Convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$ by Asymptotic Comparison Test.












1












$begingroup$


According to the material I have,

If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $

Then:
i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$

Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$

In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$

Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$






asymptotics improper-integrals







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    According to the material I have,

    If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $

    Then:
    i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
    ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$

    Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$

    In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$

    Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$






    asymptotics improper-integrals







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      According to the material I have,

      If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $

      Then:
      i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
      ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$

      Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$

      In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$

      Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$






      asymptotics improper-integrals







      share|cite|improve this question









      $endgroup$




      According to the material I have,

      If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $

      Then:
      i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
      ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$

      Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$

      In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$

      Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$






      asymptotics improper-integrals




      asymptotics improper-integrals






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 14:23









      Turan NəsibliTuran Nəsibli

      294




      294






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.



          Rewrite, for $x>1$,
          $$
          sqrt{ln x} = sqrt{ln(1+(x-1))},.
          $$
          This seems silly, but now, recall that
          $$
          lim_{uto 0} frac{ln(1+u)}{u} = 1
          $$
          so that
          $$
          lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
          $$



          Can you conclude?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:42












          • $begingroup$
            You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
            $endgroup$
            – Clement C.
            Jan 8 at 15:44












          • $begingroup$
            I'm trying to fix the problem,and thank you for the help.
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:48










          • $begingroup$
            @TuranNəsibli You're welcome! (for the Latex issue: us a^{-1}, with braces.)
            $endgroup$
            – Clement C.
            Jan 8 at 15:48











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          1 Answer
          1






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          active

          oldest

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          2












          $begingroup$

          An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.



          Rewrite, for $x>1$,
          $$
          sqrt{ln x} = sqrt{ln(1+(x-1))},.
          $$
          This seems silly, but now, recall that
          $$
          lim_{uto 0} frac{ln(1+u)}{u} = 1
          $$
          so that
          $$
          lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
          $$



          Can you conclude?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:42












          • $begingroup$
            You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
            $endgroup$
            – Clement C.
            Jan 8 at 15:44












          • $begingroup$
            I'm trying to fix the problem,and thank you for the help.
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:48










          • $begingroup$
            @TuranNəsibli You're welcome! (for the Latex issue: us a^{-1}, with braces.)
            $endgroup$
            – Clement C.
            Jan 8 at 15:48
















          2












          $begingroup$

          An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.



          Rewrite, for $x>1$,
          $$
          sqrt{ln x} = sqrt{ln(1+(x-1))},.
          $$
          This seems silly, but now, recall that
          $$
          lim_{uto 0} frac{ln(1+u)}{u} = 1
          $$
          so that
          $$
          lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
          $$



          Can you conclude?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:42












          • $begingroup$
            You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
            $endgroup$
            – Clement C.
            Jan 8 at 15:44












          • $begingroup$
            I'm trying to fix the problem,and thank you for the help.
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:48










          • $begingroup$
            @TuranNəsibli You're welcome! (for the Latex issue: us a^{-1}, with braces.)
            $endgroup$
            – Clement C.
            Jan 8 at 15:48














          2












          2








          2





          $begingroup$

          An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.



          Rewrite, for $x>1$,
          $$
          sqrt{ln x} = sqrt{ln(1+(x-1))},.
          $$
          This seems silly, but now, recall that
          $$
          lim_{uto 0} frac{ln(1+u)}{u} = 1
          $$
          so that
          $$
          lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
          $$



          Can you conclude?






          share|cite|improve this answer









          $endgroup$



          An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.



          Rewrite, for $x>1$,
          $$
          sqrt{ln x} = sqrt{ln(1+(x-1))},.
          $$
          This seems silly, but now, recall that
          $$
          lim_{uto 0} frac{ln(1+u)}{u} = 1
          $$
          so that
          $$
          lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
          $$



          Can you conclude?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 14:49









          Clement C.Clement C.

          49.8k33886




          49.8k33886












          • $begingroup$
            Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:42












          • $begingroup$
            You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
            $endgroup$
            – Clement C.
            Jan 8 at 15:44












          • $begingroup$
            I'm trying to fix the problem,and thank you for the help.
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:48










          • $begingroup$
            @TuranNəsibli You're welcome! (for the Latex issue: us a^{-1}, with braces.)
            $endgroup$
            – Clement C.
            Jan 8 at 15:48


















          • $begingroup$
            Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:42












          • $begingroup$
            You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
            $endgroup$
            – Clement C.
            Jan 8 at 15:44












          • $begingroup$
            I'm trying to fix the problem,and thank you for the help.
            $endgroup$
            – Turan Nəsibli
            Jan 8 at 15:48










          • $begingroup$
            @TuranNəsibli You're welcome! (for the Latex issue: us a^{-1}, with braces.)
            $endgroup$
            – Clement C.
            Jan 8 at 15:48
















          $begingroup$
          Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
          $endgroup$
          – Turan Nəsibli
          Jan 8 at 15:42






          $begingroup$
          Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
          $endgroup$
          – Turan Nəsibli
          Jan 8 at 15:42














          $begingroup$
          You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
          $endgroup$
          – Clement C.
          Jan 8 at 15:44






          $begingroup$
          You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
          $endgroup$
          – Clement C.
          Jan 8 at 15:44














          $begingroup$
          I'm trying to fix the problem,and thank you for the help.
          $endgroup$
          – Turan Nəsibli
          Jan 8 at 15:48




          $begingroup$
          I'm trying to fix the problem,and thank you for the help.
          $endgroup$
          – Turan Nəsibli
          Jan 8 at 15:48












          $begingroup$
          @TuranNəsibli You're welcome! (for the Latex issue: us a^{-1}, with braces.)
          $endgroup$
          – Clement C.
          Jan 8 at 15:48




          $begingroup$
          @TuranNəsibli You're welcome! (for the Latex issue: us a^{-1}, with braces.)
          $endgroup$
          – Clement C.
          Jan 8 at 15:48


















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