Denote $f(x)=int_x^{x+1}cos t^2 {rm d}t.$Prove $limlimits_{x to +infty}f(x)=0.$












3












$begingroup$


Problem



Denote $$f(x)=int_x^{x+1}cos t^2 {rm d}t.$$Prove $limlimits_{x to +infty}f(x)=0.$



Proof



Assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
begin{align*}
f(x)&=int_x^{x+1}cos t^2 {rm d}t\
&=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
&=frac{1}{2sqrt{xi}}int_{x^2}^{(x+1)^2}cos u{rm d}u\
&=frac{sin(x+1)^2-sin x^2}{2sqrt{xi}},
end{align*}
where $x^2 leq xileq (x+1)^2$. Further,
$$0leq |f(x)|leq frac{|sin(x+1)^2|+|sin x^2|}{2sqrt{xi}}leq frac{1}{sqrt{xi}}leq frac{1}{x}to 0(x to +infty),$$
which implies $f(x)to 0(x to +infty)$, according to the squeeze theorem.






proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    The equality $f(x)=...=frac{1}{sqrtxi}int...$ should be $leq$.
    $endgroup$
    – Surb
    Dec 26 '18 at 9:34








  • 3




    $begingroup$
    The proof is not right, because $cos u$ under integral does not save sign
    $endgroup$
    – Minz
    Dec 26 '18 at 9:40










  • $begingroup$
    @MinzMin yes，you're right！ but can we add a absolute value mark from the beginning？
    $endgroup$
    – mengdie1982
    Dec 26 '18 at 9:45










  • $begingroup$
    @mengdie1982 OK it changes things
    $endgroup$
    – Minz
    Dec 26 '18 at 9:57
















3












$begingroup$


Problem



Denote $$f(x)=int_x^{x+1}cos t^2 {rm d}t.$$Prove $limlimits_{x to +infty}f(x)=0.$



Proof



Assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
begin{align*}
f(x)&=int_x^{x+1}cos t^2 {rm d}t\
&=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
&=frac{1}{2sqrt{xi}}int_{x^2}^{(x+1)^2}cos u{rm d}u\
&=frac{sin(x+1)^2-sin x^2}{2sqrt{xi}},
end{align*}
where $x^2 leq xileq (x+1)^2$. Further,
$$0leq |f(x)|leq frac{|sin(x+1)^2|+|sin x^2|}{2sqrt{xi}}leq frac{1}{sqrt{xi}}leq frac{1}{x}to 0(x to +infty),$$
which implies $f(x)to 0(x to +infty)$, according to the squeeze theorem.






proof-verification







share|cite|improve this question









$endgroup$












  • $begingroup$
    The equality $f(x)=...=frac{1}{sqrtxi}int...$ should be $leq$.
    $endgroup$
    – Surb
    Dec 26 '18 at 9:34








  • 3




    $begingroup$
    The proof is not right, because $cos u$ under integral does not save sign
    $endgroup$
    – Minz
    Dec 26 '18 at 9:40










  • $begingroup$
    @MinzMin yes，you're right！ but can we add a absolute value mark from the beginning？
    $endgroup$
    – mengdie1982
    Dec 26 '18 at 9:45










  • $begingroup$
    @mengdie1982 OK it changes things
    $endgroup$
    – Minz
    Dec 26 '18 at 9:57














3












3








3


1



$begingroup$


Problem



Denote $$f(x)=int_x^{x+1}cos t^2 {rm d}t.$$Prove $limlimits_{x to +infty}f(x)=0.$



Proof



Assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
begin{align*}
f(x)&=int_x^{x+1}cos t^2 {rm d}t\
&=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
&=frac{1}{2sqrt{xi}}int_{x^2}^{(x+1)^2}cos u{rm d}u\
&=frac{sin(x+1)^2-sin x^2}{2sqrt{xi}},
end{align*}
where $x^2 leq xileq (x+1)^2$. Further,
$$0leq |f(x)|leq frac{|sin(x+1)^2|+|sin x^2|}{2sqrt{xi}}leq frac{1}{sqrt{xi}}leq frac{1}{x}to 0(x to +infty),$$
which implies $f(x)to 0(x to +infty)$, according to the squeeze theorem.






