Prove that integral of $ 1-cosleft(frac{1}{x^2}right) $ is finite












1














I need to prove that
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$



My attempt:
$$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
Using L'Hôpital's rule I can show that:
$$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.



So I think that, there exist $N in mathbb{N} $ such that:
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$



but I'm not sure. I would appreciate any tips or hints.






calculus integration proof-verification







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    1














    I need to prove that
    $$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$



    My attempt:
    $$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
    Using L'Hôpital's rule I can show that:
    $$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
    Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.



    So I think that, there exist $N in mathbb{N} $ such that:
    $$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$



    but I'm not sure. I would appreciate any tips or hints.






    calculus integration proof-verification







    share|cite|improve this question



























      1












      1








      1







      I need to prove that
      $$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$



      My attempt:
      $$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
      Using L'Hôpital's rule I can show that:
      $$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
      Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.



      So I think that, there exist $N in mathbb{N} $ such that:
      $$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$



      but I'm not sure. I would appreciate any tips or hints.






      calculus integration proof-verification







      share|cite|improve this question















      I need to prove that
      $$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$



      My attempt:
      $$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
      Using L'Hôpital's rule I can show that:
      $$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
      Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.



      So I think that, there exist $N in mathbb{N} $ such that:
      $$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$



      but I'm not sure. I would appreciate any tips or hints.






      calculus integration proof-verification




      calculus integration proof-verification






      share|cite|improve this question















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      edited Dec 2 '18 at 14:34









      Larry

      1,7232823




      1,7232823










      asked Dec 2 '18 at 14:01









      Wywana

      535




      535






















          4 Answers
          4






          active

          oldest

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          2














          Use the substitution $1/x=t$, so the integral becomes
          $$
          int_0^{infty} frac{1-cos(t^2)}{t^2},dt
          $$
          Since
          $$
          lim_{tto0}frac{1-cos(t^2)}{t^2}=0
          $$
          convergence is not an issue at $0$. Since
          $$
          0le 1-cos(t^2)le 2
          $$
          we have that, for $tge1$,
          $$
          0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
          $$
          Now show convergence of
          $$
          int_1^infty frac{2}{t^2},dt
          $$






          share|cite|improve this answer





























            3














            $$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
            where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:



            $$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
            leads to $I=color{red}{sqrt{frac{pi}{2}}}$.






            share|cite|improve this answer





























              1














              $int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$






              share|cite|improve this answer





























                1














                One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get



                begin{multline}
                int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
                \ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
                end{multline}



                Since both integrals are of bounded functions over a finite range, they must be finite.






                share|cite|improve this answer





















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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  2














                  Use the substitution $1/x=t$, so the integral becomes
                  $$
                  int_0^{infty} frac{1-cos(t^2)}{t^2},dt
                  $$
                  Since
                  $$
                  lim_{tto0}frac{1-cos(t^2)}{t^2}=0
                  $$
                  convergence is not an issue at $0$. Since
                  $$
                  0le 1-cos(t^2)le 2
                  $$
                  we have that, for $tge1$,
                  $$
                  0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
                  $$
                  Now show convergence of
                  $$
                  int_1^infty frac{2}{t^2},dt
                  $$






                  share|cite|improve this answer


























                    2














                    Use the substitution $1/x=t$, so the integral becomes
                    $$
                    int_0^{infty} frac{1-cos(t^2)}{t^2},dt
                    $$
                    Since
                    $$
                    lim_{tto0}frac{1-cos(t^2)}{t^2}=0
                    $$
                    convergence is not an issue at $0$. Since
                    $$
                    0le 1-cos(t^2)le 2
                    $$
                    we have that, for $tge1$,
                    $$
                    0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
                    $$
                    Now show convergence of
                    $$
                    int_1^infty frac{2}{t^2},dt
                    $$






                    share|cite|improve this answer
























                      2












                      2








                      2






                      Use the substitution $1/x=t$, so the integral becomes
                      $$
                      int_0^{infty} frac{1-cos(t^2)}{t^2},dt
                      $$
                      Since
                      $$
                      lim_{tto0}frac{1-cos(t^2)}{t^2}=0
                      $$
                      convergence is not an issue at $0$. Since
                      $$
                      0le 1-cos(t^2)le 2
                      $$
                      we have that, for $tge1$,
                      $$
                      0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
                      $$
                      Now show convergence of
                      $$
                      int_1^infty frac{2}{t^2},dt
                      $$






                      share|cite|improve this answer












                      Use the substitution $1/x=t$, so the integral becomes
                      $$
                      int_0^{infty} frac{1-cos(t^2)}{t^2},dt
                      $$
                      Since
                      $$
                      lim_{tto0}frac{1-cos(t^2)}{t^2}=0
                      $$
                      convergence is not an issue at $0$. Since
                      $$
                      0le 1-cos(t^2)le 2
                      $$
                      we have that, for $tge1$,
                      $$
                      0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
                      $$
                      Now show convergence of
                      $$
                      int_1^infty frac{2}{t^2},dt
                      $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 2 '18 at 15:22









                      egreg

                      178k1484201




                      178k1484201























                          3














                          $$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
                          where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:



                          $$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
                          leads to $I=color{red}{sqrt{frac{pi}{2}}}$.






                          share|cite|improve this answer


























                            3














                            $$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
                            where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:



                            $$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
                            leads to $I=color{red}{sqrt{frac{pi}{2}}}$.






                            share|cite|improve this answer
























                              3












                              3








                              3






                              $$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
                              where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:



                              $$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
                              leads to $I=color{red}{sqrt{frac{pi}{2}}}$.






                              share|cite|improve this answer












                              $$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
                              where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:



                              $$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
                              leads to $I=color{red}{sqrt{frac{pi}{2}}}$.







                              share|cite|improve this answer












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                              share|cite|improve this answer










                              answered Dec 2 '18 at 17:24









                              Jack D'Aurizio

                              287k33280657




                              287k33280657























                                  1














                                  $int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$






                                  share|cite|improve this answer


























                                    1














                                    $int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$






                                    share|cite|improve this answer
























                                      1












                                      1








                                      1






                                      $int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$






                                      share|cite|improve this answer












                                      $int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$







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                                      answered Dec 2 '18 at 14:22









                                      John_Wick

                                      1,381111




                                      1,381111























                                          1














                                          One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get



                                          begin{multline}
                                          int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
                                          \ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
                                          end{multline}



                                          Since both integrals are of bounded functions over a finite range, they must be finite.






                                          share|cite|improve this answer


























                                            1














                                            One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get



                                            begin{multline}
                                            int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
                                            \ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
                                            end{multline}



                                            Since both integrals are of bounded functions over a finite range, they must be finite.






                                            share|cite|improve this answer
























                                              1












                                              1








                                              1






                                              One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get



                                              begin{multline}
                                              int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
                                              \ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
                                              end{multline}



                                              Since both integrals are of bounded functions over a finite range, they must be finite.






                                              share|cite|improve this answer












                                              One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get



                                              begin{multline}
                                              int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
                                              \ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
                                              end{multline}



                                              Since both integrals are of bounded functions over a finite range, they must be finite.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 2 '18 at 14:49









                                              eyeballfrog

                                              6,048629




                                              6,048629






























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                                                  Marschland

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                                                  Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si...
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                                                  Dieringhausen

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                                                  Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7 ...
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