Convergence/Divergence of Improper Integrals












1












$begingroup$


We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
begin{equation}
int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
end{equation}
I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated






calculus analysis convergence improper-integrals divergence







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
    begin{equation}
    int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
    end{equation}
    I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated






    calculus analysis convergence improper-integrals divergence







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
      begin{equation}
      int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
      end{equation}
      I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated






      calculus analysis convergence improper-integrals divergence







      share|cite|improve this question









      $endgroup$




      We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
      begin{equation}
      int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
      end{equation}
      I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated






      calculus analysis convergence improper-integrals divergence




      calculus analysis convergence improper-integrals divergence






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      share|cite|improve this question











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      asked 1 hour ago









      kareem bokaikareem bokai

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      444






















          1 Answer
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          5












          $begingroup$

          Hint



          $$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
            $endgroup$
            – kareem bokai
            1 hour ago










          • $begingroup$
            No smashing, please! You're welcome :).
            $endgroup$
            – StackTD
            1 hour ago










          • $begingroup$
            Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
            $endgroup$
            – imranfat
            8 mins ago












          • $begingroup$
            I know, this was only a hint. The original function has no problem at $x=0$.
            $endgroup$
            – StackTD
            7 mins ago










          • $begingroup$
            @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
            $endgroup$
            – imranfat
            5 mins ago











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Hint



          $$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
            $endgroup$
            – kareem bokai
            1 hour ago










          • $begingroup$
            No smashing, please! You're welcome :).
            $endgroup$
            – StackTD
            1 hour ago










          • $begingroup$
            Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
            $endgroup$
            – imranfat
            8 mins ago












          • $begingroup$
            I know, this was only a hint. The original function has no problem at $x=0$.
            $endgroup$
            – StackTD
            7 mins ago










          • $begingroup$
            @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
            $endgroup$
            – imranfat
            5 mins ago
















          5












          $begingroup$

          Hint



          $$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
            $endgroup$
            – kareem bokai
            1 hour ago










          • $begingroup$
            No smashing, please! You're welcome :).
            $endgroup$
            – StackTD
            1 hour ago










          • $begingroup$
            Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
            $endgroup$
            – imranfat
            8 mins ago












          • $begingroup$
            I know, this was only a hint. The original function has no problem at $x=0$.
            $endgroup$
            – StackTD
            7 mins ago










          • $begingroup$
            @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
            $endgroup$
            – imranfat
            5 mins ago














          5












          5








          5





          $begingroup$

          Hint



          $$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






          share|cite|improve this answer









          $endgroup$



          Hint



          $$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          StackTDStackTD

          22.6k2049




          22.6k2049








          • 2




            $begingroup$
            Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
            $endgroup$
            – kareem bokai
            1 hour ago










          • $begingroup$
            No smashing, please! You're welcome :).
            $endgroup$
            – StackTD
            1 hour ago










          • $begingroup$
            Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
            $endgroup$
            – imranfat
            8 mins ago












          • $begingroup$
            I know, this was only a hint. The original function has no problem at $x=0$.
            $endgroup$
            – StackTD
            7 mins ago










          • $begingroup$
            @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
            $endgroup$
            – imranfat
            5 mins ago














          • 2




            $begingroup$
            Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
            $endgroup$
            – kareem bokai
            1 hour ago










          • $begingroup$
            No smashing, please! You're welcome :).
            $endgroup$
            – StackTD
            1 hour ago










          • $begingroup$
            Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
            $endgroup$
            – imranfat
            8 mins ago












          • $begingroup$
            I know, this was only a hint. The original function has no problem at $x=0$.
            $endgroup$
            – StackTD
            7 mins ago










          • $begingroup$
            @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
            $endgroup$
            – imranfat
            5 mins ago








          2




          2




          $begingroup$
          Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
          $endgroup$
          – kareem bokai
          1 hour ago




          $begingroup$
          Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
          $endgroup$
          – kareem bokai
          1 hour ago












          $begingroup$
          No smashing, please! You're welcome :).
          $endgroup$
          – StackTD
          1 hour ago




          $begingroup$
          No smashing, please! You're welcome :).
          $endgroup$
          – StackTD
          1 hour ago












          $begingroup$
          Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
          $endgroup$
          – imranfat
          8 mins ago






          $begingroup$
          Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
          $endgroup$
          – imranfat
          8 mins ago














          $begingroup$
          I know, this was only a hint. The original function has no problem at $x=0$.
          $endgroup$
          – StackTD
          7 mins ago




          $begingroup$
          I know, this was only a hint. The original function has no problem at $x=0$.
          $endgroup$
          – StackTD
          7 mins ago












          $begingroup$
          @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
          $endgroup$
          – imranfat
          5 mins ago




          $begingroup$
          @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
          $endgroup$
          – imranfat
          5 mins ago


















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