Is $int_0^{+infty}frac{sin ln x}{sqrt x},dx$ convergent or absolutely convergent?












0












$begingroup$


Is
$$int_0^inftyfrac{sinln x}{sqrt x},dx$$
convergent? Absolutely convergent?



I found, that in fact it converges, but the only way I found is to say that $$int_0^{+infty}frac{sin ln x}{sqrt x} , dx leq int_0^{+infty}frac{1}{sqrt x },dx,$$ but this integral doesn't converge, so this way I can't say something about convergence of this integral.






convergence improper-integrals







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$endgroup$












  • $begingroup$
    If $displaystyle int_0^{+infty} dfrac{sinln x}{sqrt x} , dx le int_0^{+infty} dfrac1{sqrt x}, dx$ and $displaystyleint_0^{+infty} dfrac1{sqrt x}, dx$ doesn't converge, then you can't conclude anything about $displaystyleint_0^{+infty} dfrac{sinln x}{sqrt x} , dx$. It sounds like you might know that, but you also said you found that the given integral does converge. How'd you determine that?
    $endgroup$
    – tilper
    May 10 '17 at 13:59












  • $begingroup$
    @tipler, yes, of course I understand this, but I didn't found any another ways, so, i'm trying to say, that I tried to use this method, but this method not right way
    $endgroup$
    – Eugene Korotkov
    May 10 '17 at 14:01










  • $begingroup$
    Why do you think the integral converges?
    $endgroup$
    – zhw.
    May 10 '17 at 14:23
















0












$begingroup$


Is
$$int_0^inftyfrac{sinln x}{sqrt x},dx$$
convergent? Absolutely convergent?



I found, that in fact it converges, but the only way I found is to say that $$int_0^{+infty}frac{sin ln x}{sqrt x} , dx leq int_0^{+infty}frac{1}{sqrt x },dx,$$ but this integral doesn't converge, so this way I can't say something about convergence of this integral.






convergence improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $displaystyle int_0^{+infty} dfrac{sinln x}{sqrt x} , dx le int_0^{+infty} dfrac1{sqrt x}, dx$ and $displaystyleint_0^{+infty} dfrac1{sqrt x}, dx$ doesn't converge, then you can't conclude anything about $displaystyleint_0^{+infty} dfrac{sinln x}{sqrt x} , dx$. It sounds like you might know that, but you also said you found that the given integral does converge. How'd you determine that?
    $endgroup$
    – tilper
    May 10 '17 at 13:59












  • $begingroup$
    @tipler, yes, of course I understand this, but I didn't found any another ways, so, i'm trying to say, that I tried to use this method, but this method not right way
    $endgroup$
    – Eugene Korotkov
    May 10 '17 at 14:01










  • $begingroup$
    Why do you think the integral converges?
    $endgroup$
    – zhw.
    May 10 '17 at 14:23














0












0








0


1



$begingroup$


Is
$$int_0^inftyfrac{sinln x}{sqrt x},dx$$
convergent? Absolutely convergent?



I found, that in fact it converges, but the only way I found is to say that $$int_0^{+infty}frac{sin ln x}{sqrt x} , dx leq int_0^{+infty}frac{1}{sqrt x },dx,$$ but this integral doesn't converge, so this way I can't say something about convergence of this integral.






convergence improper-integrals







share|cite|improve this question











$endgroup$




Is
$$int_0^inftyfrac{sinln x}{sqrt x},dx$$
convergent? Absolutely convergent?



I found, that in fact it converges, but the only way I found is to say that $$int_0^{+infty}frac{sin ln x}{sqrt x} , dx leq int_0^{+infty}frac{1}{sqrt x },dx,$$ but this integral doesn't converge, so this way I can't say something about convergence of this integral.






