Limit $lim_{ttoinfty}int_{[1,+infty)}e^{-tx}frac{sin x}{x} dlambda(x)$












0














$f(t)=int_{[1,+infty)}e^{-tx}frac{sin x}{x} dlambda(x)$. $tin [1,+infty)$




How can I argue that $lim_{ttoinfty}f(t)=int_{[1,+infty)}lim_{ttoinfty}e^{-tx}frac{sin x}{x} dlambda(x)$? (switching limit and integral)




I know that $|e^{-tx}frac{sin x}{x}|le e^{-x}$ for $tge1$ and $int_1^infty e^{-x}<infty$ so I thought about using dominated convergence theorem dct
but what is my sequence $f_n$ in this case?






real-analysis integration analysis







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  • In the notation of wikipedia we have $g=e^{-x},f=e^{-tx}frac{sin x}{x}$ but what is my $f_n$?
    – conrad
    Nov 25 '18 at 10:56










  • Not sure I fully understand where the problem is... You know that, for every $x$, $$left|frac{sin x}{x}right|leqslant 1$$ hence $$|f(t)|leqslantint_1^inftyleft|e^{-tx}frac{sin x}{x}right|dxleqslantint_1^infty e^{-tx}dxleqslantint_0^infty e^{-tx}dx=frac1tto0$$
    – Did
    Nov 25 '18 at 12:05


















0














$f(t)=int_{[1,+infty)}e^{-tx}frac{sin x}{x} dlambda(x)$. $tin [1,+infty)$




How can I argue that $lim_{ttoinfty}f(t)=int_{[1,+infty)}lim_{ttoinfty}e^{-tx}frac{sin x}{x} dlambda(x)$? (switching limit and integral)




I know that $|e^{-tx}frac{sin x}{x}|le e^{-x}$ for $tge1$ and $int_1^infty e^{-x}<infty$ so I thought about using dominated convergence theorem dct
but what is my sequence $f_n$ in this case?






real-analysis integration analysis







share|cite|improve this question
























  • In the notation of wikipedia we have $g=e^{-x},f=e^{-tx}frac{sin x}{x}$ but what is my $f_n$?
    – conrad
    Nov 25 '18 at 10:56










  • Not sure I fully understand where the problem is... You know that, for every $x$, $$left|frac{sin x}{x}right|leqslant 1$$ hence $$|f(t)|leqslantint_1^inftyleft|e^{-tx}frac{sin x}{x}right|dxleqslantint_1^infty e^{-tx}dxleqslantint_0^infty e^{-tx}dx=frac1tto0$$
    – Did
    Nov 25 '18 at 12:05
















0












0








0







$f(t)=int_{[1,+infty)}e^{-tx}frac{sin x}{x} dlambda(x)$. $tin [1,+infty)$




How can I argue that $lim_{ttoinfty}f(t)=int_{[1,+infty)}lim_{ttoinfty}e^{-tx}frac{sin x}{x} dlambda(x)$? (switching limit and integral)




I know that $|e^{-tx}frac{sin x}{x}|le e^{-x}$ for $tge1$ and $int_1^infty e^{-x}<infty$ so I thought about using dominated convergence theorem dct
but what is my sequence $f_n$ in this case?






real-analysis integration analysis







share|cite|improve this question















$f(t)=int_{[1,+infty)}e^{-tx}frac{sin x}{x} dlambda(x)$. $tin [1,+infty)$




How can I argue that $lim_{ttoinfty}f(t)=int_{[1,+infty)}lim_{ttoinfty}e^{-tx}frac{sin x}{x} dlambda(x)$? (switching limit and integral)




I know that $|e^{-tx}frac{sin x}{x}|le e^{-x}$ for $tge1$ and $int_1^infty e^{-x}<infty$ so I thought about using dominated convergence theorem dct
but what is my sequence $f_n$ in this case?






real-analysis integration analysis




real-analysis integration analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 25 '18 at 12:18









