test the integral $int_{0}^{infty} frac {x}{3x^4 + 5x^2 +1}dx$ for convergence












0












$begingroup$


test the integral $int_{0}^{infty} frac {x}{3x^4 + 5x^2 +1}dx$ for convergence.



My thought



Can I compare it with 1/(3x^4)?



Any hints for the solution are appreciated!






real-analysis calculus convergence improper-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is $3x^4+5x^2+1>x^4$?
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 6:16






  • 2




    $begingroup$
    you can try $$frac{x}{3x^4+5x^2+1}leq frac{1}{x^3},,,,xgeq 0$$
    $endgroup$
    – Lau
    Dec 7 '18 at 6:19












  • $begingroup$
    @Lau but I have to prove this fact first ....... is it proved here on math stack ?
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:40






  • 1




    $begingroup$
    This can be easily seen:$forall xgeq0, frac x{3x^4+5x^2+1}leqfrac x{3x^4}leqfrac x{x^4} because$ the terms we ignore in the denominator are all positive. By ignoring them, we make the denominator smaller and thus the value of the ratio larger.
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 6:44












  • $begingroup$
    I am not speaking about this @Lau I am speaking about proving the convergence of an integral that is different from the required one (the initial point differes)
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:55
















0












$begingroup$


test the integral $int_{0}^{infty} frac {x}{3x^4 + 5x^2 +1}dx$ for convergence.



My thought



Can I compare it with 1/(3x^4)?



Any hints for the solution are appreciated!






real-analysis calculus convergence improper-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is $3x^4+5x^2+1>x^4$?
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 6:16






  • 2




    $begingroup$
    you can try $$frac{x}{3x^4+5x^2+1}leq frac{1}{x^3},,,,xgeq 0$$
    $endgroup$
    – Lau
    Dec 7 '18 at 6:19












  • $begingroup$
    @Lau but I have to prove this fact first ....... is it proved here on math stack ?
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:40






  • 1




    $begingroup$
    This can be easily seen:$forall xgeq0, frac x{3x^4+5x^2+1}leqfrac x{3x^4}leqfrac x{x^4} because$ the terms we ignore in the denominator are all positive. By ignoring them, we make the denominator smaller and thus the value of the ratio larger.
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 6:44












  • $begingroup$
    I am not speaking about this @Lau I am speaking about proving the convergence of an integral that is different from the required one (the initial point differes)
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:55














0












0








0





$begingroup$


test the integral $int_{0}^{infty} frac {x}{3x^4 + 5x^2 +1}dx$ for convergence.



My thought



Can I compare it with 1/(3x^4)?



Any hints for the solution are appreciated!






real-analysis calculus convergence improper-integrals







share|cite|improve this question









$endgroup$




test the integral $int_{0}^{infty} frac {x}{3x^4 + 5x^2 +1}dx$ for convergence.



My thought



Can I compare it with 1/(3x^4)?



Any hints for the solution are appreciated!






real-analysis calculus convergence improper-integrals




real-analysis calculus convergence improper-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 6:10









hopefullyhopefully

290113




290113












  • $begingroup$
    Is $3x^4+5x^2+1>x^4$?
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 6:16






  • 2




    $begingroup$
    you can try $$frac{x}{3x^4+5x^2+1}leq frac{1}{x^3},,,,xgeq 0$$
    $endgroup$
    – Lau
    Dec 7 '18 at 6:19












  • $begingroup$
    @Lau but I have to prove this fact first ....... is it proved here on math stack ?
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:40






  • 1




    $begingroup$
    This can be easily seen:$forall xgeq0, frac x{3x^4+5x^2+1}leqfrac x{3x^4}leqfrac x{x^4} because$ the terms we ignore in the denominator are all positive. By ignoring them, we make the denominator smaller and thus the value of the ratio larger.
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 6:44












  • $begingroup$
    I am not speaking about this @Lau I am speaking about proving the convergence of an integral that is different from the required one (the initial point differes)
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:55


















  • $begingroup$
    Is $3x^4+5x^2+1>x^4$?
    $endgroup$
    – John Wayland Bales
    Dec 7 '18 at 6:16






  • 2




    $begingroup$
    you can try $$frac{x}{3x^4+5x^2+1}leq frac{1}{x^3},,,,xgeq 0$$
    $endgroup$
    – Lau
    Dec 7 '18 at 6:19












  • $begingroup$
    @Lau but I have to prove this fact first ....... is it proved here on math stack ?
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:40






  • 1




    $begingroup$
    This can be easily seen:$forall xgeq0, frac x{3x^4+5x^2+1}leqfrac x{3x^4}leqfrac x{x^4} because$ the terms we ignore in the denominator are all positive. By ignoring them, we make the denominator smaller and thus the value of the ratio larger.
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 6:44












  • $begingroup$
    I am not speaking about this @Lau I am speaking about proving the convergence of an integral that is different from the required one (the initial point differes)
    $endgroup$
    – hopefully
    Dec 7 '18 at 6:55
















$begingroup$
Is $3x^4+5x^2+1>x^4$?
$endgroup$
– John Wayland Bales
Dec 7 '18 at 6:16