convergence improper-integrals




convergence improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 8:26









Rócherz

2,7762721




2,7762721










asked May 10 '17 at 13:55









Eugene KorotkovEugene Korotkov

1689




1689












  • $begingroup$
    If $displaystyle int_0^{+infty} dfrac{sinln x}{sqrt x} , dx le int_0^{+infty} dfrac1{sqrt x}, dx$ and $displaystyleint_0^{+infty} dfrac1{sqrt x}, dx$ doesn't converge, then you can't conclude anything about $displaystyleint_0^{+infty} dfrac{sinln x}{sqrt x} , dx$. It sounds like you might know that, but you also said you found that the given integral does converge. How'd you determine that?
    $endgroup$
    – tilper
    May 10 '17 at 13:59












  • $begingroup$
    @tipler, yes, of course I understand this, but I didn't found any another ways, so, i'm trying to say, that I tried to use this method, but this method not right way
    $endgroup$
    – Eugene Korotkov
    May 10 '17 at 14:01










  • $begingroup$
    Why do you think the integral converges?
    $endgroup$
    – zhw.
    May 10 '17 at 14:23


















  • $begingroup$
    If $displaystyle int_0^{+infty} dfrac{sinln x}{sqrt x} , dx le int_0^{+infty} dfrac1{sqrt x}, dx$ and $displaystyleint_0^{+infty} dfrac1{sqrt x}, dx$ doesn't converge, then you can't conclude anything about $displaystyleint_0^{+infty} dfrac{sinln x}{sqrt x} , dx$. It sounds like you might know that, but you also said you found that the given integral does converge. How'd you determine that?
    $endgroup$
    – tilper
    May 10 '17 at 13:59












  • $begingroup$
    @tipler, yes, of course I understand this, but I didn't found any another ways, so, i'm trying to say, that I tried to use this method, but this method not right way
    $endgroup$
    – Eugene Korotkov
    May 10 '17 at 14:01










  • $begingroup$
    Why do you think the integral converges?
    $endgroup$
    – zhw.
    May 10 '17 at 14:23
















$begingroup$
If $displaystyle int_0^{+infty} dfrac{sinln x}{sqrt x} , dx le int_0^{+infty} dfrac1{sqrt x}, dx$ and $displaystyleint_0^{+infty} dfrac1{sqrt x}, dx$ doesn't converge, then you can't conclude anything about $displaystyleint_0^{+infty} dfrac{sinln x}{sqrt x} , dx$. It sounds like you might know that, but you also said you found that the given integral does converge. How'd you determine that?
$endgroup$
– tilper
May 10 '17 at 13:59






$begingroup$
If $displaystyle int_0^{+infty} dfrac{sinln x}{sqrt x} , dx le int_0^{+infty} dfrac1{sqrt x}, dx$ and $displaystyleint_0^{+infty} dfrac1{sqrt x}, dx$ doesn't converge, then you can't conclude anything about $displaystyleint_0^{+infty} dfrac{sinln x}{sqrt x} , dx$. It sounds like you might know that, but you also said you found that the given integral does converge. How'd you determine that?
$endgroup$
– tilper
May 10 '17 at 13:59














$begingroup$
@tipler, yes, of course I understand this, but I didn't found any another ways, so, i'm trying to say, that I tried to use this method, but this method not right way
$endgroup$
– Eugene Korotkov
May 10 '17 at 14:01




$begingroup$
@tipler, yes, of course I understand this, but I didn't found any another ways, so, i'm trying to say, that I tried to use this method, but this method not right way
$endgroup$
– Eugene Korotkov
May 10 '17 at 14:01












$begingroup$
Why do you think the integral converges?
$endgroup$
– zhw.
May 10 '17 at 14:23




$begingroup$
Why do you think the integral converges?
$endgroup$
– zhw.
May 10 '17 at 14:23










3 Answers
3






active

oldest

votes


















2












$begingroup$

Use substitution $x=e^u$ to get



$$I=int_0^{+infty}frac{sinln x}{sqrt x},dx = int_{-infty}^{+infty}frac{sin u}{e^{u/2}}e^u,du = int_{-infty}^{+infty}e^{u/2}sin u,du$$



and then use partial integration:



begin{align}I &=int_{-infty}^{+infty}e^{x/2}sin x,dx \ &=left[begin{array}{ll} u = sin x,&du=cos x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
&= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2int_{-infty}^{+infty}e^{x/2}cos x,dx \
&= left[begin{array}{ll}u = cos x,&du=-sin x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
&= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2left(left.2e^{x/2}cos x,right|_{-infty}^{+infty} + 2int_{-infty}^{+infty}e^{x/2}sin x,dxright) \
&= left.2e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}-4Iend{align}



