Need help with calculus, show that: $frac{d}{dc} left[ int_{c}^infty x f(x) dx right] = -cf(c)$












0












$begingroup$



Given that $f(x)$ is the pdf of a continuous RV and $f(x)$ is positive everywhere, show that:



$$frac{d}{dc} left[ int_{c}^infty x f(x) dx right] = -cf(c)$$




Apparently you can solve this directly using the FTC but I can't see it.



So instead I tried using integration by parts:



$u = x quad quad dv = f(x) dx$



$du = dx quad quad v = F(x)$



So integration by parts yields



$$frac{d}{dc} left[ xF(x) biggvert_c^infty - int_c^infty F(x) dx right]$$



$$=frac{d}{dc} left[ infty F(infty) - cF(c) - G(infty) + G(c)right]$$



where I'm calling $G$ the integral of $F$, and then taking the derivative with respect to $c$ I get



$$=-F(c) - cf(c) + F(c)$$



$$= -cf(c)$$



which in theory is the right answer but I feel like I broke many rules.



My questions:




  1. How to solve this directly using FTC?


  2. Can you point out what is wrong with my attempt?







calculus probability algebra-precalculus







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    Given that $f(x)$ is the pdf of a continuous RV and $f(x)$ is positive everywhere, show that:



    $$frac{d}{dc} left[ int_{c}^infty x f(x) dx right] = -cf(c)$$




    Apparently you can solve this directly using the FTC but I can't see it.



    So instead I tried using integration by parts:



    $u = x quad quad dv = f(x) dx$



    $du = dx quad quad v = F(x)$



    So integration by parts yields



    $$frac{d}{dc} left[ xF(x) biggvert_c^infty - int_c^infty F(x) dx right]$$



    $$=frac{d}{dc} left[ infty F(infty) - cF(c) - G(infty) + G(c)right]$$



    where I'm calling $G$ the integral of $F$, and then taking the derivative with respect to $c$ I get



    $$=-F(c) - cf(c) + F(c)$$



    $$= -cf(c)$$



    which in theory is the right answer but I feel like I broke many rules.



    My questions:




    1. How to solve this directly using FTC?


    2. Can you point out what is wrong with my attempt?







    calculus probability algebra-precalculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Given that $f(x)$ is the pdf of a continuous RV and $f(x)$ is positive everywhere, show that:



      $$frac{d}{dc} left[ int_{c}^infty x f(x) dx right] = -cf(c)$$




      Apparently you can solve this directly using the FTC but I can't see it.



      So instead I tried using integration by parts:



      $u = x quad quad dv = f(x) dx$



      $du = dx quad quad v = F(x)$



      So integration by parts yields



      $$frac{d}{dc} left[ xF(x) biggvert_c^infty - int_c^infty F(x) dx right]$$



      $$=frac{d}{dc} left[ infty F(infty) - cF(c) - G(infty) + G(c)right]$$



      where I'm calling $G$ the integral of $F$, and then taking the derivative with respect to $c$ I get



      $$=-F(c) - cf(c) + F(c)$$



      $$= -cf(c)$$



      which in theory is the right answer but I feel like I broke many rules.



      My questions:




      1. How to solve this directly using FTC?


      2. Can you point out what is wrong with my attempt?







      calculus probability algebra-precalculus







      share|cite|improve this question









      $endgroup$





      Given that $f(x)$ is the pdf of a continuous RV and $f(x)$ is positive everywhere, show that:



      $$frac{d}{dc} left[ int_{c}^infty x f(x) dx right] = -cf(c)$$




      Apparently you can solve this directly using the FTC but I can't see it.



      So instead I tried using integration by parts:



      $u = x quad quad dv = f(x) dx$



      $du = dx quad quad v = F(x)$



      So integration by parts yields



      $$frac{d}{dc} left[ xF(x) biggvert_c^infty - int_c^infty F(x) dx right]$$



      $$=frac{d}{dc} left[ infty F(infty) - cF(c) - G(infty) + G(c)right]$$



      where I'm calling $G$ the integral of $F$, and then taking the derivative with respect to $c$ I get



      $$=-F(c) - cf(c) + F(c)$$



      $$= -cf(c)$$



      which in theory is the right answer but I feel like I broke many rules.



