minimum value of $I(l)$ in definite integration












2












$begingroup$


If $displaystyle I(l)=int^{infty}_{0}frac{x^l}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx.$ for $0<l<5.$



Then value of $l$ for which $I(l)$ is minimum



What i tried



Put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$



$displaystyle I(l)=int^{infty}_{0}frac{t^{4-l}}{2t^6+4t^5+3t^4+5t^3+3t^2+4t+2}dt=int^{infty}_{0}frac{x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



$displaystyle 2I(l)=int^{infty}_{0}frac{x^l+x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



How do i solve it.Help me



Thanks in advance






definite-integrals







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    If $displaystyle I(l)=int^{infty}_{0}frac{x^l}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx.$ for $0<l<5.$



    Then value of $l$ for which $I(l)$ is minimum



    What i tried



    Put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$



    $displaystyle I(l)=int^{infty}_{0}frac{t^{4-l}}{2t^6+4t^5+3t^4+5t^3+3t^2+4t+2}dt=int^{infty}_{0}frac{x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



    $displaystyle 2I(l)=int^{infty}_{0}frac{x^l+x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



    How do i solve it.Help me



    Thanks in advance






    definite-integrals







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      If $displaystyle I(l)=int^{infty}_{0}frac{x^l}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx.$ for $0<l<5.$



      Then value of $l$ for which $I(l)$ is minimum



      What i tried



      Put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$



      $displaystyle I(l)=int^{infty}_{0}frac{t^{4-l}}{2t^6+4t^5+3t^4+5t^3+3t^2+4t+2}dt=int^{infty}_{0}frac{x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



      $displaystyle 2I(l)=int^{infty}_{0}frac{x^l+x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



      How do i solve it.Help me



      Thanks in advance






      definite-integrals







      share|cite|improve this question









      $endgroup$




      If $displaystyle I(l)=int^{infty}_{0}frac{x^l}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx.$ for $0<l<5.$



      Then value of $l$ for which $I(l)$ is minimum



      What i tried



      Put $displaystyle x=frac{1}{t}$ and $displaystyle dx=-frac{1}{t^2}dt$



      $displaystyle I(l)=int^{infty}_{0}frac{t^{4-l}}{2t^6+4t^5+3t^4+5t^3+3t^2+4t+2}dt=int^{infty}_{0}frac{x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



      $displaystyle 2I(l)=int^{infty}_{0}frac{x^l+x^{4-l}}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx$



      How do i solve it.Help me



      Thanks in advance






      definite-integrals




      definite-integrals






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      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 28 '18 at 6:35









      jackyjacky

      1,188715




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          1 Answer
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          $begingroup$

          Denote $P(x)$ the polynomial of denominator, $$I'(l)=int_0^inftyfrac{x^lln x}{P(x)}dx\
          =int_0^1frac{x^lln x}{P(x)}dx+int_1^inftyfrac{x^lln x}{P(x)}dx\
          =int_0^1frac{x^lln x}{P(x)}dx-int_0^1frac{x^{4-l}ln x}{P(x)}dxtext{ (Sub $xmapsto 1/x$)}\
          =int_0^1frac{(x^l-x^{4-l})ln x}{P(x)}dx$$
          $x^l-x^{4-l}>0$ if $0<l<2$,
          $x^l-x^{4-l}<0$ if $2<l<5$,
          $x^l-x^{4-l}=0$ if $l=2$.

          Notice $frac{ln x}{P(x)}<0$ if $0<x<1$, can you continue?

          Can you prove it is valid to differentiate under the integral sign?






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            active

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            2












            $begingroup$

            Denote $P(x)$ the polynomial of denominator, $$I'(l)=int_0^inftyfrac{x^lln x}{P(x)}dx\
            =int_0^1frac{x^lln x}{P(x)}dx+int_1^inftyfrac{x^lln x}{P(x)}dx\
            =int_0^1frac{x^lln x}{P(x)}dx-int_0^1frac{x^{4-l}ln x}{P(x)}dxtext{ (Sub $xmapsto 1/x$)}\
            =int_0^1frac{(x^l-x^{4-l})ln x}{P(x)}dx$$
            $x^l-x^{4-l}>0$ if $0<l<2$,
            $x^l-x^{4-l}<0$ if $2<l<5$,
            $x^l-x^{4-l}=0$ if $l=2$.

            Notice $frac{ln x}{P(x)}<0$ if $0<x<1$, can you continue?

            Can you prove it is valid to differentiate under the integral sign?






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Denote $P(x)$ the polynomial of denominator, $$I'(l)=int_0^inftyfrac{x^lln x}{P(x)}dx\
              =int_0^1frac{x^lln x}{P(x)}dx+int_1^inftyfrac{x^lln x}{P(x)}dx\
              =int_0^1frac{x^lln x}{P(x)}dx-int_0^1frac{x^{4-l}ln x}{P(x)}dxtext{ (Sub $xmapsto 1/x$)}\
              =int_0^1frac{(x^l-x^{4-l})ln x}{P(x)}dx$$
              $x^l-x^{4-l}>0$ if $0<l<2$,
              $x^l-x^{4-l}<0$ if $2<l<5$,
              $x^l-x^{4-l}=0$ if $l=2$.

              Notice $frac{ln x}{P(x)}<0$ if $0<x<1$, can you continue?

              Can you prove it is valid to differentiate under the integral sign?






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Denote $P(x)$ the polynomial of denominator, $$I'(l)=int_0^inftyfrac{x^lln x}{P(x)}dx\
                =int_0^1frac{x^lln x}{P(x)}dx+int_1^inftyfrac{x^lln x}{P(x)}dx\
                =int_0^1frac{x^lln x}{P(x)}dx-int_0^1frac{x^{4-l}ln x}{P(x)}dxtext{ (Sub $xmapsto 1/x$)}\
                =int_0^1frac{(x^l-x^{4-l})ln x}{P(x)}dx$$
                $x^l-x^{4-l}>0$ if $0<l<2$,
                $x^l-x^{4-l}<0$ if $2<l<5$,
                $x^l-x^{4-l}=0$ if $l=2$.

                Notice $frac{ln x}{P(x)}<0$ if $0<x<1$, can you continue?

                Can you prove it is valid to differentiate under the integral sign?






                share|cite|improve this answer









                $endgroup$



                Denote $P(x)$ the polynomial of denominator, $$I'(l)=int_0^inftyfrac{x^lln x}{P(x)}dx\
                =int_0^1frac{x^lln x}{P(x)}dx+int_1^inftyfrac{x^lln x}{P(x)}dx\
                =int_0^1frac{x^lln x}{P(x)}dx-int_0^1frac{x^{4-l}ln x}{P(x)}dxtext{ (Sub $xmapsto 1/x$)}\
                =int_0^1frac{(x^l-x^{4-l})ln x}{P(x)}dx$$
                $x^l-x^{4-l}>0$ if $0<l<2$,
                $x^l-x^{4-l}<0$ if $2<l<5$,
                $x^l-x^{4-l}=0$ if $l=2$.

                Notice $frac{ln x}{P(x)}<0$ if $0<x<1$, can you continue?

                Can you prove it is valid to differentiate under the integral sign?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '18 at 6:52









                Kemono ChenKemono Chen

                3,1991844




                3,1991844






























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