We conclude that $I = frac 25 left.e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}$ which doesn't converge.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint: For $n=1,2,dots,$ estimate the integral over $[e^{2pi n +pi/4},e^{2pi n +pi/2}].$ Note that $sin (ln x) ge 1/sqrt 2$ on this interval.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You could solve the indefinite integral,
      begin{align*}
      intfrac{sin(ln(x))}{sqrt x},dx &= intfrac{sqrt x cdotsin(ln(x))}{x},dx \ &= int e^{lnsqrt x}sin(ln x),d(ln x) \ &= int e^{frac12 ln x}sin(ln x),d(ln x) \ &= frac25sqrt xleft(2cos(ln x))-sin(ln x))right)\
      end{align*}
      and then evaluate the limits.






      share|cite|improve this answer











      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Use substitution $x=e^u$ to get



        $$I=int_0^{+infty}frac{sinln x}{sqrt x},dx = int_{-infty}^{+infty}frac{sin u}{e^{u/2}}e^u,du = int_{-infty}^{+infty}e^{u/2}sin u,du$$



        and then use partial integration:



        begin{align}I &=int_{-infty}^{+infty}e^{x/2}sin x,dx \ &=left[begin{array}{ll} u = sin x,&du=cos x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
        &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2int_{-infty}^{+infty}e^{x/2}cos x,dx \
        &= left[begin{array}{ll}u = cos x,&du=-sin x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
        &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2left(left.2e^{x/2}cos x,right|_{-infty}^{+infty} + 2int_{-infty}^{+infty}e^{x/2}sin x,dxright) \
        &= left.2e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}-4Iend{align}



        We conclude that $I = frac 25 left.e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}$ which doesn't converge.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Use substitution $x=e^u$ to get



          $$I=int_0^{+infty}frac{sinln x}{sqrt x},dx = int_{-infty}^{+infty}frac{sin u}{e^{u/2}}e^u,du = int_{-infty}^{+infty}e^{u/2}sin u,du$$



          and then use partial integration:



          begin{align}I &=int_{-infty}^{+infty}e^{x/2}sin x,dx \ &=left[begin{array}{ll} u = sin x,&du=cos x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
          &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2int_{-infty}^{+infty}e^{x/2}cos x,dx \
          &= left[begin{array}{ll}u = cos x,&du=-sin x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
          &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2left(left.2e^{x/2}cos x,right|_{-infty}^{+infty} + 2int_{-infty}^{+infty}e^{x/2}sin x,dxright) \
          &= left.2e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}-4Iend{align}



          We conclude that $I = frac 25 left.e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}$ which doesn't converge.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Use substitution $x=e^u$ to get



            $$I=int_0^{+infty}frac{sinln x}{sqrt x},dx = int_{-infty}^{+infty}frac{sin u}{e^{u/2}}e^u,du = int_{-infty}^{+infty}e^{u/2}sin u,du$$



            and then use partial integration:



            begin{align}I &=int_{-infty}^{+infty}e^{x/2}sin x,dx \ &=left[begin{array}{ll} u = sin x,&du=cos x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
            &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2int_{-infty}^{+infty}e^{x/2}cos x,dx \
            &= left[begin{array}{ll}u = cos x,&du=-sin x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
            &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2left(left.2e^{x/2}cos x,right|_{-infty}^{+infty} + 2int_{-infty}^{+infty}e^{x/2}sin x,dxright) \
            &= left.2e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}-4Iend{align}



            We conclude that $I = frac 25 left.e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}$ which doesn't converge.