      My questions:




      1. How to solve this directly using FTC?


      2. Can you point out what is wrong with my attempt?







      calculus probability algebra-precalculus




      calculus probability algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 13 at 20:36









      HJ_beginnerHJ_beginner

      8861415




      8861415






















          1 Answer
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          active

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          1












          $begingroup$

          You are right to feel breaking many rules!! Note that both $xF(x) biggvert_c^infty $ and $ int_c^infty F(x) dx$ diverge for bounded $c$ and the result becomes ambiguous and inarguable. To fix this, consider $F(x)-1$ instead of $F(x)$.



          It's easy to show that using FTC without bothering or caring about breaking laws. According to the definition of derivative we have:$$Ltriangleqfrac{d}{dc} left[ int_{c}^infty x f(x) dx right]{=lim_{hto 0}{int_{c+h}^infty x f(x) dx-int_{c}^infty x f(x) dxover h}\=lim_{hto 0}{-int_{c}^{c+h} x f(x) dxover h}}$$also note that$$-(c+h)int_{c}^{c+h} f(x) dxle -int_{c}^{c+h} x f(x) dxle -cint_{c}^{c+h} f(x) dx$$or by using the definition of CDF$$-(c+h)Big(F(c+h)-F(c)Big)le -int_{c}^{c+h} x f(x) dxle-cBig(F(c+h)-F(c)Big)$$from which we obtain$$lim_{hto 0}{-(c+h)Big(F(c+h)-F(c)Big)over h}le Lle lim_{hto 0}{-cBig(F(c+h)-F(c)Big)over h}$$since $f(c)=lim_{hto 0}{F(c+h)-F(c)over h}$ we obtain $$-cf(c)le Lle -cf(c)$$therefore $$L=-cf(c)$$ and the proof is complete.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much for your help. I appreciate the formal proof and will spend time to digest it. Is there also a shortcut way to solve the problem... for example by inspection? I got that impression from the solution guide.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:04






          • 1




            $begingroup$
            You're welcome. Well...May be after finding the limits, their value can be determined using L^Hopital's rule if we have already agreed to accept that :) That makes a shortcut, though we must make sure of ambiguities being fulfilled. I also suspect to Markov's inequality (en.wikipedia.org/wiki/Markov%27s_inequality) to make the proof easier. Oh..... and I don't mind to answer questions if there were anymore anytime.......
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:11










          • $begingroup$
            Thanks again for helping me (I am a newb). I find your reasoning understandable and logically satisfying. What about this... define $g(x) = xf(x)$ and $G$ to be the integral of $g$ and then we have $$L = frac{d}{dc} left[ G(infty) - G(c) right]= frac{d}{dc} (-G(c)) = -cf(c) $$ I know this is not rigorous like your answer but does this thought process technically work because of the FTC? Or is this still rule breaking because of divergence and ambiguousness that you mentioned? I think this does not work but want to confirm.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:17








          • 1




            $begingroup$
            I think that the function $G(x)$ may not necessarily converge to a real number as $x to infty$ (or if this is not true i.e. $G(x)$ finally converges, it should be proved though honestly it is not clear to me whether this is the case or not :) ) . If not, then $G(infty)$ is either $infty$ or undefined because of bad oscillations of $xf(x)$....
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:22










          • $begingroup$
            Thanks your reasoning helps me understand why that approach is tainted. Thanks again, you are doing God's work.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:26











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          active

          oldest

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          1












          $begingroup$

          You are right to feel breaking many rules!! Note that both $xF(x) biggvert_c^infty $ and $ int_c^infty F(x) dx$ diverge for bounded $c$ and the result becomes ambiguous and inarguable. To fix this, consider $F(x)-1$ instead of $F(x)$.



