Need help with integration by substitution for $int_x^{x+1}left(sin t^2right)dt$












1












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I am doing exer 13 of baby Rudin Ch 6. I learned integration by substitution and by parts only recently, hence having trouble doing it.




Exercise: Define $$f(x)=int_x^{x+1}left(sin t^2right)dt .$$ Prove that $|f(x)|<frac{1}{x}$ if $x>0.$




Hint says to substitute $t^2=u$, which, I think means that substitute $t=sqrt u$ or $t=-sqrt u$. Which one should I substitute, and why?



Also, the solution I have says that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.






real-analysis integration inequality







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  • $begingroup$
    HINT: en.wikipedia.org/wiki/Fresnel_integral
    $endgroup$
    – user150203
    Dec 26 '18 at 8:13
















1












$begingroup$


I am doing exer 13 of baby Rudin Ch 6. I learned integration by substitution and by parts only recently, hence having trouble doing it.




Exercise: Define $$f(x)=int_x^{x+1}left(sin t^2right)dt .$$ Prove that $|f(x)|<frac{1}{x}$ if $x>0.$




Hint says to substitute $t^2=u$, which, I think means that substitute $t=sqrt u$ or $t=-sqrt u$. Which one should I substitute, and why?



Also, the solution I have says that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.






real-analysis integration inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    HINT: en.wikipedia.org/wiki/Fresnel_integral
    $endgroup$
    – user150203
    Dec 26 '18 at 8:13














1












1








1


1



$begingroup$


I am doing exer 13 of baby Rudin Ch 6. I learned integration by substitution and by parts only recently, hence having trouble doing it.




Exercise: Define $$f(x)=int_x^{x+1}left(sin t^2right)dt .$$ Prove that $|f(x)|<frac{1}{x}$ if $x>0.$




Hint says to substitute $t^2=u$, which, I think means that substitute $t=sqrt u$ or $t=-sqrt u$. Which one should I substitute, and why?



Also, the solution I have says that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.






real-analysis integration inequality







share|cite|improve this question









$endgroup$




I am doing exer 13 of baby Rudin Ch 6. I learned integration by substitution and by parts only recently, hence having trouble doing it.




Exercise: Define $$f(x)=int_x^{x+1}left(sin t^2right)dt .$$ Prove that $|f(x)|<frac{1}{x}$ if $x>0.$




Hint says to substitute $t^2=u$, which, I think means that substitute $t=sqrt u$ or $t=-sqrt u$. Which one should I substitute, and why?



Also, the solution I have says that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.






real-analysis integration inequality




real-analysis integration inequality






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share|cite|improve this question











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share|cite|improve this question










asked Dec 26 '18 at 8:02









SilentSilent

2,90032152




2,90032152












  • $begingroup$
    HINT: en.wikipedia.org/wiki/Fresnel_integral
    $endgroup$
    – user150203
    Dec 26 '18 at 8:13


















  • $begingroup$
    HINT: en.wikipedia.org/wiki/Fresnel_integral
    $endgroup$
    – user150203
    Dec 26 '18 at 8:13
















$begingroup$
HINT: en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– user150203
Dec 26 '18 at 8:13




$begingroup$
HINT: en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– user150203
Dec 26 '18 at 8:13










2 Answers
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$begingroup$

Since $vertsin t^2vertle 1$ you always have that $vert f(x)vertleint_x^{x+1}1,dt=1$. To get strict inequality, take any $t_0in (x,x+1)$ such that $vertsin t^2_0vert<1$. Then by continuity you can find that $vertsin t^2vert<1-varepsilon$ in a small interval $(t_0-delta,t_0+delta)$ contained in $(x,x+1)$. So now
begin{align}vert f(x)vert&leint_x^{x+1}vert sin t^2vert,dt
=int_{(t_0-delta,t_0+delta)}vert sin t^2vert,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}vert sin t^2vert,dt\&le (1-varepsilon)int_{(t_0-delta,t_0+delta)} 1,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}1,dt<1.
end{align}
So if $0<xle 1$ you have $1le frac1x$.
As for the change of variables, either one is admissible, but the one with the plus is simpler.






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    $begingroup$


    Which one should I substitute, and why?




    Whichever is convenient. It'll work out either way; it just changes a few signs around in a way that makes no net difference.



    I wouldn't follow the hint myself; I'd go with an integration by parts instead - or, if I followed the hint, I'd probably follow up with an integration by parts. The key to this integral is that the rapid oscillation will lead to things mostly canceling out. We need to get a handle on that, either with something like an alternating series estimate or by converting it into an absolutely convergent form.