proof-verification







share|cite|improve this question









$endgroup$




Problem



Denote $$f(x)=int_x^{x+1}cos t^2 {rm d}t.$$Prove $limlimits_{x to +infty}f(x)=0.$



Proof



Assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
begin{align*}
f(x)&=int_x^{x+1}cos t^2 {rm d}t\
&=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
&=frac{1}{2sqrt{xi}}int_{x^2}^{(x+1)^2}cos u{rm d}u\
&=frac{sin(x+1)^2-sin x^2}{2sqrt{xi}},
end{align*}
where $x^2 leq xileq (x+1)^2$. Further,
$$0leq |f(x)|leq frac{|sin(x+1)^2|+|sin x^2|}{2sqrt{xi}}leq frac{1}{sqrt{xi}}leq frac{1}{x}to 0(x to +infty),$$
which implies $f(x)to 0(x to +infty)$, according to the squeeze theorem.






proof-verification




proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 9:06









mengdie1982mengdie1982

4,974618




4,974618












  • $begingroup$
    The equality $f(x)=...=frac{1}{sqrtxi}int...$ should be $leq$.
    $endgroup$
    – Surb
    Dec 26 '18 at 9:34








  • 3




    $begingroup$
    The proof is not right, because $cos u$ under integral does not save sign
    $endgroup$
    – Minz
    Dec 26 '18 at 9:40










  • $begingroup$
    @MinzMin yes，you're right！ but can we add a absolute value mark from the beginning？
    $endgroup$
    – mengdie1982
    Dec 26 '18 at 9:45










  • $begingroup$
    @mengdie1982 OK it changes things
    $endgroup$
    – Minz
    Dec 26 '18 at 9:57


















  • $begingroup$
    The equality $f(x)=...=frac{1}{sqrtxi}int...$ should be $leq$.
    $endgroup$
    – Surb
    Dec 26 '18 at 9:34








  • 3




    $begingroup$
    The proof is not right, because $cos u$ under integral does not save sign
    $endgroup$
    – Minz
    Dec 26 '18 at 9:40










  • $begingroup$
    @MinzMin yes，you're right！ but can we add a absolute value mark from the beginning？
    $endgroup$
    – mengdie1982
    Dec 26 '18 at 9:45










  • $begingroup$
    @mengdie1982 OK it changes things
    $endgroup$
    – Minz
    Dec 26 '18 at 9:57
















$begingroup$
The equality $f(x)=...=frac{1}{sqrtxi}int...$ should be $leq$.
$endgroup$
– Surb
Dec 26 '18 at 9:34






$begingroup$
The equality $f(x)=...=frac{1}{sqrtxi}int...$ should be $leq$.
$endgroup$
– Surb
Dec 26 '18 at 9:34






3




3




$begingroup$
The proof is not right, because $cos u$ under integral does not save sign
$endgroup$
– Minz
Dec 26 '18 at 9:40




$begingroup$
The proof is not right, because $cos u$ under integral does not save sign
$endgroup$
– Minz
Dec 26 '18 at 9:40












$begingroup$
@MinzMin yes，you're right！ but can we add a absolute value mark from the beginning？
$endgroup$
– mengdie1982
Dec 26 '18 at 9:45




$begingroup$
@MinzMin yes，you're right！ but can we add a absolute value mark from the beginning？
$endgroup$
– mengdie1982
Dec 26 '18 at 9:45












$begingroup$
@mengdie1982 OK it changes things
$endgroup$
– Minz
Dec 26 '18 at 9:57




$begingroup$
@mengdie1982 OK it changes things
$endgroup$
– Minz
Dec 26 '18 at 9:57










2 Answers
2






active

oldest

votes


















1












$begingroup$

Alternative solution:

Denote $$g(x)=int_x^infty cos t^2dt (x>0),$$
we can actually prove this definition is valid (integral converges) using substitution $t^2=u$ and Dirichlet's test.