$begingroup$
Is $3x^4+5x^2+1>x^4$?
$endgroup$
– John Wayland Bales
Dec 7 '18 at 6:16




2




2




$begingroup$
you can try $$frac{x}{3x^4+5x^2+1}leq frac{1}{x^3},,,,xgeq 0$$
$endgroup$
– Lau
Dec 7 '18 at 6:19






$begingroup$
you can try $$frac{x}{3x^4+5x^2+1}leq frac{1}{x^3},,,,xgeq 0$$
$endgroup$
– Lau
Dec 7 '18 at 6:19














$begingroup$
@Lau but I have to prove this fact first ....... is it proved here on math stack ?
$endgroup$
– hopefully
Dec 7 '18 at 6:40




$begingroup$
@Lau but I have to prove this fact first ....... is it proved here on math stack ?
$endgroup$
– hopefully
Dec 7 '18 at 6:40




1




1




$begingroup$
This can be easily seen:$forall xgeq0, frac x{3x^4+5x^2+1}leqfrac x{3x^4}leqfrac x{x^4} because$ the terms we ignore in the denominator are all positive. By ignoring them, we make the denominator smaller and thus the value of the ratio larger.
$endgroup$
– Shubham Johri
Dec 7 '18 at 6:44






$begingroup$
This can be easily seen:$forall xgeq0, frac x{3x^4+5x^2+1}leqfrac x{3x^4}leqfrac x{x^4} because$ the terms we ignore in the denominator are all positive. By ignoring them, we make the denominator smaller and thus the value of the ratio larger.
$endgroup$
– Shubham Johri
Dec 7 '18 at 6:44














$begingroup$
I am not speaking about this @Lau I am speaking about proving the convergence of an integral that is different from the required one (the initial point differes)
$endgroup$
– hopefully
Dec 7 '18 at 6:55




$begingroup$
I am not speaking about this @Lau I am speaking about proving the convergence of an integral that is different from the required one (the initial point differes)
$endgroup$
– hopefully
Dec 7 '18 at 6:55










1 Answer
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$begingroup$

$I=int_0^infty frac x{3x^4 + 5x^2 +1}dx$



$forall x>0, 0leqfrac x{3x^4 + 5x^2 +1}leqfrac x{x^4}=frac1{x^3}$



$implies 0leq I=int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac x{3x^4 + 5x^2 +1}dx\ leqint_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+frac12$



Since $frac x{3x^4 + 5x^2 +1}$ is continuous on $[0,1]$ (the denominator has no real zeroes), it is bounded there and so is the proper integral $int_0^1 frac x{3x^4 + 5x^2 +1}dx$.






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    $begingroup$

    $I=int_0^infty frac x{3x^4 + 5x^2 +1}dx$



    $forall x>0, 0leqfrac x{3x^4 + 5x^2 +1}leqfrac x{x^4}=frac1{x^3}$



    $implies 0leq I=int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac x{3x^4 + 5x^2 +1}dx\ leqint_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+frac12$



    Since $frac x{3x^4 + 5x^2 +1}$ is continuous on $[0,1]$ (the denominator has no real zeroes), it is bounded there and so is the proper integral $int_0^1 frac x{3x^4 + 5x^2 +1}dx$.






    share|cite|improve this answer











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      2












      $begingroup$

      $I=int_0^infty frac x{3x^4 + 5x^2 +1}dx$



      $forall x>0, 0leqfrac x{3x^4 + 5x^2 +1}leqfrac x{x^4}=frac1{x^3}$



      $implies 0leq I=int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac x{3x^4 + 5x^2 +1}dx\ leqint_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+frac12$



      Since $frac x{3x^4 + 5x^2 +1}$ is continuous on $[0,1]$ (the denominator has no real zeroes), it is bounded there and so is the proper integral $int_0^1 frac x{3x^4 + 5x^2 +1}dx$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $I=int_0^infty frac x{3x^4 + 5x^2 +1}dx$



        $forall x>0, 0leqfrac x{3x^4 + 5x^2 +1}leqfrac x{x^4}=frac1{x^3}$



        $implies 0leq I=int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac x{3x^4 + 5x^2 +1}dx\ leqint_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+frac12$



        Since $frac x{3x^4 + 5x^2 +1}$ is continuous on $[0,1]$ (the denominator has no real zeroes), it is bounded there and so is the proper integral $int_0^1 frac x{3x^4 + 5x^2 +1}dx$.






        share|cite|improve this answer











        $endgroup$



        $I=int_0^infty frac x{3x^4 + 5x^2 +1}dx$



        $forall x>0, 0leqfrac x{3x^4 + 5x^2 +1}leqfrac x{x^4}=frac1{x^3}$



        $implies 0leq I=int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac x{3x^4 + 5x^2 +1}dx\ leqint_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+int_1^infty frac 1{x^3}dx\ =int_0^1 frac x{3x^4 + 5x^2 +1}dx+frac12$



        Since $frac x{3x^4 + 5x^2 +1}$ is continuous on $[0,1]$ (the denominator has no real zeroes), it is bounded there and so is the proper integral $int_0^1 frac x{3x^4 + 5x^2 +1}dx$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 6:46

























        answered Dec 7 '18 at 6:39









        Shubham JohriShubham Johri

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