Dominated convergence theorem on $F(x) = int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt$












2












$begingroup$


Consider the function:
$$F(x) = int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt$$
Here $x in mathbb{R}^{+}$. I want to use the dominated convergence theorem to show that:
$$lim_{x to infty} F(x) = lim_{x to infty} int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt = int_{0}^{infty} lim_{x to infty} e^{-xt}frac{sin(t)}{t}dt = 0$$



So to get the limit in the integral, there are $2$ conditions. First, the sequence of integrable functions $f_x(t) = e^{-xt}frac{sin(t)}{t} $ needs to converge to $f(t) = 0$ for $x to infty$, which is satisfied. Secondly, you need to estimate the function $f_x(t)$ from above so that:
$$|f_x(t)| leq g(t)$$
Here $g: mathbb{R} to [0, infty]$ is a integrable function. In this case we get:
$$|f_x(t)| = |e^{-xt}frac{sin(t)}{t}| leq |e^{-xt}| leq g(t) $$
The problem is, is that I can't find a function $g(t)$ which fulfills this condition and is independent of $x$. If such a $g(t)$ exists, then we can use the dominated convergence theorem.






real-analysis convergence







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Consider the function:
    $$F(x) = int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt$$
    Here $x in mathbb{R}^{+}$. I want to use the dominated convergence theorem to show that:
    $$lim_{x to infty} F(x) = lim_{x to infty} int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt = int_{0}^{infty} lim_{x to infty} e^{-xt}frac{sin(t)}{t}dt = 0$$



    So to get the limit in the integral, there are $2$ conditions. First, the sequence of integrable functions $f_x(t) = e^{-xt}frac{sin(t)}{t} $ needs to converge to $f(t) = 0$ for $x to infty$, which is satisfied. Secondly, you need to estimate the function $f_x(t)$ from above so that:
    $$|f_x(t)| leq g(t)$$
    Here $g: mathbb{R} to [0, infty]$ is a integrable function. In this case we get:
    $$|f_x(t)| = |e^{-xt}frac{sin(t)}{t}| leq |e^{-xt}| leq g(t) $$
    The problem is, is that I can't find a function $g(t)$ which fulfills this condition and is independent of $x$. If such a $g(t)$ exists, then we can use the dominated convergence theorem.






    real-analysis convergence







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Consider the function:
      $$F(x) = int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt$$
      Here $x in mathbb{R}^{+}$. I want to use the dominated convergence theorem to show that:
      $$lim_{x to infty} F(x) = lim_{x to infty} int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt = int_{0}^{infty} lim_{x to infty} e^{-xt}frac{sin(t)}{t}dt = 0$$



      So to get the limit in the integral, there are $2$ conditions. First, the sequence of integrable functions $f_x(t) = e^{-xt}frac{sin(t)}{t} $ needs to converge to $f(t) = 0$ for $x to infty$, which is satisfied. Secondly, you need to estimate the function $f_x(t)$ from above so that:
      $$|f_x(t)| leq g(t)$$
      Here $g: mathbb{R} to [0, infty]$ is a integrable function. In this case we get:
      $$|f_x(t)| = |e^{-xt}frac{sin(t)}{t}| leq |e^{-xt}| leq g(t) $$
      The problem is, is that I can't find a function $g(t)$ which fulfills this condition and is independent of $x$. If such a $g(t)$ exists, then we can use the dominated convergence theorem.






      real-analysis convergence







      share|cite|improve this question









      $endgroup$




      Consider the function:
      $$F(x) = int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt$$
      Here $x in mathbb{R}^{+}$. I want to use the dominated convergence theorem to show that:
      $$lim_{x to infty} F(x) = lim_{x to infty} int_{0}^{infty} e^{-xt}frac{sin(t)}{t}dt = int_{0}^{infty} lim_{x to infty} e^{-xt}frac{sin(t)}{t}dt = 0$$



      So to get the limit in the integral, there are $2$ conditions. First, the sequence of integrable functions $f_x(t) = e^{-xt}frac{sin(t)}{t} $ needs to converge to $f(t) = 0$ for $x to infty$, which is satisfied. Secondly, you need to estimate the function $f_x(t)$ from above so that:
      $$|f_x(t)| leq g(t)$$
      Here $g: mathbb{R} to [0, infty]$ is a integrable function. In this case we get:
      $$|f_x(t)| = |e^{-xt}frac{sin(t)}{t}| leq |e^{-xt}| leq g(t) $$
      The problem is, is that I can't find a function $g(t)$ which fulfills this condition and is independent of $x$. If such a $g(t)$ exists, then we can use the dominated convergence theorem.






      real-analysis convergence




      real-analysis convergence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 23 at 16:32









      Belgium_PhysicsBelgium_Physics

      325110




      325110






















          1 Answer
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          active

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          1












          $begingroup$

          Hint: Since we are taking limit as $xto infty$, we can assume that $xge 1$. Observe that
          $$
          e^{-xt}leq e^{-t}
          $$           for all $xgeq 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see this is true, but I thought that the condition for the estimation needed to be for every x>0
            $endgroup$
            – Belgium_Physics
            Jan 23 at 16:50










          • $begingroup$
            @Belgium_Physics And in fact, since $suplimits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$.
            $endgroup$
            – Song
            Jan 23 at 16:52