According to the definition of improper integral: $$lim_{ytoinfty}int_x^ycos t^2dt=int_x^inftycos t^2dt,$$
we can deduce $$lim_{xtoinfty}g(x)=int_a^inftycos t^2dt-lim_{xtoinfty}int_a^xcos t^2dt=int_a^inftycos t^2dt-int_a^inftycos t^2dt=0.$$
Therefore $lim_{xtoinfty}f(x)=lim_{xtoinfty}g(x+1)-g(x)=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't get it: it seems to be that for you it is obvious the left limit in the middle $;(int_x^y;)$ is zero. Why? What I mean: from what exactly you can deduce that $;lim g(x)=0;$ ?
    $endgroup$
    – DonAntonio
    Dec 26 '18 at 10:02












  • $begingroup$
    @DonAntonio The remainder (tail) of a convergent improper integral goes to $0$. That's a well-known result that Kemono re-proved.
    $endgroup$
    – Gabriel Romon
    Dec 26 '18 at 11:10





















1












$begingroup$

Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.



Another Proof



Likewise, assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
begin{align*}
f(x)&=int_x^{x+1}cos t^2 {rm d}t\
&=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
&=int_{x^2}^{(x+1)^2}frac{1}{2sqrt{u}}{rm d}(sin u)\
&=frac{sin u}{2sqrt{u}}bigg|_{x^2}^{(x+1)^2}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u\
&=frac{sin(x+1)^2}{2(x+1)}-frac{sin x^2}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u.\
end{align*}
Thus
begin{align*}
|f(x)|&leq frac{|sin(x+1)^2|}{2(x+1)}+frac{|sin x^2|}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}|sin u|{rm d}u\
&leq frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}{rm d}u\
&=frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}cdot left[-frac{2}{sqrt{u}}right]_{x^2}^{(x+1)^2}\
&=frac{1}{x}to 0(x to +infty),
end{align*}
which implies $|f(x)|to 0$ according to the squeeze theorem. It follows that $f(x) to 0$.






share|cite|improve this answer











$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Alternative solution:

    Denote $$g(x)=int_x^infty cos t^2dt (x>0),$$
    we can actually prove this definition is valid (integral converges) using substitution $t^2=u$ and Dirichlet's test.

    According to the definition of improper integral: $$lim_{ytoinfty}int_x^ycos t^2dt=int_x^inftycos t^2dt,$$
    we can deduce $$lim_{xtoinfty}g(x)=int_a^inftycos t^2dt-lim_{xtoinfty}int_a^xcos t^2dt=int_a^inftycos t^2dt-int_a^inftycos t^2dt=0.$$
    Therefore $lim_{xtoinfty}f(x)=lim_{xtoinfty}g(x+1)-g(x)=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't get it: it seems to be that for you it is obvious the left limit in the middle $;(int_x^y;)$ is zero. Why? What I mean: from what exactly you can deduce that $;lim g(x)=0;$ ?
      $endgroup$
      – DonAntonio
      Dec 26 '18 at 10:02












    • $begingroup$
      @DonAntonio The remainder (tail) of a convergent improper integral goes to $0$. That's a well-known result that Kemono re-proved.
      $endgroup$
      – Gabriel Romon
      Dec 26 '18 at 11:10


















    1












    $begingroup$

    Alternative solution:

    Denote $$g(x)=int_x^infty cos t^2dt (x>0),$$
    we can actually prove this definition is valid (integral converges) using substitution $t^2=u$ and Dirichlet's test.

    According to the definition of improper integral: $$lim_{ytoinfty}int_x^ycos t^2dt=int_x^inftycos t^2dt,$$
    we can deduce $$lim_{xtoinfty}g(x)=int_a^inftycos t^2dt-lim_{xtoinfty}int_a^xcos t^2dt=int_a^inftycos t^2dt-int_a^inftycos t^2dt=0.$$
    Therefore $lim_{xtoinfty}f(x)=lim_{xtoinfty}g(x+1)-g(x)=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't get it: it seems to be that for you it is obvious the left limit in the middle $;(int_x^y;)$ is zero. Why? What I mean: from what exactly you can deduce that $;lim g(x)=0;$ ?
      $endgroup$
      – DonAntonio
      Dec 26 '18 at 10:02












    • $begingroup$
      @DonAntonio The remainder (tail) of a convergent improper integral goes to $0$. That's a well-known result that Kemono re-proved.
      $endgroup$
      – Gabriel Romon
      Dec 26 '18 at 11:10
















    1












    1








    1





    $begingroup$

    Alternative solution:

    Denote $$g(x)=int_x^infty cos t^2dt (x>0),$$
    we can actually prove this definition is valid (integral converges) using substitution $t^2=u$ and Dirichlet's test.