Martin Sleziak

44.7k7115270




44.7k7115270










asked Nov 25 '18 at 10:40









conrad

757




757












  • In the notation of wikipedia we have $g=e^{-x},f=e^{-tx}frac{sin x}{x}$ but what is my $f_n$?
    – conrad
    Nov 25 '18 at 10:56










  • Not sure I fully understand where the problem is... You know that, for every $x$, $$left|frac{sin x}{x}right|leqslant 1$$ hence $$|f(t)|leqslantint_1^inftyleft|e^{-tx}frac{sin x}{x}right|dxleqslantint_1^infty e^{-tx}dxleqslantint_0^infty e^{-tx}dx=frac1tto0$$
    – Did
    Nov 25 '18 at 12:05




















  • In the notation of wikipedia we have $g=e^{-x},f=e^{-tx}frac{sin x}{x}$ but what is my $f_n$?
    – conrad
    Nov 25 '18 at 10:56










  • Not sure I fully understand where the problem is... You know that, for every $x$, $$left|frac{sin x}{x}right|leqslant 1$$ hence $$|f(t)|leqslantint_1^inftyleft|e^{-tx}frac{sin x}{x}right|dxleqslantint_1^infty e^{-tx}dxleqslantint_0^infty e^{-tx}dx=frac1tto0$$
    – Did
    Nov 25 '18 at 12:05


















In the notation of wikipedia we have $g=e^{-x},f=e^{-tx}frac{sin x}{x}$ but what is my $f_n$?
– conrad
Nov 25 '18 at 10:56




In the notation of wikipedia we have $g=e^{-x},f=e^{-tx}frac{sin x}{x}$ but what is my $f_n$?
– conrad
Nov 25 '18 at 10:56












Not sure I fully understand where the problem is... You know that, for every $x$, $$left|frac{sin x}{x}right|leqslant 1$$ hence $$|f(t)|leqslantint_1^inftyleft|e^{-tx}frac{sin x}{x}right|dxleqslantint_1^infty e^{-tx}dxleqslantint_0^infty e^{-tx}dx=frac1tto0$$
– Did
Nov 25 '18 at 12:05






Not sure I fully understand where the problem is... You know that, for every $x$, $$left|frac{sin x}{x}right|leqslant 1$$ hence $$|f(t)|leqslantint_1^inftyleft|e^{-tx}frac{sin x}{x}right|dxleqslantint_1^infty e^{-tx}dxleqslantint_0^infty e^{-tx}dx=frac1tto0$$
– Did
Nov 25 '18 at 12:05












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Theorem:




A functional limit $lim_{xto a}f(x)$ converges to some limit $L$ if and only if for every sequence ${x_n}_{ninBbb N}$ such that $x_nneq a$ for all $ninBbb N$ and that converges to $a$ the sequence ${f(x_n)}_{ninBbb N}$ converges to $L$.




Hence



$$lim_{ttoinfty}int_X f(x,t), dx=Liff lim_{ntoinfty}int_X f(x,t_n), dx=Ltag1$$



for every sequence ${t_n}_{ninBbb N}$ that converges to infinity. If we set $f_n(x):=f(x,t_n)$ then we have the equivalent expression



$$lim_{ttoinfty}int_X f(x,t), dx=Lifflim_{ntoinfty}int_X f_n(x), dx=Ltag2$$



for every sequence ${t_n}_{ninBbb N}toinfty$. Now using the dominated convergence theorem it is easy to check that the RHS of $(2)$ converges to $L=0$ for any chosen sequence ${t_n}_{ninBbb N}toinfty$.






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    Theorem:




    A functional limit $lim_{xto a}f(x)$ converges to some limit $L$ if and only if for every sequence ${x_n}_{ninBbb N}$ such that $x_nneq a$ for all $ninBbb N$ and that converges to $a$ the sequence ${f(x_n)}_{ninBbb N}$ converges to $L$.