            share|cite|improve this answer









            $endgroup$



            Use substitution $x=e^u$ to get



            $$I=int_0^{+infty}frac{sinln x}{sqrt x},dx = int_{-infty}^{+infty}frac{sin u}{e^{u/2}}e^u,du = int_{-infty}^{+infty}e^{u/2}sin u,du$$



            and then use partial integration:



            begin{align}I &=int_{-infty}^{+infty}e^{x/2}sin x,dx \ &=left[begin{array}{ll} u = sin x,&du=cos x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
            &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2int_{-infty}^{+infty}e^{x/2}cos x,dx \
            &= left[begin{array}{ll}u = cos x,&du=-sin x\ v =2e^{x/2},&dv = e^{x/2}end{array}right] \
            &= left.2e^{x/2}sin x,right|_{-infty}^{+infty} - 2left(left.2e^{x/2}cos x,right|_{-infty}^{+infty} + 2int_{-infty}^{+infty}e^{x/2}sin x,dxright) \
            &= left.2e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}-4Iend{align}



            We conclude that $I = frac 25 left.e^{x/2}(sin x-2cos x),right|_{-infty}^{+infty}$ which doesn't converge.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 10 '17 at 14:58









            EnnarEnnar

            14.5k32443




            14.5k32443























                2












                $begingroup$

                Hint: For $n=1,2,dots,$ estimate the integral over $[e^{2pi n +pi/4},e^{2pi n +pi/2}].$ Note that $sin (ln x) ge 1/sqrt 2$ on this interval.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Hint: For $n=1,2,dots,$ estimate the integral over $[e^{2pi n +pi/4},e^{2pi n +pi/2}].$ Note that $sin (ln x) ge 1/sqrt 2$ on this interval.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Hint: For $n=1,2,dots,$ estimate the integral over $[e^{2pi n +pi/4},e^{2pi n +pi/2}].$ Note that $sin (ln x) ge 1/sqrt 2$ on this interval.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: For $n=1,2,dots,$ estimate the integral over $[e^{2pi n +pi/4},e^{2pi n +pi/2}].$ Note that $sin (ln x) ge 1/sqrt 2$ on this interval.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered May 10 '17 at 14:22









                    zhw.zhw.

                    73.3k43175




                    73.3k43175























                        0












                        $begingroup$

                        You could solve the indefinite integral,
                        begin{align*}
                        intfrac{sin(ln(x))}{sqrt x},dx &= intfrac{sqrt x cdotsin(ln(x))}{x},dx \ &= int e^{lnsqrt x}sin(ln x),d(ln x) \ &= int e^{frac12 ln x}sin(ln x),d(ln x) \ &= frac25sqrt xleft(2cos(ln x))-sin(ln x))right)\
                        end{align*}
                        and then evaluate the limits.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          You could solve the indefinite integral,
                          begin{align*}
                          intfrac{sin(ln(x))}{sqrt x},dx &= intfrac{sqrt x cdotsin(ln(x))}{x},dx \ &= int e^{lnsqrt x}sin(ln x),d(ln x) \ &= int e^{frac12 ln x}sin(ln x),d(ln x) \ &= frac25sqrt xleft(2cos(ln x))-sin(ln x))right)\
                          end{align*}
                          and then evaluate the limits.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You could solve the indefinite integral,
                            begin{align*}
                            intfrac{sin(ln(x))}{sqrt x},dx &= intfrac{sqrt x cdotsin(ln(x))}{x},dx \ &= int e^{lnsqrt x}sin(ln x),d(ln x) \ &= int e^{frac12 ln x}sin(ln x),d(ln x) \ &= frac25sqrt xleft(2cos(ln x))-sin(ln x))right)\
                            end{align*}
                            and then evaluate the limits.






                            share|cite|improve this answer











                            $endgroup$



                            You could solve the indefinite integral,
                            begin{align*}
                            intfrac{sin(ln(x))}{sqrt x},dx &= intfrac{sqrt x cdotsin(ln(x))}{x},dx \ &= int e^{lnsqrt x}sin(ln x),d(ln x) \ &= int e^{frac12 ln x}sin(ln x),d(ln x) \ &= frac25sqrt xleft(2cos(ln x))-sin(ln x))right)\
                            end{align*}
                            and then evaluate the limits.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 28 '18 at 8:31









                            Rócherz

                            2,7762721




                            2,7762721










                            answered May 10 '17 at 14:47







                            user409521





































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                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
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