    ... that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.




    Just estimate that $sin$ by $1$:
    $$|f(x)|le int_x^{x+1} 1,dt = 1 le frac1x$$
    As for why we'd have to treat that case separately? The suggested substitution introduces a factor of $frac1{sqrt{u}}$ which blows up if $x$ is too small. Similarly, my integration by parts introduces a factor of $frac1t$. The inequalities we want to use to estimate $f$ elsewhere fail for small $x$, so we use a different method to estimate it there.






    share|cite|improve this answer









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    • $begingroup$
      Thank you very much for this explanation.
      $endgroup$
      – Silent
      Dec 26 '18 at 9:16












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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    3












    $begingroup$

    Since $vertsin t^2vertle 1$ you always have that $vert f(x)vertleint_x^{x+1}1,dt=1$. To get strict inequality, take any $t_0in (x,x+1)$ such that $vertsin t^2_0vert<1$. Then by continuity you can find that $vertsin t^2vert<1-varepsilon$ in a small interval $(t_0-delta,t_0+delta)$ contained in $(x,x+1)$. So now
    begin{align}vert f(x)vert&leint_x^{x+1}vert sin t^2vert,dt
    =int_{(t_0-delta,t_0+delta)}vert sin t^2vert,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}vert sin t^2vert,dt\&le (1-varepsilon)int_{(t_0-delta,t_0+delta)} 1,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}1,dt<1.
    end{align}
    So if $0<xle 1$ you have $1le frac1x$.
    As for the change of variables, either one is admissible, but the one with the plus is simpler.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Since $vertsin t^2vertle 1$ you always have that $vert f(x)vertleint_x^{x+1}1,dt=1$. To get strict inequality, take any $t_0in (x,x+1)$ such that $vertsin t^2_0vert<1$. Then by continuity you can find that $vertsin t^2vert<1-varepsilon$ in a small interval $(t_0-delta,t_0+delta)$ contained in $(x,x+1)$. So now
      begin{align}vert f(x)vert&leint_x^{x+1}vert sin t^2vert,dt
      =int_{(t_0-delta,t_0+delta)}vert sin t^2vert,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}vert sin t^2vert,dt\&le (1-varepsilon)int_{(t_0-delta,t_0+delta)} 1,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}1,dt<1.
      end{align}
      So if $0<xle 1$ you have $1le frac1x$.
      As for the change of variables, either one is admissible, but the one with the plus is simpler.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Since $vertsin t^2vertle 1$ you always have that $vert f(x)vertleint_x^{x+1}1,dt=1$. To get strict inequality, take any $t_0in (x,x+1)$ such that $vertsin t^2_0vert<1$. Then by continuity you can find that $vertsin t^2vert<1-varepsilon$ in a small interval $(t_0-delta,t_0+delta)$ contained in $(x,x+1)$. So now
        begin{align}vert f(x)vert&leint_x^{x+1}vert sin t^2vert,dt
        =int_{(t_0-delta,t_0+delta)}vert sin t^2vert,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}vert sin t^2vert,dt\&le (1-varepsilon)int_{(t_0-delta,t_0+delta)} 1,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}1,dt<1.
        end{align}
        So if $0<xle 1$ you have $1le frac1x$.
        As for the change of variables, either one is admissible, but the one with the plus is simpler.






        share|cite|improve this answer









        $endgroup$



        Since $vertsin t^2vertle 1$ you always have that $vert f(x)vertleint_x^{x+1}1,dt=1$. To get strict inequality, take any $t_0in (x,x+1)$ such that $vertsin t^2_0vert<1$. Then by continuity you can find that $vertsin t^2vert<1-varepsilon$ in a small interval $(t_0-delta,t_0+delta)$ contained in $(x,x+1)$. So now
        begin{align}vert f(x)vert&leint_x^{x+1}vert sin t^2vert,dt
        =int_{(t_0-delta,t_0+delta)}vert sin t^2vert,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}vert sin t^2vert,dt\&le (1-varepsilon)int_{(t_0-delta,t_0+delta)} 1,dt+int_{(x,x+1)setminus(t_0-delta,t_0+delta)}1,dt<1.
        end{align}
        So if $0<xle 1$ you have $1le frac1x$.
        As for the change of variables, either one is admissible, but the one with the plus is simpler.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 26 '18 at 8:18









        Gio67Gio67

        12.8k1627




        12.8k1627























            2












            $begingroup$


            Which one should I substitute, and why?




            Whichever is convenient. It'll work out either way; it just changes a few signs around in a way that makes no net difference.



