Convergence/Divergence of $int_0^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x$












1












$begingroup$


Let's call $f(x)= frac {sin x}{xln^2(x^2+2)}$ and split our integral: $$int_0^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x={int_0^2 {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x}+int_2^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x$$

Then I should consider behavior of $f(x)$ as $x$ approaches $0$ and $+infty$

Now let's call $I=int_2^{+infty}frac{1}{xln^beta(x)}$ and if we consider behavior of $f(x)$ at $+infty$
$|f(x)|=frac{|sin x|}{xln^2(x^2+2)}leqfrac{1}{xln^2(x^2+2)}$ and second integral converges because $I$ converges when $betagt1$

As for $x$ approaching $0$,$f(x)$ behaves like $frac{1}{ln^2(x^2+2)}$.So,my question is how can we study convergence/divergence of this one?






calculus improper-integrals







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let's call $f(x)= frac {sin x}{xln^2(x^2+2)}$ and split our integral: $$int_0^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x={int_0^2 {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x}+int_2^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x$$

    Then I should consider behavior of $f(x)$ as $x$ approaches $0$ and $+infty$

    Now let's call $I=int_2^{+infty}frac{1}{xln^beta(x)}$ and if we consider behavior of $f(x)$ at $+infty$
    $|f(x)|=frac{|sin x|}{xln^2(x^2+2)}leqfrac{1}{xln^2(x^2+2)}$ and second integral converges because $I$ converges when $betagt1$

    As for $x$ approaching $0$,$f(x)$ behaves like $frac{1}{ln^2(x^2+2)}$.So,my question is how can we study convergence/divergence of this one?






    calculus improper-integrals







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let's call $f(x)= frac {sin x}{xln^2(x^2+2)}$ and split our integral: $$int_0^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x={int_0^2 {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x}+int_2^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x$$

      Then I should consider behavior of $f(x)$ as $x$ approaches $0$ and $+infty$

      Now let's call $I=int_2^{+infty}frac{1}{xln^beta(x)}$ and if we consider behavior of $f(x)$ at $+infty$
      $|f(x)|=frac{|sin x|}{xln^2(x^2+2)}leqfrac{1}{xln^2(x^2+2)}$ and second integral converges because $I$ converges when $betagt1$

      As for $x$ approaching $0$,$f(x)$ behaves like $frac{1}{ln^2(x^2+2)}$.So,my question is how can we study convergence/divergence of this one?






      calculus improper-integrals







      share|cite|improve this question









      $endgroup$




      Let's call $f(x)= frac {sin x}{xln^2(x^2+2)}$ and split our integral: $$int_0^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x={int_0^2 {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x}+int_2^{+infty} {frac {sin x}{xln^2 (x^2+2)}} mathrm{d}x$$

      Then I should consider behavior of $f(x)$ as $x$ approaches $0$ and $+infty$

      Now let's call $I=int_2^{+infty}frac{1}{xln^beta(x)}$ and if we consider behavior of $f(x)$ at $+infty$
      $|f(x)|=frac{|sin x|}{xln^2(x^2+2)}leqfrac{1}{xln^2(x^2+2)}$ and second integral converges because $I$ converges when $betagt1$

      As for $x$ approaching $0$,$f(x)$ behaves like $frac{1}{ln^2(x^2+2)}$.So,my question is how can we study convergence/divergence of this one?






      calculus improper-integrals




      calculus improper-integrals






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 13 at 14:25









      Turan NəsibliTuran Nəsibli

      726




      726






















          1 Answer
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          $begingroup$

          Hint. Note that $f(x)=frac{sin(x)}{xln^2(x^2+2)}$ is continuous in $(0,2]$ and its limit at $0^+$ is $1/ln^2(2)$. So it can be extended to a continuous function in $[0,2]$. Moreover, as you already remarked, for $xin [2,+infty)$,
          $$|f(x)|=frac{|sin(x)|}{xln^2(x^2+2)}leq frac{1}{4xln^2(x)}$$
          and the integral of the right-hand side is convergent:
          $$int_2^{+infty}frac{1}{4xln^2(x)},dx=left[-frac{1}{4ln(x)}right]_2^{+infty}=frac{1}{4ln(2)}.$$
          What may we conclude?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            By a numerical method we get $$approx 1.676060785605060001$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 13 at 14:35










          • $begingroup$
            Well,yes $f(x)$ is bounded in the right neighbourhood of $0$ and because the interval we are considered is also bounded,we can concluse that the integral is convergent,is that right?
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:41










          • $begingroup$
            Yes, you are correct!
            $endgroup$
            – Robert Z
            Jan 13 at 14:42










          • $begingroup$
            Thank you both for the help!
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:44











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          1 Answer
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          active

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          active

          oldest

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          1












          $begingroup$

          Hint. Note that $f(x)=frac{sin(x)}{xln^2(x^2+2)}$ is continuous in $(0,2]$ and its limit at $0^+$ is $1/ln^2(2)$. So it can be extended to a continuous function in $[0,2]$. Moreover, as you already remarked, for $xin [2,+infty)$,
          $$|f(x)|=frac{|sin(x)|}{xln^2(x^2+2)}leq frac{1}{4xln^2(x)}$$
          and the integral of the right-hand side is convergent:
          $$int_2^{+infty}frac{1}{4xln^2(x)},dx=left[-frac{1}{4ln(x)}right]_2^{+infty}=frac{1}{4ln(2)}.$$
          What may we conclude?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            By a numerical method we get $$approx 1.676060785605060001$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 13 at 14:35










          • $begingroup$
            Well,yes $f(x)$ is bounded in the right neighbourhood of $0$ and because the interval we are considered is also bounded,we can concluse that the integral is convergent,is that right?
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:41










