      – André Nicolas
      Nov 19 '13 at 3:07
















    $begingroup$
    You are welcome.
    $endgroup$
    – André Nicolas
    Nov 19 '13 at 3:07




    $begingroup$
    You are welcome.
    $endgroup$
    – André Nicolas
    Nov 19 '13 at 3:07











    0












    $begingroup$

    Recall the Taylor series for $e^x$ at $x = 0$ is given by:



    begin{equation}
    e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
    end{equation}



    Thus,



    begin{equation}
    e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
    end{equation}



    Thus,



    begin{align}
    int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
    &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
    end{align}



    We now apply the Ratio Test:



    begin{align}
    R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
    &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
    end{align}



    Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Recall the Taylor series for $e^x$ at $x = 0$ is given by:



      begin{equation}
      e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
      end{equation}



      Thus,



      begin{equation}
      e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
      end{equation}



      Thus,



      begin{align}
      int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
      &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
      end{align}



      We now apply the Ratio Test:



      begin{align}
      R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
      &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
      end{align}



      Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Recall the Taylor series for $e^x$ at $x = 0$ is given by:



        begin{equation}
        e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
        end{equation}



        Thus,



        begin{equation}
        e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
        end{equation}



        Thus,



        begin{align}
        int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
        &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
        end{align}



        We now apply the Ratio Test:



        begin{align}
        R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
        &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
        end{align}



        Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.






        share|cite|improve this answer









        $endgroup$



        Recall the Taylor series for $e^x$ at $x = 0$ is given by:



        begin{equation}
        e^{x} = sum_{n = 0}^{infty} frac{x^n}{n!}
        end{equation}



        Thus,



        begin{equation}
        e^{-x^4} = sum_{n = 0}^{infty} frac{(-1)^n x^{4n}}{n!}
        end{equation}



        Thus,



        begin{align}
        int_{0}^{t} e^{-x^4} :dx &= sum_{n = 0}^{infty} frac{(-1)^n }{n!}int_{0}^{t}x^{4n}:dx = sum_{n = 0}^{infty} frac{(-1)^n }{n!} left[frac{x^{4n + 1}}{4n + 1} right]_{0}^{t} \
        &= sum_{n = 0}^{infty} frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1} = sum_{n = 0}^{infty} b_n(t)
        end{align}



        We now apply the Ratio Test:



        begin{align}
        R &= lim_{n rightarrow infty} left|frac{b_{n + 1}}{b_{n}}right| = lim_{n rightarrow infty} left|frac{frac{(-1)^{n+1} }{left(n + 1right)!} frac{t^{4left(n + 1right) + 1}}{4left(n + 1right) + 1}}{frac{(-1)^n }{n!} frac{t^{4n + 1}}{4n + 1}}right| \
        &= lim_{n rightarrow infty} left| frac{1}{n + 1} cdot frac{4n + 1}{4n + 5}cdot t^4right|
        end{align}



        Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 2:28







        user150203





































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