$int_0^infty e^{-x}x^{n-1}dx$ is convergent for $n>0$












1












$begingroup$


I have something to ask regarding convergence of gamma function. I have done the proof as below. Please tell me if it is correct.




$int_0^infty e^{-x}x^{n-1}dx$ is convergent for $n>0$




Proof: For $nin(0,1],~int_0^infty e^{-x}x^{n-1}dx$ is convergent since $int_0^infty e^{-x}x^{n-1}dx=int_0^1 e^{-x}x^{n-1}dx+int_1^infty e^{-x}x^{n-1}dxleint_0^1 x^{n-1}dx+int_1^infty e^{-x}dx.$



On the other hand, $int e^{-x}x^{(n+1)-1}dx=-x^n.e^{-x}+nint e^{-x}x^{n-1}dxcdots(1)$



Since, $limlimits_{xtoinfty}x^n.e^{-x}=0,$ by successive application of $(1)$ $n$ can be put in $(0,1],$ whence the result follows.






improper-integrals







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  • $begingroup$
    The proof is flawed. For $x>1$, $x^{n-1} not le 1$ when $n>1$
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:30










  • $begingroup$
    Instead of integrating by parts, you can pick $a(n) in (0,1),c(n)$ such that $x^{n-1} < c(n) e^{a(n) x}$ for $x > 1$
    $endgroup$
    – reuns
    Dec 14 '18 at 3:28


















1












$begingroup$


I have something to ask regarding convergence of gamma function. I have done the proof as below. Please tell me if it is correct.




$int_0^infty e^{-x}x^{n-1}dx$ is convergent for $n>0$




Proof: For $nin(0,1],~int_0^infty e^{-x}x^{n-1}dx$ is convergent since $int_0^infty e^{-x}x^{n-1}dx=int_0^1 e^{-x}x^{n-1}dx+int_1^infty e^{-x}x^{n-1}dxleint_0^1 x^{n-1}dx+int_1^infty e^{-x}dx.$



On the other hand, $int e^{-x}x^{(n+1)-1}dx=-x^n.e^{-x}+nint e^{-x}x^{n-1}dxcdots(1)$



Since, $limlimits_{xtoinfty}x^n.e^{-x}=0,$ by successive application of $(1)$ $n$ can be put in $(0,1],$ whence the result follows.






improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    The proof is flawed. For $x>1$, $x^{n-1} not le 1$ when $n>1$
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:30










  • $begingroup$
    Instead of integrating by parts, you can pick $a(n) in (0,1),c(n)$ such that $x^{n-1} < c(n) e^{a(n) x}$ for $x > 1$
    $endgroup$
    – reuns
    Dec 14 '18 at 3:28
















1












1








1





$begingroup$


I have something to ask regarding convergence of gamma function. I have done the proof as below. Please tell me if it is correct.




$int_0^infty e^{-x}x^{n-1}dx$ is convergent for $n>0$




Proof: For $nin(0,1],~int_0^infty e^{-x}x^{n-1}dx$ is convergent since $int_0^infty e^{-x}x^{n-1}dx=int_0^1 e^{-x}x^{n-1}dx+int_1^infty e^{-x}x^{n-1}dxleint_0^1 x^{n-1}dx+int_1^infty e^{-x}dx.$



On the other hand, $int e^{-x}x^{(n+1)-1}dx=-x^n.e^{-x}+nint e^{-x}x^{n-1}dxcdots(1)$



Since, $limlimits_{xtoinfty}x^n.e^{-x}=0,$ by successive application of $(1)$ $n$ can be put in $(0,1],$ whence the result follows.






improper-integrals







share|cite|improve this question











$endgroup$




I have something to ask regarding convergence of gamma function. I have done the proof as below. Please tell me if it is correct.




$int_0^infty e^{-x}x^{n-1}dx$ is convergent for $n>0$




Proof: For $nin(0,1],~int_0^infty e^{-x}x^{n-1}dx$ is convergent since $int_0^infty e^{-x}x^{n-1}dx=int_0^1 e^{-x}x^{n-1}dx+int_1^infty e^{-x}x^{n-1}dxleint_0^1 x^{n-1}dx+int_1^infty e^{-x}dx.$



On the other hand, $int e^{-x}x^{(n+1)-1}dx=-x^n.e^{-x}+nint e^{-x}x^{n-1}dxcdots(1)$



Since, $limlimits_{xtoinfty}x^n.e^{-x}=0,$ by successive application of $(1)$ $n$ can be put in $(0,1],$ whence the result follows.






improper-integrals




improper-integrals






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share|cite|improve this question













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edited Dec 14 '18 at 2:31









TonyK

42.6k355134




42.6k355134










asked Dec 14 '18 at 2:24









JaveJave

472114




472114












  • $begingroup$
    The proof is flawed. For $x>1$, $x^{n-1} not le 1$ when $n>1$
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:30










  • $begingroup$
    Instead of integrating by parts, you can pick $a(n) in (0,1),c(n)$ such that $x^{n-1} < c(n) e^{a(n) x}$ for $x > 1$
    $endgroup$
    – reuns
    Dec 14 '18 at 3:28




















  • $begingroup$
    The proof is flawed. For $x>1$, $x^{n-1} not le 1$ when $n>1$
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:30










  • $begingroup$
    Instead of integrating by parts, you can pick $a(n) in (0,1),c(n)$ such that $x^{n-1} < c(n) e^{a(n) x}$ for $x > 1$
    $endgroup$
    – reuns
    Dec 14 '18 at 3:28


