          It's easy to show that using FTC without bothering or caring about breaking laws. According to the definition of derivative we have:$$Ltriangleqfrac{d}{dc} left[ int_{c}^infty x f(x) dx right]{=lim_{hto 0}{int_{c+h}^infty x f(x) dx-int_{c}^infty x f(x) dxover h}\=lim_{hto 0}{-int_{c}^{c+h} x f(x) dxover h}}$$also note that$$-(c+h)int_{c}^{c+h} f(x) dxle -int_{c}^{c+h} x f(x) dxle -cint_{c}^{c+h} f(x) dx$$or by using the definition of CDF$$-(c+h)Big(F(c+h)-F(c)Big)le -int_{c}^{c+h} x f(x) dxle-cBig(F(c+h)-F(c)Big)$$from which we obtain$$lim_{hto 0}{-(c+h)Big(F(c+h)-F(c)Big)over h}le Lle lim_{hto 0}{-cBig(F(c+h)-F(c)Big)over h}$$since $f(c)=lim_{hto 0}{F(c+h)-F(c)over h}$ we obtain $$-cf(c)le Lle -cf(c)$$therefore $$L=-cf(c)$$ and the proof is complete.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much for your help. I appreciate the formal proof and will spend time to digest it. Is there also a shortcut way to solve the problem... for example by inspection? I got that impression from the solution guide.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:04






          • 1




            $begingroup$
            You're welcome. Well...May be after finding the limits, their value can be determined using L^Hopital's rule if we have already agreed to accept that :) That makes a shortcut, though we must make sure of ambiguities being fulfilled. I also suspect to Markov's inequality (en.wikipedia.org/wiki/Markov%27s_inequality) to make the proof easier. Oh..... and I don't mind to answer questions if there were anymore anytime.......
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:11










          • $begingroup$
            Thanks again for helping me (I am a newb). I find your reasoning understandable and logically satisfying. What about this... define $g(x) = xf(x)$ and $G$ to be the integral of $g$ and then we have $$L = frac{d}{dc} left[ G(infty) - G(c) right]= frac{d}{dc} (-G(c)) = -cf(c) $$ I know this is not rigorous like your answer but does this thought process technically work because of the FTC? Or is this still rule breaking because of divergence and ambiguousness that you mentioned? I think this does not work but want to confirm.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:17








          • 1




            $begingroup$
            I think that the function $G(x)$ may not necessarily converge to a real number as $x to infty$ (or if this is not true i.e. $G(x)$ finally converges, it should be proved though honestly it is not clear to me whether this is the case or not :) ) . If not, then $G(infty)$ is either $infty$ or undefined because of bad oscillations of $xf(x)$....
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:22










          • $begingroup$
            Thanks your reasoning helps me understand why that approach is tainted. Thanks again, you are doing God's work.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:26
















          1












          $begingroup$

          You are right to feel breaking many rules!! Note that both $xF(x) biggvert_c^infty $ and $ int_c^infty F(x) dx$ diverge for bounded $c$ and the result becomes ambiguous and inarguable. To fix this, consider $F(x)-1$ instead of $F(x)$.



          It's easy to show that using FTC without bothering or caring about breaking laws. According to the definition of derivative we have:$$Ltriangleqfrac{d}{dc} left[ int_{c}^infty x f(x) dx right]{=lim_{hto 0}{int_{c+h}^infty x f(x) dx-int_{c}^infty x f(x) dxover h}\=lim_{hto 0}{-int_{c}^{c+h} x f(x) dxover h}}$$also note that$$-(c+h)int_{c}^{c+h} f(x) dxle -int_{c}^{c+h} x f(x) dxle -cint_{c}^{c+h} f(x) dx$$or by using the definition of CDF$$-(c+h)Big(F(c+h)-F(c)Big)le -int_{c}^{c+h} x f(x) dxle-cBig(F(c+h)-F(c)Big)$$from which we obtain$$lim_{hto 0}{-(c+h)Big(F(c+h)-F(c)Big)over h}le Lle lim_{hto 0}{-cBig(F(c+h)-F(c)Big)over h}$$since $f(c)=lim_{hto 0}{F(c+h)-F(c)over h}$ we obtain $$-cf(c)le Lle -cf(c)$$therefore $$L=-cf(c)$$ and the proof is complete.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much for your help. I appreciate the formal proof and will spend time to digest it. Is there also a shortcut way to solve the problem... for example by inspection? I got that impression from the solution guide.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:04






