Determining if this integral converge












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Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus







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  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40


















0












$begingroup$


Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40
















0












0








0


1



$begingroup$


Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus







share|cite|improve this question









$endgroup$




Does the following integral converge:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx $$



my attempt:



$$ int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx = int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx ge int_1^infty frac{-x}{xsqrt{x^2+1}}dx = int_1^infty frac{-1}{sqrt{x^2+1}}dx ge -1 cdot int_1^infty frac{1}{sqrt{x^2+x^2}}dx = -frac{1}{sqrt{2}} int_1^infty frac{1}{sqrt{x^2}} = -frac{1}{sqrt{2}} int_1^infty frac{1}{x} $$



and since I know $ frac{1}{x} $ doesn't converge, the integral also doesn't??






calculus




calculus






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asked Jan 9 at 14:36









bm1125bm1125

68516




68516








  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40
















  • 1




    $begingroup$
    No. You've shown the value of the integral is $ge-infty$ which doesn't help.
    $endgroup$
    – saulspatz
    Jan 9 at 14:40










1




1




$begingroup$
No. You've shown the value of the integral is $ge-infty$ which doesn't help.
$endgroup$
– saulspatz
Jan 9 at 14:40






$begingroup$
No. You've shown the value of the integral is $ge-infty$ which doesn't help.
$endgroup$
– saulspatz
Jan 9 at 14:40












2 Answers
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$begingroup$

begin{eqnarray*}
int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
& = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
& = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
& stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
end{eqnarray*}



So, it is convergent.






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    2












    $begingroup$

    For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






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      2 Answers
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      2 Answers
      2






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      1












      $begingroup$

      begin{eqnarray*}
      int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
      & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
      & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
      & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
      end{eqnarray*}



      So, it is convergent.






      share|cite|improve this answer









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        1












        $begingroup$

        begin{eqnarray*}
        int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
        & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
        & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
        & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
        end{eqnarray*}



        So, it is convergent.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          begin{eqnarray*}
          int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
          & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
          & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
          & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
          end{eqnarray*}



          So, it is convergent.






          share|cite|improve this answer









          $endgroup$



          begin{eqnarray*}
          int_1^infty left( frac{1}{x} - frac{1}{sqrt{x^2+1}} right)dx
          & = & int_1^infty frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}}dx \
          & = & int_1^infty frac{1}{xsqrt{x^2+1}(sqrt{x^2+1} + x)}dx \
          & stackrel{sqrt{x^2+1}geq x}{leq} & int_1^infty frac{1}{xcdot x (x+x)}dx = int_1^infty frac{1}{2x^3}dx < infty
          end{eqnarray*}



          So, it is convergent.







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          answered Jan 9 at 14:58









          trancelocationtrancelocation

          13.6k1828




          13.6k1828























              2












              $begingroup$

              For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$






                  share|cite|improve this answer









                  $endgroup$



                  For large $x$, we see by Taylor expansion that $$sqrt{x^2 + 1} = xsqrt{1 + frac 1 {x^2}} approx xleft(1 + frac 1 {2x^2}right).$$ Thus $$sqrt{x^2 + 1} - x approx frac1{2x},$$ so as $x to infty$, we have $$frac{sqrt{x^2+1}-x}{xsqrt{x^2+1}} approx frac{1/(2x)}{x^2}approx frac{1}{2x^{3}}.$$ Hence the integral converges by comparison with $$int^infty_{1}frac{dx}{2x^{3}}.$$







                  share|cite|improve this answer












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                  answered Jan 9 at 14:50









                  User8128User8128

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