$begingroup$
The proof is flawed. For $x>1$, $x^{n-1} not le 1$ when $n>1$
$endgroup$
– Mark Viola
Dec 14 '18 at 2:30




$begingroup$
The proof is flawed. For $x>1$, $x^{n-1} not le 1$ when $n>1$
$endgroup$
– Mark Viola
Dec 14 '18 at 2:30












$begingroup$
Instead of integrating by parts, you can pick $a(n) in (0,1),c(n)$ such that $x^{n-1} < c(n) e^{a(n) x}$ for $x > 1$
$endgroup$
– reuns
Dec 14 '18 at 3:28






$begingroup$
Instead of integrating by parts, you can pick $a(n) in (0,1),c(n)$ such that $x^{n-1} < c(n) e^{a(n) x}$ for $x > 1$
$endgroup$
– reuns
Dec 14 '18 at 3:28












1 Answer
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$begingroup$

The proof in the OP fails when $n>1$ since for $x>1$, $x^{n-1}>1$ or $n>1$.



But we can assert that $e^xge frac{x^{lfloor nrfloor+1}}{(lfloor nrfloor +1)!}$ for $x ge 1$. So, for $Lge1$



$$begin{align}
left|int_1^L e^{-x}x^{n-1},dxright|&le left(lfloor nrfloor+1right) ! int_1^L x^{n-lfloor nrfloor -2},dx\\
&=frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
end{align}$$



Since $n-lfloor nrfloor -1<0$, $lim_{Ltoinfty}left(frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
right)=frac{left(lfloor nrfloor+1right) !}{1-left(n-lfloor nrfloorright) }$, and the integral on the left-hand side of $(1)$ converges. And we are done!






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    $begingroup$

    The proof in the OP fails when $n>1$ since for $x>1$, $x^{n-1}>1$ or $n>1$.



    But we can assert that $e^xge frac{x^{lfloor nrfloor+1}}{(lfloor nrfloor +1)!}$ for $x ge 1$. So, for $Lge1$



    $$begin{align}
    left|int_1^L e^{-x}x^{n-1},dxright|&le left(lfloor nrfloor+1right) ! int_1^L x^{n-lfloor nrfloor -2},dx\\
    &=frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
    end{align}$$



    Since $n-lfloor nrfloor -1<0$, $lim_{Ltoinfty}left(frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
    right)=frac{left(lfloor nrfloor+1right) !}{1-left(n-lfloor nrfloorright) }$    , and the integral on the left-hand side of $(1)$ converges. And we are done!






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      The proof in the OP fails when $n>1$ since for $x>1$, $x^{n-1}>1$ or $n>1$.



      But we can assert that $e^xge frac{x^{lfloor nrfloor+1}}{(lfloor nrfloor +1)!}$ for $x ge 1$. So, for $Lge1$



      $$begin{align}
      left|int_1^L e^{-x}x^{n-1},dxright|&le left(lfloor nrfloor+1right) ! int_1^L x^{n-lfloor nrfloor -2},dx\\
      &=frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
      end{align}$$



      Since $n-lfloor nrfloor -1<0$, $lim_{Ltoinfty}left(frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
      right)=frac{left(lfloor nrfloor+1right) !}{1-left(n-lfloor nrfloorright) }$      , and the integral on the left-hand side of $(1)$ converges. And we are done!






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        The proof in the OP fails when $n>1$ since for $x>1$, $x^{n-1}>1$ or $n>1$.



        But we can assert that $e^xge frac{x^{lfloor nrfloor+1}}{(lfloor nrfloor +1)!}$ for $x ge 1$. So, for $Lge1$



        $$begin{align}
        left|int_1^L e^{-x}x^{n-1},dxright|&le left(lfloor nrfloor+1right) ! int_1^L x^{n-lfloor nrfloor -2},dx\\
        &=frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
        end{align}$$



        Since $n-lfloor nrfloor -1<0$, $lim_{Ltoinfty}left(frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
        right)=frac{left(lfloor nrfloor+1right) !}{1-left(n-lfloor nrfloorright) }$        , and the integral on the left-hand side of $(1)$ converges. And we are done!






        share|cite|improve this answer











        $endgroup$



        The proof in the OP fails when $n>1$ since for $x>1$, $x^{n-1}>1$ or $n>1$.



        But we can assert that $e^xge frac{x^{lfloor nrfloor+1}}{(lfloor nrfloor +1)!}$ for $x ge 1$. So, for $Lge1$



        $$begin{align}
        left|int_1^L e^{-x}x^{n-1},dxright|&le left(lfloor nrfloor+1right) ! int_1^L x^{n-lfloor nrfloor -2},dx\\
        &=frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
        end{align}$$



        Since $n-lfloor nrfloor -1<0$, $lim_{Ltoinfty}left(frac{left(lfloor nrfloor+1right) !}{left(n-lfloor nrfloor-1right) }left(L^{left(n-lfloor nrfloor-1right) }-1right)
        right)=frac{left(lfloor nrfloor+1right) !}{1-left(n-lfloor nrfloorright) }$        , and the integral on the left-hand side of $(1)$ converges. And we are done!







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 3:51

























        answered Dec 14 '18 at 3:45









        Mark ViolaMark Viola

        132k1275173




        132k1275173






























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