          • 1




            $begingroup$
            You're welcome. Well...May be after finding the limits, their value can be determined using L^Hopital's rule if we have already agreed to accept that :) That makes a shortcut, though we must make sure of ambiguities being fulfilled. I also suspect to Markov's inequality (en.wikipedia.org/wiki/Markov%27s_inequality) to make the proof easier. Oh..... and I don't mind to answer questions if there were anymore anytime.......
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:11










          • $begingroup$
            Thanks again for helping me (I am a newb). I find your reasoning understandable and logically satisfying. What about this... define $g(x) = xf(x)$ and $G$ to be the integral of $g$ and then we have $$L = frac{d}{dc} left[ G(infty) - G(c) right]= frac{d}{dc} (-G(c)) = -cf(c) $$ I know this is not rigorous like your answer but does this thought process technically work because of the FTC? Or is this still rule breaking because of divergence and ambiguousness that you mentioned? I think this does not work but want to confirm.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:17








          • 1




            $begingroup$
            I think that the function $G(x)$ may not necessarily converge to a real number as $x to infty$ (or if this is not true i.e. $G(x)$ finally converges, it should be proved though honestly it is not clear to me whether this is the case or not :) ) . If not, then $G(infty)$ is either $infty$ or undefined because of bad oscillations of $xf(x)$....
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:22










          • $begingroup$
            Thanks your reasoning helps me understand why that approach is tainted. Thanks again, you are doing God's work.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:26














          1












          1








          1





          $begingroup$

          You are right to feel breaking many rules!! Note that both $xF(x) biggvert_c^infty $ and $ int_c^infty F(x) dx$ diverge for bounded $c$ and the result becomes ambiguous and inarguable. To fix this, consider $F(x)-1$ instead of $F(x)$.



          It's easy to show that using FTC without bothering or caring about breaking laws. According to the definition of derivative we have:$$Ltriangleqfrac{d}{dc} left[ int_{c}^infty x f(x) dx right]{=lim_{hto 0}{int_{c+h}^infty x f(x) dx-int_{c}^infty x f(x) dxover h}\=lim_{hto 0}{-int_{c}^{c+h} x f(x) dxover h}}$$also note that$$-(c+h)int_{c}^{c+h} f(x) dxle -int_{c}^{c+h} x f(x) dxle -cint_{c}^{c+h} f(x) dx$$or by using the definition of CDF$$-(c+h)Big(F(c+h)-F(c)Big)le -int_{c}^{c+h} x f(x) dxle-cBig(F(c+h)-F(c)Big)$$from which we obtain$$lim_{hto 0}{-(c+h)Big(F(c+h)-F(c)Big)over h}le Lle lim_{hto 0}{-cBig(F(c+h)-F(c)Big)over h}$$since $f(c)=lim_{hto 0}{F(c+h)-F(c)over h}$ we obtain $$-cf(c)le Lle -cf(c)$$therefore $$L=-cf(c)$$ and the proof is complete.






          share|cite|improve this answer









          $endgroup$



          You are right to feel breaking many rules!! Note that both $xF(x) biggvert_c^infty $ and $ int_c^infty F(x) dx$ diverge for bounded $c$ and the result becomes ambiguous and inarguable. To fix this, consider $F(x)-1$ instead of $F(x)$.