    According to the definition of improper integral: $$lim_{ytoinfty}int_x^ycos t^2dt=int_x^inftycos t^2dt,$$
    we can deduce $$lim_{xtoinfty}g(x)=int_a^inftycos t^2dt-lim_{xtoinfty}int_a^xcos t^2dt=int_a^inftycos t^2dt-int_a^inftycos t^2dt=0.$$
    Therefore $lim_{xtoinfty}f(x)=lim_{xtoinfty}g(x+1)-g(x)=0$.






    share|cite|improve this answer











    $endgroup$



    Alternative solution:

    Denote $$g(x)=int_x^infty cos t^2dt (x>0),$$
    we can actually prove this definition is valid (integral converges) using substitution $t^2=u$ and Dirichlet's test.

    According to the definition of improper integral: $$lim_{ytoinfty}int_x^ycos t^2dt=int_x^inftycos t^2dt,$$
    we can deduce $$lim_{xtoinfty}g(x)=int_a^inftycos t^2dt-lim_{xtoinfty}int_a^xcos t^2dt=int_a^inftycos t^2dt-int_a^inftycos t^2dt=0.$$
    Therefore $lim_{xtoinfty}f(x)=lim_{xtoinfty}g(x+1)-g(x)=0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 26 '18 at 10:08

























    answered Dec 26 '18 at 9:52









    Kemono ChenKemono Chen

    3,1991844




    3,1991844












    • $begingroup$
      I don't get it: it seems to be that for you it is obvious the left limit in the middle $;(int_x^y;)$ is zero. Why? What I mean: from what exactly you can deduce that $;lim g(x)=0;$ ?
      $endgroup$
      – DonAntonio
      Dec 26 '18 at 10:02












    • $begingroup$
      @DonAntonio The remainder (tail) of a convergent improper integral goes to $0$. That's a well-known result that Kemono re-proved.
      $endgroup$
      – Gabriel Romon
      Dec 26 '18 at 11:10




















    • $begingroup$
      I don't get it: it seems to be that for you it is obvious the left limit in the middle $;(int_x^y;)$ is zero. Why? What I mean: from what exactly you can deduce that $;lim g(x)=0;$ ?
      $endgroup$
      – DonAntonio
      Dec 26 '18 at 10:02












    • $begingroup$
      @DonAntonio The remainder (tail) of a convergent improper integral goes to $0$. That's a well-known result that Kemono re-proved.
      $endgroup$
      – Gabriel Romon
      Dec 26 '18 at 11:10


















    $begingroup$
    I don't get it: it seems to be that for you it is obvious the left limit in the middle $;(int_x^y;)$ is zero. Why? What I mean: from what exactly you can deduce that $;lim g(x)=0;$ ?
    $endgroup$
    – DonAntonio
    Dec 26 '18 at 10:02






    $begingroup$
    I don't get it: it seems to be that for you it is obvious the left limit in the middle $;(int_x^y;)$ is zero. Why? What I mean: from what exactly you can deduce that $;lim g(x)=0;$ ?
    $endgroup$
    – DonAntonio
    Dec 26 '18 at 10:02














    $begingroup$
    @DonAntonio The remainder (tail) of a convergent improper integral goes to $0$. That's a well-known result that Kemono re-proved.
    $endgroup$
    – Gabriel Romon
    Dec 26 '18 at 11:10






    $begingroup$
    @DonAntonio The remainder (tail) of a convergent improper integral goes to $0$. That's a well-known result that Kemono re-proved.
    $endgroup$
    – Gabriel Romon
    Dec 26 '18 at 11:10













    1












    $begingroup$

    Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.



    Another Proof



    Likewise, assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
    begin{align*}
    f(x)&=int_x^{x+1}cos t^2 {rm d}t\
    &=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
    &=int_{x^2}^{(x+1)^2}frac{1}{2sqrt{u}}{rm d}(sin u)\
    &=frac{sin u}{2sqrt{u}}bigg|_{x^2}^{(x+1)^2}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u\
    &=frac{sin(x+1)^2}{2(x+1)}-frac{sin x^2}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u.\
    end{align*}
    Thus
    begin{align*}
    |f(x)|&leq frac{|sin(x+1)^2|}{2(x+1)}+frac{|sin x^2|}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}|sin u|{rm d}u\
    &leq frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}{rm d}u\
    &=frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}cdot left[-frac{2}{sqrt{u}}right]_{x^2}^{(x+1)^2}\
    &=frac{1}{x}to 0(x to +infty),
    end{align*}
    which implies $|f(x)|to 0$ according to the squeeze theorem. It follows that $f(x) to 0$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.