          • $begingroup$
            Yes, you are correct!
            $endgroup$
            – Robert Z
            Jan 13 at 14:42










          • $begingroup$
            Thank you both for the help!
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:44
















          1












          $begingroup$

          Hint. Note that $f(x)=frac{sin(x)}{xln^2(x^2+2)}$ is continuous in $(0,2]$ and its limit at $0^+$ is $1/ln^2(2)$. So it can be extended to a continuous function in $[0,2]$. Moreover, as you already remarked, for $xin [2,+infty)$,
          $$|f(x)|=frac{|sin(x)|}{xln^2(x^2+2)}leq frac{1}{4xln^2(x)}$$
          and the integral of the right-hand side is convergent:
          $$int_2^{+infty}frac{1}{4xln^2(x)},dx=left[-frac{1}{4ln(x)}right]_2^{+infty}=frac{1}{4ln(2)}.$$
          What may we conclude?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            By a numerical method we get $$approx 1.676060785605060001$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 13 at 14:35










          • $begingroup$
            Well,yes $f(x)$ is bounded in the right neighbourhood of $0$ and because the interval we are considered is also bounded,we can concluse that the integral is convergent,is that right?
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:41










          • $begingroup$
            Yes, you are correct!
            $endgroup$
            – Robert Z
            Jan 13 at 14:42










          • $begingroup$
            Thank you both for the help!
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:44














          1












          1








          1





          $begingroup$

          Hint. Note that $f(x)=frac{sin(x)}{xln^2(x^2+2)}$ is continuous in $(0,2]$ and its limit at $0^+$ is $1/ln^2(2)$. So it can be extended to a continuous function in $[0,2]$. Moreover, as you already remarked, for $xin [2,+infty)$,
          $$|f(x)|=frac{|sin(x)|}{xln^2(x^2+2)}leq frac{1}{4xln^2(x)}$$
          and the integral of the right-hand side is convergent:
          $$int_2^{+infty}frac{1}{4xln^2(x)},dx=left[-frac{1}{4ln(x)}right]_2^{+infty}=frac{1}{4ln(2)}.$$
          What may we conclude?






          share|cite|improve this answer











          $endgroup$



          Hint. Note that $f(x)=frac{sin(x)}{xln^2(x^2+2)}$ is continuous in $(0,2]$ and its limit at $0^+$ is $1/ln^2(2)$. So it can be extended to a continuous function in $[0,2]$. Moreover, as you already remarked, for $xin [2,+infty)$,
          $$|f(x)|=frac{|sin(x)|}{xln^2(x^2+2)}leq frac{1}{4xln^2(x)}$$
          and the integral of the right-hand side is convergent:
          $$int_2^{+infty}frac{1}{4xln^2(x)},dx=left[-frac{1}{4ln(x)}right]_2^{+infty}=frac{1}{4ln(2)}.$$
          What may we conclude?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 13 at 14:50

























          answered Jan 13 at 14:34









          Robert ZRobert Z

          95.9k1065136




          95.9k1065136








          • 1




            $begingroup$
            By a numerical method we get $$approx 1.676060785605060001$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 13 at 14:35










          • $begingroup$
            Well,yes $f(x)$ is bounded in the right neighbourhood of $0$ and because the interval we are considered is also bounded,we can concluse that the integral is convergent,is that right?
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:41










          • $begingroup$
            Yes, you are correct!
            $endgroup$
            – Robert Z
            Jan 13 at 14:42










          • $begingroup$
            Thank you both for the help!
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:44














          • 1




            $begingroup$
            By a numerical method we get $$approx 1.676060785605060001$$
            $endgroup$
            – Dr. Sonnhard Graubner
            Jan 13 at 14:35










          • $begingroup$
            Well,yes $f(x)$ is bounded in the right neighbourhood of $0$ and because the interval we are considered is also bounded,we can concluse that the integral is convergent,is that right?
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:41










          • $begingroup$
            Yes, you are correct!
            $endgroup$
            – Robert Z
            Jan 13 at 14:42










          • $begingroup$
            Thank you both for the help!
            $endgroup$
            – Turan Nəsibli
            Jan 13 at 14:44








          1




          1




          $begingroup$
          By a numerical method we get $$approx 1.676060785605060001$$
          $endgroup$
          – Dr. Sonnhard Graubner
          Jan 13 at 14:35




          $begingroup$
          By a numerical method we get $$approx 1.676060785605060001$$
          $endgroup$
          – Dr. Sonnhard Graubner
          Jan 13 at 14:35












          $begingroup$
          Well,yes $f(x)$ is bounded in the right neighbourhood of $0$ and because the interval we are considered is also bounded,we can concluse that the integral is convergent,is that right?
          $endgroup$
          – Turan Nəsibli
          Jan 13 at 14:41




          $begingroup$
          Well,yes $f(x)$ is bounded in the right neighbourhood of $0$ and because the interval we are considered is also bounded,we can concluse that the integral is convergent,is that right?
          $endgroup$
          – Turan Nəsibli
          Jan 13 at 14:41












          $begingroup$
          Yes, you are correct!
          $endgroup$
          – Robert Z
          Jan 13 at 14:42




          $begingroup$
          Yes, you are correct!
          $endgroup$
          – Robert Z
          Jan 13 at 14:42












          $begingroup$
          Thank you both for the help!
          $endgroup$
          – Turan Nəsibli
          Jan 13 at 14:44




          $begingroup$
          Thank you both for the help!
          $endgroup$
          – Turan Nəsibli
          Jan 13 at 14:44


















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