          It's easy to show that using FTC without bothering or caring about breaking laws. According to the definition of derivative we have:$$Ltriangleqfrac{d}{dc} left[ int_{c}^infty x f(x) dx right]{=lim_{hto 0}{int_{c+h}^infty x f(x) dx-int_{c}^infty x f(x) dxover h}\=lim_{hto 0}{-int_{c}^{c+h} x f(x) dxover h}}$$also note that$$-(c+h)int_{c}^{c+h} f(x) dxle -int_{c}^{c+h} x f(x) dxle -cint_{c}^{c+h} f(x) dx$$or by using the definition of CDF$$-(c+h)Big(F(c+h)-F(c)Big)le -int_{c}^{c+h} x f(x) dxle-cBig(F(c+h)-F(c)Big)$$from which we obtain$$lim_{hto 0}{-(c+h)Big(F(c+h)-F(c)Big)over h}le Lle lim_{hto 0}{-cBig(F(c+h)-F(c)Big)over h}$$since $f(c)=lim_{hto 0}{F(c+h)-F(c)over h}$ we obtain $$-cf(c)le Lle -cf(c)$$therefore $$L=-cf(c)$$ and the proof is complete.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 20:52









          Mostafa AyazMostafa Ayaz

          15.3k3939




          15.3k3939












          • $begingroup$
            Thanks so much for your help. I appreciate the formal proof and will spend time to digest it. Is there also a shortcut way to solve the problem... for example by inspection? I got that impression from the solution guide.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:04






          • 1




            $begingroup$
            You're welcome. Well...May be after finding the limits, their value can be determined using L^Hopital's rule if we have already agreed to accept that :) That makes a shortcut, though we must make sure of ambiguities being fulfilled. I also suspect to Markov's inequality (en.wikipedia.org/wiki/Markov%27s_inequality) to make the proof easier. Oh..... and I don't mind to answer questions if there were anymore anytime.......
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:11










          • $begingroup$
            Thanks again for helping me (I am a newb). I find your reasoning understandable and logically satisfying. What about this... define $g(x) = xf(x)$ and $G$ to be the integral of $g$ and then we have $$L = frac{d}{dc} left[ G(infty) - G(c) right]= frac{d}{dc} (-G(c)) = -cf(c) $$ I know this is not rigorous like your answer but does this thought process technically work because of the FTC? Or is this still rule breaking because of divergence and ambiguousness that you mentioned? I think this does not work but want to confirm.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:17








          • 1




            $begingroup$
            I think that the function $G(x)$ may not necessarily converge to a real number as $x to infty$ (or if this is not true i.e. $G(x)$ finally converges, it should be proved though honestly it is not clear to me whether this is the case or not :) ) . If not, then $G(infty)$ is either $infty$ or undefined because of bad oscillations of $xf(x)$....
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:22










          • $begingroup$
            Thanks your reasoning helps me understand why that approach is tainted. Thanks again, you are doing God's work.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:26


















          • $begingroup$
            Thanks so much for your help. I appreciate the formal proof and will spend time to digest it. Is there also a shortcut way to solve the problem... for example by inspection? I got that impression from the solution guide.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:04






          • 1




            $begingroup$
            You're welcome. Well...May be after finding the limits, their value can be determined using L^Hopital's rule if we have already agreed to accept that :) That makes a shortcut, though we must make sure of ambiguities being fulfilled. I also suspect to Markov's inequality (en.wikipedia.org/wiki/Markov%27s_inequality) to make the proof easier. Oh..... and I don't mind to answer questions if there were anymore anytime.......
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:11










          • $begingroup$
            Thanks again for helping me (I am a newb). I find your reasoning understandable and logically satisfying. What about this... define $g(x) = xf(x)$ and $G$ to be the integral of $g$ and then we have $$L = frac{d}{dc} left[ G(infty) - G(c) right]= frac{d}{dc} (-G(c)) = -cf(c) $$ I know this is not rigorous like your answer but does this thought process technically work because of the FTC? Or is this still rule breaking because of divergence and ambiguousness that you mentioned? I think this does not work but want to confirm.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:17








          • 1




            $begingroup$
            I think that the function $G(x)$ may not necessarily converge to a real number as $x to infty$ (or if this is not true i.e. $G(x)$ finally converges, it should be proved though honestly it is not clear to me whether this is the case or not :) ) . If not, then $G(infty)$ is either $infty$ or undefined because of bad oscillations of $xf(x)$....
            $endgroup$
            – Mostafa Ayaz
            Jan 13 at 21:22










          • $begingroup$
            Thanks your reasoning helps me understand why that approach is tainted. Thanks again, you are doing God's work.
            $endgroup$
            – HJ_beginner
            Jan 13 at 21:26
