      Another Proof



      Likewise, assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
      begin{align*}
      f(x)&=int_x^{x+1}cos t^2 {rm d}t\
      &=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
      &=int_{x^2}^{(x+1)^2}frac{1}{2sqrt{u}}{rm d}(sin u)\
      &=frac{sin u}{2sqrt{u}}bigg|_{x^2}^{(x+1)^2}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u\
      &=frac{sin(x+1)^2}{2(x+1)}-frac{sin x^2}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u.\
      end{align*}
      Thus
      begin{align*}
      |f(x)|&leq frac{|sin(x+1)^2|}{2(x+1)}+frac{|sin x^2|}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}|sin u|{rm d}u\
      &leq frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}{rm d}u\
      &=frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}cdot left[-frac{2}{sqrt{u}}right]_{x^2}^{(x+1)^2}\
      &=frac{1}{x}to 0(x to +infty),
      end{align*}
      which implies $|f(x)|to 0$ according to the squeeze theorem. It follows that $f(x) to 0$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.



        Another Proof



        Likewise, assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
        begin{align*}
        f(x)&=int_x^{x+1}cos t^2 {rm d}t\
        &=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
        &=int_{x^2}^{(x+1)^2}frac{1}{2sqrt{u}}{rm d}(sin u)\
        &=frac{sin u}{2sqrt{u}}bigg|_{x^2}^{(x+1)^2}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u\
        &=frac{sin(x+1)^2}{2(x+1)}-frac{sin x^2}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u.\
        end{align*}
        Thus
        begin{align*}
        |f(x)|&leq frac{|sin(x+1)^2|}{2(x+1)}+frac{|sin x^2|}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}|sin u|{rm d}u\
        &leq frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}{rm d}u\
        &=frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}cdot left[-frac{2}{sqrt{u}}right]_{x^2}^{(x+1)^2}\
        &=frac{1}{x}to 0(x to +infty),
        end{align*}
        which implies $|f(x)|to 0$ according to the squeeze theorem. It follows that $f(x) to 0$.






        share|cite|improve this answer











        $endgroup$



        Thanks to @MinzMin, I find there exists a fatal mistake in my former proof. Since $cos u$ does not keep its sign over the integral interval $[x^2,(x+1)^2]$, we can not apply such a kind of integral mean value theorem. Now, I modify that and give another one.



        Another Proof



        Likewise, assume $x>0$. Making a substitution $t=sqrt{u}$，we have ${rm d}t=dfrac{1}{2sqrt{u}}{rm d}u.$ Therefore,
        begin{align*}
        f(x)&=int_x^{x+1}cos t^2 {rm d}t\
        &=int_{x^2}^{(x+1)^2} frac{cos u}{2sqrt{u}}{rm d}u\
        &=int_{x^2}^{(x+1)^2}frac{1}{2sqrt{u}}{rm d}(sin u)\
        &=frac{sin u}{2sqrt{u}}bigg|_{x^2}^{(x+1)^2}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u\
        &=frac{sin(x+1)^2}{2(x+1)}-frac{sin x^2}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}sin u{rm d}u.\
        end{align*}
        Thus
        begin{align*}
        |f(x)|&leq frac{|sin(x+1)^2|}{2(x+1)}+frac{|sin x^2|}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}|sin u|{rm d}u\
        &leq frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}int_{x^2}^{(x+1)^2}u^{-frac{3}{2}}{rm d}u\
        &=frac{1}{2(x+1)}+frac{1}{2x}+frac{1}{4}cdot left[-frac{2}{sqrt{u}}right]_{x^2}^{(x+1)^2}\
        &=frac{1}{x}to 0(x to +infty),
        end{align*}
        which implies $|f(x)|to 0$ according to the squeeze theorem. It follows that $f(x) to 0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 26 '18 at 12:51

























        answered Dec 26 '18 at 10:28









        mengdie1982mengdie1982

        4,974618




        4,974618






























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