          • $begingroup$
            @Belgium_Physics Then you use DCT on the set, say, $[pi/2, +infty)$ and you simply compute $lim_{xto +infty} vert int_0^{pi/2} mathrm e^{-xt} sin t cdot t^{-1} mathrm dtvert $ by basic estimates.
            $endgroup$
            – xbh
            Jan 23 at 17:05











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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint: Since we are taking limit as $xto infty$, we can assume that $xge 1$. Observe that
          $$
          e^{-xt}leq e^{-t}
          $$           for all $xgeq 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see this is true, but I thought that the condition for the estimation needed to be for every x>0
            $endgroup$
            – Belgium_Physics
            Jan 23 at 16:50










          • $begingroup$
            @Belgium_Physics And in fact, since $suplimits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$.
            $endgroup$
            – Song
            Jan 23 at 16:52












          • $begingroup$
            @Belgium_Physics Then you use DCT on the set, say, $[pi/2, +infty)$ and you simply compute $lim_{xto +infty} vert int_0^{pi/2} mathrm e^{-xt} sin t cdot t^{-1} mathrm dtvert $ by basic estimates.
            $endgroup$
            – xbh
            Jan 23 at 17:05
















          1












          $begingroup$

          Hint: Since we are taking limit as $xto infty$, we can assume that $xge 1$. Observe that
          $$
          e^{-xt}leq e^{-t}
          $$           for all $xgeq 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see this is true, but I thought that the condition for the estimation needed to be for every x>0
            $endgroup$
            – Belgium_Physics
            Jan 23 at 16:50










          • $begingroup$
            @Belgium_Physics And in fact, since $suplimits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$.
            $endgroup$
            – Song
            Jan 23 at 16:52












          • $begingroup$
            @Belgium_Physics Then you use DCT on the set, say, $[pi/2, +infty)$ and you simply compute $lim_{xto +infty} vert int_0^{pi/2} mathrm e^{-xt} sin t cdot t^{-1} mathrm dtvert $ by basic estimates.
            $endgroup$
            – xbh
            Jan 23 at 17:05














          1












          1








          1





          $begingroup$

          Hint: Since we are taking limit as $xto infty$, we can assume that $xge 1$. Observe that
          $$
          e^{-xt}leq e^{-t}
          $$           for all $xgeq 1$.






          share|cite|improve this answer









          $endgroup$



          Hint: Since we are taking limit as $xto infty$, we can assume that $xge 1$. Observe that
          $$
          e^{-xt}leq e^{-t}
          $$           for all $xgeq 1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 16:42









          SongSong

          16.5k1741




          16.5k1741












          • $begingroup$
            I see this is true, but I thought that the condition for the estimation needed to be for every x>0
            $endgroup$
            – Belgium_Physics
            Jan 23 at 16:50










          • $begingroup$
            @Belgium_Physics And in fact, since $suplimits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$.
            $endgroup$
            – Song
            Jan 23 at 16:52












          • $begingroup$
            @Belgium_Physics Then you use DCT on the set, say, $[pi/2, +infty)$ and you simply compute $lim_{xto +infty} vert int_0^{pi/2} mathrm e^{-xt} sin t cdot t^{-1} mathrm dtvert $ by basic estimates.
            $endgroup$
            – xbh
            Jan 23 at 17:05


















          • $begingroup$
            I see this is true, but I thought that the condition for the estimation needed to be for every x>0
            $endgroup$
            – Belgium_Physics
            Jan 23 at 16:50










          • $begingroup$
            @Belgium_Physics And in fact, since $suplimits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$.
            $endgroup$
            – Song
            Jan 23 at 16:52












          • $begingroup$
            @Belgium_Physics Then you use DCT on the set, say, $[pi/2, +infty)$ and you simply compute $lim_{xto +infty} vert int_0^{pi/2} mathrm e^{-xt} sin t cdot t^{-1} mathrm dtvert $ by basic estimates.
            $endgroup$
            – xbh
            Jan 23 at 17:05
















          $begingroup$
          I see this is true, but I thought that the condition for the estimation needed to be for every x>0
          $endgroup$
          – Belgium_Physics
          Jan 23 at 16:50




          $begingroup$
          I see this is true, but I thought that the condition for the estimation needed to be for every x>0
          $endgroup$
          – Belgium_Physics
          Jan 23 at 16:50












          $begingroup$
          @Belgium_Physics And in fact, since $suplimits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$.
          $endgroup$
          – Song
          Jan 23 at 16:52






          $begingroup$
          @Belgium_Physics And in fact, since $suplimits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$.
          $endgroup$
          – Song
          Jan 23 at 16:52














          $begingroup$
          @Belgium_Physics Then you use DCT on the set, say, $[pi/2, +infty)$ and you simply compute $lim_{xto +infty} vert int_0^{pi/2} mathrm e^{-xt} sin t cdot t^{-1} mathrm dtvert $ by basic estimates.
          $endgroup$
          – xbh
          Jan 23 at 17:05




          $begingroup$
          @Belgium_Physics Then you use DCT on the set, say, $[pi/2, +infty)$ and you simply compute $lim_{xto +infty} vert int_0^{pi/2} mathrm e^{-xt} sin t cdot t^{-1} mathrm dtvert $ by basic estimates.
          $endgroup$
          – xbh
          Jan 23 at 17:05


















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