    Hence



    $$lim_{ttoinfty}int_X f(x,t), dx=Liff lim_{ntoinfty}int_X f(x,t_n), dx=Ltag1$$



    for every sequence ${t_n}_{ninBbb N}$ that converges to infinity. If we set $f_n(x):=f(x,t_n)$ then we have the equivalent expression



    $$lim_{ttoinfty}int_X f(x,t), dx=Lifflim_{ntoinfty}int_X f_n(x), dx=Ltag2$$



    for every sequence ${t_n}_{ninBbb N}toinfty$. Now using the dominated convergence theorem it is easy to check that the RHS of $(2)$ converges to $L=0$ for any chosen sequence ${t_n}_{ninBbb N}toinfty$.






    share|cite|improve this answer




























      0














      Theorem:




      A functional limit $lim_{xto a}f(x)$ converges to some limit $L$ if and only if for every sequence ${x_n}_{ninBbb N}$ such that $x_nneq a$ for all $ninBbb N$ and that converges to $a$ the sequence ${f(x_n)}_{ninBbb N}$ converges to $L$.




      Hence



      $$lim_{ttoinfty}int_X f(x,t), dx=Liff lim_{ntoinfty}int_X f(x,t_n), dx=Ltag1$$



      for every sequence ${t_n}_{ninBbb N}$ that converges to infinity. If we set $f_n(x):=f(x,t_n)$ then we have the equivalent expression



      $$lim_{ttoinfty}int_X f(x,t), dx=Lifflim_{ntoinfty}int_X f_n(x), dx=Ltag2$$



      for every sequence ${t_n}_{ninBbb N}toinfty$. Now using the dominated convergence theorem it is easy to check that the RHS of $(2)$ converges to $L=0$ for any chosen sequence ${t_n}_{ninBbb N}toinfty$.






      share|cite|improve this answer


























        0












        0








        0






        Theorem:




        A functional limit $lim_{xto a}f(x)$ converges to some limit $L$ if and only if for every sequence ${x_n}_{ninBbb N}$ such that $x_nneq a$ for all $ninBbb N$ and that converges to $a$ the sequence ${f(x_n)}_{ninBbb N}$ converges to $L$.




        Hence



        $$lim_{ttoinfty}int_X f(x,t), dx=Liff lim_{ntoinfty}int_X f(x,t_n), dx=Ltag1$$



        for every sequence ${t_n}_{ninBbb N}$ that converges to infinity. If we set $f_n(x):=f(x,t_n)$ then we have the equivalent expression



        $$lim_{ttoinfty}int_X f(x,t), dx=Lifflim_{ntoinfty}int_X f_n(x), dx=Ltag2$$



        for every sequence ${t_n}_{ninBbb N}toinfty$. Now using the dominated convergence theorem it is easy to check that the RHS of $(2)$ converges to $L=0$ for any chosen sequence ${t_n}_{ninBbb N}toinfty$.






        share|cite|improve this answer














        Theorem:




        A functional limit $lim_{xto a}f(x)$ converges to some limit $L$ if and only if for every sequence ${x_n}_{ninBbb N}$ such that $x_nneq a$ for all $ninBbb N$ and that converges to $a$ the sequence ${f(x_n)}_{ninBbb N}$ converges to $L$.




        Hence



        $$lim_{ttoinfty}int_X f(x,t), dx=Liff lim_{ntoinfty}int_X f(x,t_n), dx=Ltag1$$



        for every sequence ${t_n}_{ninBbb N}$ that converges to infinity. If we set $f_n(x):=f(x,t_n)$ then we have the equivalent expression



        $$lim_{ttoinfty}int_X f(x,t), dx=Lifflim_{ntoinfty}int_X f_n(x), dx=Ltag2$$



        for every sequence ${t_n}_{ninBbb N}toinfty$. Now using the dominated convergence theorem it is easy to check that the RHS of $(2)$ converges to $L=0$ for any chosen sequence ${t_n}_{ninBbb N}toinfty$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 '18 at 12:16

























        answered Nov 25 '18 at 12:01









        Masacroso

        12.9k41746




        12.9k41746






























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