Show that there exists $T>0$ such that...












1












$begingroup$


Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function



$F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.



I calculated
$$lim_{trightarrow infty} F(t)=0.$$
Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$



It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.



Notice that



$$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
So, $f$ is decreasing.
Therefore, it suffices to show that there exists $T>0$ such that
$$f(T)<0.$$



Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$






real-analysis calculus analysis multivariable-calculus inequality







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function



    $F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.



    I calculated
    $$lim_{trightarrow infty} F(t)=0.$$
    Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$



    It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.



    Notice that



    $$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
    So, $f$ is decreasing.
    Therefore, it suffices to show that there exists $T>0$ such that
    $$f(T)<0.$$



    Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$






    real-analysis calculus analysis multivariable-calculus inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function



      $F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.



      I calculated
      $$lim_{trightarrow infty} F(t)=0.$$
      Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$



      It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.



      Notice that



      $$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
      So, $f$ is decreasing.
      Therefore, it suffices to show that there exists $T>0$ such that
      $$f(T)<0.$$



      Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$






      real-analysis calculus analysis multivariable-calculus inequality







      share|cite|improve this question











      $endgroup$




      Let $alphain,]0,1[$. I want to show that there exists $T>0$ such that the positive function



      $F(t):=e^{-t}int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi$ decays monotonically to zero.



      I calculated
      $$lim_{trightarrow infty} F(t)=0.$$
      Also, $$F^{prime}(t)=frac{f(t)}{e^t}.$$



      It remains then to show that there exists $T>0$ such that $f(t):=frac{e^{t}}{t^{1-alpha}}-int_{0}^{t}frac{e^{xi}}{xi^{1-alpha}}dxi<0$ for all $tgeq T$.



      Notice that



      $$f^{prime}(t)=-(1-alpha)frac{e^t}{t^{2-alpha}}<0.$$
      So, $f$ is decreasing.
      Therefore, it suffices to show that there exists $T>0$ such that
      $$f(T)<0.$$



      Testing the function $f$ numerically, it seems that there does exist $T=T(alpha)$ such that $f(T(alpha))<0$ for all $alphain,]0,1[.$






      real-analysis calculus analysis multivariable-calculus inequality




      real-analysis calculus analysis multivariable-calculus inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 11:02







      Medo

















      asked Dec 9 '18 at 4:55









      MedoMedo

      632214




      632214






















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          $begingroup$

          For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
            $endgroup$
            – Medo
            Dec 9 '18 at 16:49













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          $begingroup$

          For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
            $endgroup$
            – Medo
            Dec 9 '18 at 16:49


















          0












          $begingroup$

          For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
            $endgroup$
            – Medo
            Dec 9 '18 at 16:49
















          0












          0








          0





          $begingroup$

          For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.






          share|cite|improve this answer









          $endgroup$



          For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $tge 1$ the value $$f(t)=f(1)+int_{1}^t f’(s)dsle f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 12:40









          Alex RavskyAlex Ravsky

          40.4k32282




          40.4k32282








          • 1




            $begingroup$
            Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
            $endgroup$
            – Medo
            Dec 9 '18 at 16:49
















          • 1




            $begingroup$
            Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
            $endgroup$
            – Medo
            Dec 9 '18 at 16:49










          1




          1




          $begingroup$
          Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
          $endgroup$
          – Medo
          Dec 9 '18 at 16:49






          $begingroup$
          Great. But with a little care because $f^{primeprime}(t)= (1-alpha),frac{e^{t}}{t^{2-alpha}},(-1+frac{2-alpha}{t})$ which is negative precisely when $t>2-alpha$. Thank you for you smart answer.
          $endgroup$
          – Medo
          Dec 9 '18 at 16:49




















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