How to decide convergence of Integrals












4












$begingroup$


I got this doubt while evaluating the integrals:



$$I=int_{0}^{frac{pi}{2}}ln(sin x)sin xdx$$ and



$$J=int_{0}^{frac{pi}{4}}csc xdx$$



Now even though the integrand $f(x)=ln(sin x)sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.



But integrand in $J$ is not defined at $x=0$ and integral is infinite.



So how to identify without explicitly evaluating?






integration algebra-precalculus convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    Comparison test is definite a way to go.
    $endgroup$
    – Sangchul Lee
    11 hours ago
















4












$begingroup$


I got this doubt while evaluating the integrals:



$$I=int_{0}^{frac{pi}{2}}ln(sin x)sin xdx$$ and



$$J=int_{0}^{frac{pi}{4}}csc xdx$$



Now even though the integrand $f(x)=ln(sin x)sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.



But integrand in $J$ is not defined at $x=0$ and integral is infinite.



So how to identify without explicitly evaluating?






integration algebra-precalculus convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    Comparison test is definite a way to go.
    $endgroup$
    – Sangchul Lee
    11 hours ago














4












4








4





$begingroup$


I got this doubt while evaluating the integrals:



$$I=int_{0}^{frac{pi}{2}}ln(sin x)sin xdx$$ and



$$J=int_{0}^{frac{pi}{4}}csc xdx$$



Now even though the integrand $f(x)=ln(sin x)sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.



But integrand in $J$ is not defined at $x=0$ and integral is infinite.



So how to identify without explicitly evaluating?






integration algebra-precalculus convergence







share|cite|improve this question









$endgroup$




I got this doubt while evaluating the integrals:



$$I=int_{0}^{frac{pi}{2}}ln(sin x)sin xdx$$ and



$$J=int_{0}^{frac{pi}{4}}csc xdx$$



Now even though the integrand $f(x)=ln(sin x)sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.



But integrand in $J$ is not defined at $x=0$ and integral is infinite.



So how to identify without explicitly evaluating?






integration algebra-precalculus convergence




integration algebra-precalculus convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 11 hours ago









Umesh shankarUmesh shankar

3,04431220




3,04431220












  • $begingroup$
    Comparison test is definite a way to go.
    $endgroup$
    – Sangchul Lee
    11 hours ago


















  • $begingroup$
    Comparison test is definite a way to go.
    $endgroup$
    – Sangchul Lee
    11 hours ago
















$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
11 hours ago




$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
11 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

Notice, however, that



begin{eqnarray}
lim_{xto0^+}ln(sin x)sin x&=&lim_{xto0^+}frac{ln(sin x)}{csc x}\
&=&lim_{xto0^+}frac{cot x}{(-csc xcot x)}\
&=&lim_{xto0^+}(-sin x)\
&=&0
end{eqnarray}



Here is the graph of $y=ln(sin x)sin x$



graph of ln(sin x)(sin x)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So does it mean if limit is finite at the end points, integral exists?
    $endgroup$
    – Umesh shankar
    11 hours ago










  • $begingroup$
    In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac{1}{sqrt{x}},dx$.
    $endgroup$
    – John Wayland Bales
    10 hours ago





















0












$begingroup$

For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac{1}{sin x} = frac{1}{x}$. From the examples $int_0^1 x^{-a} dx$ we know that $a=1$ is divergent, albeit borderline so. Since
$$
sin x = x - frac{1}{3!}x^3 pm ...
$$
we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac{1}{sin x} geq frac{1}{x}$ and the integral $J$ is divergent.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Notice, however, that



    begin{eqnarray}
    lim_{xto0^+}ln(sin x)sin x&=&lim_{xto0^+}frac{ln(sin x)}{csc x}\
    &=&lim_{xto0^+}frac{cot x}{(-csc xcot x)}\
    &=&lim_{xto0^+}(-sin x)\
    &=&0
    end{eqnarray}



    Here is the graph of $y=ln(sin x)sin x$



    graph of ln(sin x)(sin x)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So does it mean if limit is finite at the end points, integral exists?
      $endgroup$
      – Umesh shankar
      11 hours ago










    • $begingroup$
      In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac{1}{sqrt{x}},dx$.
      $endgroup$
      – John Wayland Bales
      10 hours ago


















