Improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable











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Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.




I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.






real-analysis measure-theory improper-integrals lebesgue-integral







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    Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
    – Barry Cipra
    Nov 22 at 15:14












  • @BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
    – D. Brito
    Nov 22 at 16:00












  • The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
    – Barry Cipra
    Nov 22 at 16:05












  • take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
    – Masacroso
    Nov 22 at 16:22

















up vote
3
down vote

favorite
1













Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.




I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.






real-analysis measure-theory improper-integrals lebesgue-integral







share|cite|improve this question




















  • 2




    Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
    – Barry Cipra
    Nov 22 at 15:14












  • @BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
    – D. Brito
    Nov 22 at 16:00












  • The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
    – Barry Cipra
    Nov 22 at 16:05












  • take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
    – Masacroso
    Nov 22 at 16:22















up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.




I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.






real-analysis measure-theory improper-integrals lebesgue-integral







share|cite|improve this question
















Show that the improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrable.




I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.






real-analysis measure-theory improper-integrals lebesgue-integral




real-analysis measure-theory improper-integrals lebesgue-integral






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edited Nov 23 at 15:22

























asked Nov 22 at 14:59









D. Brito

360111




360111








  • 2




    Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
    – Barry Cipra
    Nov 22 at 15:14












  • @BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
    – D. Brito
    Nov 22 at 16:00












  • The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
    – Barry Cipra
    Nov 22 at 16:05












  • take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
    – Masacroso
    Nov 22 at 16:22
















  • 2




    Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
    – Barry Cipra
    Nov 22 at 15:14












  • @BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
    – D. Brito
    Nov 22 at 16:00












  • The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
    – Barry Cipra
    Nov 22 at 16:05












  • take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
    – Masacroso
    Nov 22 at 16:22










2




2




Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14






Lebesgue integrability requires $int_0^infty|cos(x^2)|,dxltinfty$.
– Barry Cipra
Nov 22 at 15:14














@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00






@BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $sum_{n=0}^{infty} int_{sqrt{npi /2}}^{sqrt{(n+1)pi /2}} |cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it.
– D. Brito
Nov 22 at 16:00














The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05






The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|cos(x^2)|ge{1over2}$ has infinite Lebesgue measure.
– Barry Cipra
Nov 22 at 16:05














take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22






take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part
– Masacroso
Nov 22 at 16:22












2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










Convergence of the improper integral.



This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$



See this post for the value of the improper integral:
https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity



See also the following questions:

- Definite integral of $cos (x)/ sqrt{x}$?

- A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$



Lebesgue integrability.



Consider the integrals
$$
int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
$$
For the second one, note that
$$
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
$$
which implies that
$$
frac{2}{a_{k+1}}leq
int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
leqfrac{2}{a_k}
$$
where $a_k = sqrt{kpi+frac{pi}{2}}$. But
$$
sum frac{1}{a_k}=infty.
$$
So one must have
$$
int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
$$
and thus
$$
int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
$$






share|cite|improve this answer






























    up vote
    0
    down vote













    The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      Convergence of the improper integral.



      This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$



      See this post for the value of the improper integral:
      https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity



      See also the following questions:

      - Definite integral of $cos (x)/ sqrt{x}$?

      - A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$



      Lebesgue integrability.



      Consider the integrals
      $$
      int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
      int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
      $$
      For the second one, note that
      $$
      int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
      $$
      which implies that
      $$
      frac{2}{a_{k+1}}leq
      int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
      leqfrac{2}{a_k}
      $$
      where $a_k = sqrt{kpi+frac{pi}{2}}$. But
      $$
      sum frac{1}{a_k}=infty.
      $$
      So one must have
      $$
      int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
      $$
      and thus
      $$
      int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
      $$






      share|cite|improve this answer



























        up vote
        4
        down vote



        accepted










        Convergence of the improper integral.



        This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$



        See this post for the value of the improper integral:
        https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity



        See also the following questions:

        - Definite integral of $cos (x)/ sqrt{x}$?

        - A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$



        Lebesgue integrability.



        Consider the integrals
        $$
        int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
        int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
        $$
        For the second one, note that
        $$
        int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
        $$
        which implies that
        $$
        frac{2}{a_{k+1}}leq
        int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
        leqfrac{2}{a_k}
        $$
        where $a_k = sqrt{kpi+frac{pi}{2}}$. But
        $$
        sum frac{1}{a_k}=infty.
        $$
        So one must have
        $$
        int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
        $$
        and thus
        $$
        int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
        $$






        share|cite|improve this answer

























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Convergence of the improper integral.



          This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$



          See this post for the value of the improper integral:
          https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity



          See also the following questions:

          - Definite integral of $cos (x)/ sqrt{x}$?

          - A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$



          Lebesgue integrability.



          Consider the integrals
          $$
          int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
          int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
          $$
          For the second one, note that
          $$
          int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
          $$
          which implies that
          $$
          frac{2}{a_{k+1}}leq
          int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
          leqfrac{2}{a_k}
          $$
          where $a_k = sqrt{kpi+frac{pi}{2}}$. But
          $$
          sum frac{1}{a_k}=infty.
          $$
          So one must have
          $$
          int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
          $$
          and thus
          $$
          int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
          $$






          share|cite|improve this answer














          Convergence of the improper integral.



          This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$



          See this post for the value of the improper integral:
          https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity



          See also the following questions:

          - Definite integral of $cos (x)/ sqrt{x}$?

          - A simple proof of the fact that $int_0^{+infty} cos(x)/sqrt{x} text{d}x neq 0$



          Lebesgue integrability.



          Consider the integrals
          $$
          int_0^inftyleftvertfrac{cos u}{sqrt{u}}rightvert du=int_0^{pi/2}leftvertfrac{cos(u)}{u}rightvert du+
          int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du.
          $$
          For the second one, note that
          $$
          int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}|cos u| du = 2,
          $$
          which implies that
          $$
          frac{2}{a_{k+1}}leq
          int_{kpi+frac{pi}{2}}^{(k+1)pi+frac{pi}{2}}leftvertfrac{cos u}{sqrt{u}}rightvert du
          leqfrac{2}{a_k}
          $$
          where $a_k = sqrt{kpi+frac{pi}{2}}$. But
          $$
          sum frac{1}{a_k}=infty.
          $$
          So one must have
          $$
          int_{pi/2}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty.
          $$
          and thus
          $$
          int_{0}^inftyleftvertfrac{cos(u)}{u}rightvert du=infty
          $$







          share|cite|improve this answer














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          share|cite|improve this answer








          edited Nov 22 at 15:52

























          answered Nov 22 at 15:36









          user587192

          1,473112




          1,473112






















              up vote
              0
              down vote













              The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.






              share|cite|improve this answer

























                up vote
                0
                down vote













                The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.






                  share|cite|improve this answer












                  The convergence of $int_{0}^{+infty}frac{cos u}{sqrt{u}},du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $cos u$ has a bounded primitive and $frac{1}{sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 20:57









                  Jack D'Aurizio

                  285k33275654




                  285k33275654






























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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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