          $begingroup$
          Thanks so much for your help. I appreciate the formal proof and will spend time to digest it. Is there also a shortcut way to solve the problem... for example by inspection? I got that impression from the solution guide.
          $endgroup$
          – HJ_beginner
          Jan 13 at 21:04




          $begingroup$
          Thanks so much for your help. I appreciate the formal proof and will spend time to digest it. Is there also a shortcut way to solve the problem... for example by inspection? I got that impression from the solution guide.
          $endgroup$
          – HJ_beginner
          Jan 13 at 21:04




          1




          1




          $begingroup$
          You're welcome. Well...May be after finding the limits, their value can be determined using L^Hopital's rule if we have already agreed to accept that :) That makes a shortcut, though we must make sure of ambiguities being fulfilled. I also suspect to Markov's inequality (en.wikipedia.org/wiki/Markov%27s_inequality) to make the proof easier. Oh..... and I don't mind to answer questions if there were anymore anytime.......
          $endgroup$
          – Mostafa Ayaz
          Jan 13 at 21:11




          $begingroup$
          You're welcome. Well...May be after finding the limits, their value can be determined using L^Hopital's rule if we have already agreed to accept that :) That makes a shortcut, though we must make sure of ambiguities being fulfilled. I also suspect to Markov's inequality (en.wikipedia.org/wiki/Markov%27s_inequality) to make the proof easier. Oh..... and I don't mind to answer questions if there were anymore anytime.......
          $endgroup$
          – Mostafa Ayaz
          Jan 13 at 21:11












          $begingroup$
          Thanks again for helping me (I am a newb). I find your reasoning understandable and logically satisfying. What about this... define $g(x) = xf(x)$ and $G$ to be the integral of $g$ and then we have $$L = frac{d}{dc} left[ G(infty) - G(c) right]= frac{d}{dc} (-G(c)) = -cf(c) $$ I know this is not rigorous like your answer but does this thought process technically work because of the FTC? Or is this still rule breaking because of divergence and ambiguousness that you mentioned? I think this does not work but want to confirm.
          $endgroup$
          – HJ_beginner
          Jan 13 at 21:17






          $begingroup$
          Thanks again for helping me (I am a newb). I find your reasoning understandable and logically satisfying. What about this... define $g(x) = xf(x)$ and $G$ to be the integral of $g$ and then we have $$L = frac{d}{dc} left[ G(infty) - G(c) right]= frac{d}{dc} (-G(c)) = -cf(c) $$ I know this is not rigorous like your answer but does this thought process technically work because of the FTC? Or is this still rule breaking because of divergence and ambiguousness that you mentioned? I think this does not work but want to confirm.
          $endgroup$
          – HJ_beginner
          Jan 13 at 21:17






          1




          1




          $begingroup$
          I think that the function $G(x)$ may not necessarily converge to a real number as $x to infty$ (or if this is not true i.e. $G(x)$ finally converges, it should be proved though honestly it is not clear to me whether this is the case or not :) ) . If not, then $G(infty)$ is either $infty$ or undefined because of bad oscillations of $xf(x)$....
          $endgroup$
          – Mostafa Ayaz
          Jan 13 at 21:22




          $begingroup$
          I think that the function $G(x)$ may not necessarily converge to a real number as $x to infty$ (or if this is not true i.e. $G(x)$ finally converges, it should be proved though honestly it is not clear to me whether this is the case or not :) ) . If not, then $G(infty)$ is either $infty$ or undefined because of bad oscillations of $xf(x)$....
          $endgroup$
          – Mostafa Ayaz
          Jan 13 at 21:22












          $begingroup$
          Thanks your reasoning helps me understand why that approach is tainted. Thanks again, you are doing God's work.
          $endgroup$
          – HJ_beginner
          Jan 13 at 21:26




          $begingroup$
          Thanks your reasoning helps me understand why that approach is tainted. Thanks again, you are doing God's work.
          $endgroup$
          – HJ_beginner
          Jan 13 at 21:26


















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