            I wouldn't follow the hint myself; I'd go with an integration by parts instead - or, if I followed the hint, I'd probably follow up with an integration by parts. The key to this integral is that the rapid oscillation will lead to things mostly canceling out. We need to get a handle on that, either with something like an alternating series estimate or by converting it into an absolutely convergent form.




            ... that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.




            Just estimate that $sin$ by $1$:
            $$|f(x)|le int_x^{x+1} 1,dt = 1 le frac1x$$
            As for why we'd have to treat that case separately? The suggested substitution introduces a factor of $frac1{sqrt{u}}$ which blows up if $x$ is too small. Similarly, my integration by parts introduces a factor of $frac1t$. The inequalities we want to use to estimate $f$ elsewhere fail for small $x$, so we use a different method to estimate it there.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much for this explanation.
              $endgroup$
              – Silent
              Dec 26 '18 at 9:16
















            2












            $begingroup$


            Which one should I substitute, and why?




            Whichever is convenient. It'll work out either way; it just changes a few signs around in a way that makes no net difference.



            I wouldn't follow the hint myself; I'd go with an integration by parts instead - or, if I followed the hint, I'd probably follow up with an integration by parts. The key to this integral is that the rapid oscillation will lead to things mostly canceling out. We need to get a handle on that, either with something like an alternating series estimate or by converting it into an absolutely convergent form.




            ... that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.




            Just estimate that $sin$ by $1$:
            $$|f(x)|le int_x^{x+1} 1,dt = 1 le frac1x$$
            As for why we'd have to treat that case separately? The suggested substitution introduces a factor of $frac1{sqrt{u}}$ which blows up if $x$ is too small. Similarly, my integration by parts introduces a factor of $frac1t$. The inequalities we want to use to estimate $f$ elsewhere fail for small $x$, so we use a different method to estimate it there.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much for this explanation.
              $endgroup$
              – Silent
              Dec 26 '18 at 9:16














            2












            2








            2





            $begingroup$


            Which one should I substitute, and why?




            Whichever is convenient. It'll work out either way; it just changes a few signs around in a way that makes no net difference.



            I wouldn't follow the hint myself; I'd go with an integration by parts instead - or, if I followed the hint, I'd probably follow up with an integration by parts. The key to this integral is that the rapid oscillation will lead to things mostly canceling out. We need to get a handle on that, either with something like an alternating series estimate or by converting it into an absolutely convergent form.




            ... that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.




            Just estimate that $sin$ by $1$:
            $$|f(x)|le int_x^{x+1} 1,dt = 1 le frac1x$$
            As for why we'd have to treat that case separately? The suggested substitution introduces a factor of $frac1{sqrt{u}}$ which blows up if $x$ is too small. Similarly, my integration by parts introduces a factor of $frac1t$. The inequalities we want to use to estimate $f$ elsewhere fail for small $x$, so we use a different method to estimate it there.






            share|cite|improve this answer









            $endgroup$




            Which one should I substitute, and why?




            Whichever is convenient. It'll work out either way; it just changes a few signs around in a way that makes no net difference.



            I wouldn't follow the hint myself; I'd go with an integration by parts instead - or, if I followed the hint, I'd probably follow up with an integration by parts. The key to this integral is that the rapid oscillation will lead to things mostly canceling out. We need to get a handle on that, either with something like an alternating series estimate or by converting it into an absolutely convergent form.




            ... that $|f(x)|<frac1x$ is obvious for $0<xle 1$. I do not see how it is obvious, or why we have to consider this as separate case.




            Just estimate that $sin$ by $1$:
            $$|f(x)|le int_x^{x+1} 1,dt = 1 le frac1x$$
            As for why we'd have to treat that case separately? The suggested substitution introduces a factor of $frac1{sqrt{u}}$ which blows up if $x$ is too small. Similarly, my integration by parts introduces a factor of $frac1t$. The inequalities we want to use to estimate $f$ elsewhere fail for small $x$, so we use a different method to estimate it there.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 26 '18 at 8:18









            jmerryjmerry

            17k11633




            17k11633












            • $begingroup$
              Thank you very much for this explanation.
              $endgroup$
              – Silent
              Dec 26 '18 at 9:16


















            • $begingroup$
              Thank you very much for this explanation.
              $endgroup$
              – Silent
              Dec 26 '18 at 9:16
















            $begingroup$
            Thank you very much for this explanation.
            $endgroup$
            – Silent
            Dec 26 '18 at 9:16




            $begingroup$
            Thank you very much for this explanation.
            $endgroup$
            – Silent
            Dec 26 '18 at 9:16


















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