    5












    $begingroup$

    Notice, however, that



    begin{eqnarray}
    lim_{xto0^+}ln(sin x)sin x&=&lim_{xto0^+}frac{ln(sin x)}{csc x}\
    &=&lim_{xto0^+}frac{cot x}{(-csc xcot x)}\
    &=&lim_{xto0^+}(-sin x)\
    &=&0
    end{eqnarray}



    Here is the graph of $y=ln(sin x)sin x$



    graph of ln(sin x)(sin x)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So does it mean if limit is finite at the end points, integral exists?
      $endgroup$
      – Umesh shankar
      11 hours ago










    • $begingroup$
      In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac{1}{sqrt{x}},dx$.
      $endgroup$
      – John Wayland Bales
      10 hours ago
















    5












    5








    5





    $begingroup$

    Notice, however, that



    begin{eqnarray}
    lim_{xto0^+}ln(sin x)sin x&=&lim_{xto0^+}frac{ln(sin x)}{csc x}\
    &=&lim_{xto0^+}frac{cot x}{(-csc xcot x)}\
    &=&lim_{xto0^+}(-sin x)\
    &=&0
    end{eqnarray}



    Here is the graph of $y=ln(sin x)sin x$



    graph of ln(sin x)(sin x)






    share|cite|improve this answer











    $endgroup$



    Notice, however, that



    begin{eqnarray}
    lim_{xto0^+}ln(sin x)sin x&=&lim_{xto0^+}frac{ln(sin x)}{csc x}\
    &=&lim_{xto0^+}frac{cot x}{(-csc xcot x)}\
    &=&lim_{xto0^+}(-sin x)\
    &=&0
    end{eqnarray}



    Here is the graph of $y=ln(sin x)sin x$



    graph of ln(sin x)(sin x)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 11 hours ago

























    answered 11 hours ago









    John Wayland BalesJohn Wayland Bales

    15k21238




    15k21238












    • $begingroup$
      So does it mean if limit is finite at the end points, integral exists?
      $endgroup$
      – Umesh shankar
      11 hours ago










    • $begingroup$
      In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac{1}{sqrt{x}},dx$.
      $endgroup$
      – John Wayland Bales
      10 hours ago




















    • $begingroup$
      So does it mean if limit is finite at the end points, integral exists?
      $endgroup$
      – Umesh shankar
      11 hours ago










    • $begingroup$
      In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac{1}{sqrt{x}},dx$.
      $endgroup$
      – John Wayland Bales
      10 hours ago


















    $begingroup$
    So does it mean if limit is finite at the end points, integral exists?
    $endgroup$
    – Umesh shankar
    11 hours ago




    $begingroup$
    So does it mean if limit is finite at the end points, integral exists?
    $endgroup$
    – Umesh shankar
    11 hours ago












    $begingroup$
    In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac{1}{sqrt{x}},dx$.
    $endgroup$
    – John Wayland Bales
    10 hours ago






    $begingroup$
    In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac{1}{sqrt{x}},dx$.
    $endgroup$
    – John Wayland Bales
    10 hours ago













    0












    $begingroup$

    For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac{1}{sin x} = frac{1}{x}$. From the examples $int_0^1 x^{-a} dx$ we know that $a=1$ is divergent, albeit borderline so. Since
    $$
    sin x = x - frac{1}{3!}x^3 pm ...
    $$
    we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac{1}{sin x} geq frac{1}{x}$ and the integral $J$ is divergent.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac{1}{sin x} = frac{1}{x}$. From the examples $int_0^1 x^{-a} dx$ we know that $a=1$ is divergent, albeit borderline so. Since
      $$
      sin x = x - frac{1}{3!}x^3 pm ...
      $$
      we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac{1}{sin x} geq frac{1}{x}$ and the integral $J$ is divergent.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac{1}{sin x} = frac{1}{x}$. From the examples $int_0^1 x^{-a} dx$ we know that $a=1$ is divergent, albeit borderline so. Since
        $$
        sin x = x - frac{1}{3!}x^3 pm ...
        $$
        we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac{1}{sin x} geq frac{1}{x}$ and the integral $J$ is divergent.






        share|cite|improve this answer











        $endgroup$



        For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac{1}{sin x} = frac{1}{x}$. From the examples $int_0^1 x^{-a} dx$ we know that $a=1$ is divergent, albeit borderline so. Since
        $$
        sin x = x - frac{1}{3!}x^3 pm ...
        $$
        we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac{1}{sin x} geq frac{1}{x}$ and the integral $J$ is divergent.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 10 hours ago

























        answered 10 hours ago









        user626368user626368

        193




        193






























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