Integral representation of a rational function












-3














I would like an integral representation for the function
$$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.



It must be a definite integral, with the range of integration independent of $x$.



My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.






real-analysis functions







share|cite|improve this question





























    -3














    I would like an integral representation for the function
    $$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
    that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.



    It must be a definite integral, with the range of integration independent of $x$.



    My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.






    real-analysis functions







    share|cite|improve this question



























      -3












      -3








      -3







      I would like an integral representation for the function
      $$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
      that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.



      It must be a definite integral, with the range of integration independent of $x$.



      My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.






      real-analysis functions







      share|cite|improve this question















      I would like an integral representation for the function
      $$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
      that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.



      It must be a definite integral, with the range of integration independent of $x$.



      My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.






      real-analysis functions




      real-analysis functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 '18 at 17:13

























      asked Nov 21 '18 at 20:21









      thedude

      670412




      670412






















          2 Answers
          2






          active

          oldest

          votes


















          2














          As an indefinite integral,



          $$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$



          works. As a definite integral,



          $$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$



          works.






          share|cite|improve this answer























          • An indefinite integral is not quite what I want, though.
            – thedude
            Nov 21 '18 at 20:27






          • 1




            Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
            – TrostAft
            Nov 21 '18 at 20:28












          • yes. @TrostAft is right. i updated my post with it for you.
            – Ekesh
            Nov 21 '18 at 20:34










          • thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
            – thedude
            Nov 21 '18 at 20:46



















          2














          Of course, there is the trivial relation



          $$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$



          For example,



          $$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$



          for any $k > -1$, or
          $$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$




          As a potentially less trivial answer, you can perform any invertible integral transform



          $$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$



          Then, we can express the original function in terms of the inverse transform



          $$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$



          As an example, using the inverse Laplace transform,



          $$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$



          we can write the function



          $$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$






          share|cite|improve this answer























          • Not helpful at all
            – thedude
            Nov 21 '18 at 20:51










          • It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
            – AlexanderJ93
            Nov 21 '18 at 20:52










          • @thedude I've updated the answer to include a less trivial solution
            – AlexanderJ93
            Nov 21 '18 at 21:03






          • 1




            Yes, that is much better, thank you
            – thedude
            Nov 21 '18 at 23:46











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          As an indefinite integral,



          $$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$



          works. As a definite integral,



          $$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$



          works.






          share|cite|improve this answer























          • An indefinite integral is not quite what I want, though.
            – thedude
            Nov 21 '18 at 20:27






          • 1




            Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
            – TrostAft
            Nov 21 '18 at 20:28












          • yes. @TrostAft is right. i updated my post with it for you.
            – Ekesh
            Nov 21 '18 at 20:34










          • thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
            – thedude
            Nov 21 '18 at 20:46
















          2














          As an indefinite integral,



          $$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$



          works. As a definite integral,



          $$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$



          works.






          share|cite|improve this answer























          • An indefinite integral is not quite what I want, though.
            – thedude
            Nov 21 '18 at 20:27






          • 1




            Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
            – TrostAft
            Nov 21 '18 at 20:28












          • yes. @TrostAft is right. i updated my post with it for you.
            – Ekesh
            Nov 21 '18 at 20:34










          • thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
            – thedude
            Nov 21 '18 at 20:46














          2












          2








          2






          As an indefinite integral,



          $$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$



          works. As a definite integral,



          $$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$



          works.






          share|cite|improve this answer














          As an indefinite integral,



          $$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$



          works. As a definite integral,



          $$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$



          works.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 '18 at 20:34

























          answered Nov 21 '18 at 20:25









          Ekesh

          5326




          5326












          • An indefinite integral is not quite what I want, though.
            – thedude
            Nov 21 '18 at 20:27






          • 1




            Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
            – TrostAft
            Nov 21 '18 at 20:28












          • yes. @TrostAft is right. i updated my post with it for you.
            – Ekesh
            Nov 21 '18 at 20:34










          • thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
            – thedude
            Nov 21 '18 at 20:46


















          • An indefinite integral is not quite what I want, though.
            – thedude
            Nov 21 '18 at 20:27






          • 1




            Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
            – TrostAft
            Nov 21 '18 at 20:28












          • yes. @TrostAft is right. i updated my post with it for you.
            – Ekesh
            Nov 21 '18 at 20:34










          • thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
            – thedude
            Nov 21 '18 at 20:46
















          An indefinite integral is not quite what I want, though.
          – thedude
          Nov 21 '18 at 20:27




          An indefinite integral is not quite what I want, though.
          – thedude
          Nov 21 '18 at 20:27




          1




          1




          Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
          – TrostAft
          Nov 21 '18 at 20:28






          Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
          – TrostAft
          Nov 21 '18 at 20:28














          yes. @TrostAft is right. i updated my post with it for you.
          – Ekesh
          Nov 21 '18 at 20:34




          yes. @TrostAft is right. i updated my post with it for you.
          – Ekesh
          Nov 21 '18 at 20:34












          thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
          – thedude
          Nov 21 '18 at 20:46




          thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
          – thedude
          Nov 21 '18 at 20:46











          2














          Of course, there is the trivial relation



          $$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$



          For example,



          $$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$



          for any $k > -1$, or
          $$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$




          As a potentially less trivial answer, you can perform any invertible integral transform



          $$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$



          Then, we can express the original function in terms of the inverse transform



          $$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$



          As an example, using the inverse Laplace transform,



          $$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$



          we can write the function



          $$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$






          share|cite|improve this answer























          • Not helpful at all
            – thedude
            Nov 21 '18 at 20:51










          • It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
            – AlexanderJ93
            Nov 21 '18 at 20:52










          • @thedude I've updated the answer to include a less trivial solution
            – AlexanderJ93
            Nov 21 '18 at 21:03






          • 1




            Yes, that is much better, thank you
            – thedude
            Nov 21 '18 at 23:46
















          2














          Of course, there is the trivial relation



          $$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$



          For example,



          $$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$



          for any $k > -1$, or
          $$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$




          As a potentially less trivial answer, you can perform any invertible integral transform



          $$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$



          Then, we can express the original function in terms of the inverse transform



          $$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$



          As an example, using the inverse Laplace transform,



          $$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$



          we can write the function



          $$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$






          share|cite|improve this answer























          • Not helpful at all
            – thedude
            Nov 21 '18 at 20:51










          • It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
            – AlexanderJ93
            Nov 21 '18 at 20:52










          • @thedude I've updated the answer to include a less trivial solution
            – AlexanderJ93
            Nov 21 '18 at 21:03






          • 1




            Yes, that is much better, thank you
            – thedude
            Nov 21 '18 at 23:46














          2












          2








          2






          Of course, there is the trivial relation



          $$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$



          For example,



          $$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$



          for any $k > -1$, or
          $$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$




          As a potentially less trivial answer, you can perform any invertible integral transform



          $$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$



          Then, we can express the original function in terms of the inverse transform



          $$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$



          As an example, using the inverse Laplace transform,



          $$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$



          we can write the function



          $$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$






          share|cite|improve this answer














          Of course, there is the trivial relation



          $$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$



          For example,



          $$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$



          for any $k > -1$, or
          $$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$




          As a potentially less trivial answer, you can perform any invertible integral transform



          $$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$



          Then, we can express the original function in terms of the inverse transform



          $$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$



          As an example, using the inverse Laplace transform,



          $$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$



          we can write the function



          $$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 '18 at 21:03

























          answered Nov 21 '18 at 20:46









          AlexanderJ93

          6,093823




          6,093823












          • Not helpful at all
            – thedude
            Nov 21 '18 at 20:51










          • It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
            – AlexanderJ93
            Nov 21 '18 at 20:52










          • @thedude I've updated the answer to include a less trivial solution
            – AlexanderJ93
            Nov 21 '18 at 21:03






          • 1




            Yes, that is much better, thank you
            – thedude
            Nov 21 '18 at 23:46


















          • Not helpful at all
            – thedude
            Nov 21 '18 at 20:51










          • It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
            – AlexanderJ93
            Nov 21 '18 at 20:52










          • @thedude I've updated the answer to include a less trivial solution
            – AlexanderJ93
            Nov 21 '18 at 21:03






          • 1




            Yes, that is much better, thank you
            – thedude
            Nov 21 '18 at 23:46
















          Not helpful at all
          – thedude
          Nov 21 '18 at 20:51




          Not helpful at all
          – thedude
          Nov 21 '18 at 20:51












          It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
          – AlexanderJ93
          Nov 21 '18 at 20:52




          It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
          – AlexanderJ93
          Nov 21 '18 at 20:52












          @thedude I've updated the answer to include a less trivial solution
          – AlexanderJ93
          Nov 21 '18 at 21:03




          @thedude I've updated the answer to include a less trivial solution
          – AlexanderJ93
          Nov 21 '18 at 21:03




          1




          1




          Yes, that is much better, thank you
          – thedude
          Nov 21 '18 at 23:46




          Yes, that is much better, thank you
          – thedude
          Nov 21 '18 at 23:46


















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