Condition on convergence of an integral












0












$begingroup$


enter image description here



I have solved the question. But I have problem with the condition on n.



This is what I think.



RHS converges if $n>1$(because of Zeta function).



For LHS to converge, the given function must tend to zero as $x$ tends to infinity. This is true for all $nin R$. Also, the given function must converge to some value when $x$ tends to $0$. When $n=2$ the function on LHS tends to $1$ when $x$ tends to $0$. When $n>2$ it tends to zero. For $n<2$ it tends to infinity. So the basic conditions for convergence of integral on LHS are $ngeq 2$ (I have not checked convergence with any test since only this condition gives me the answer).



Why do the conditions on $n$ do not match on both sides of the equation?






convergence riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
    $endgroup$
    – b00n heT
    Jan 5 at 17:36










  • $begingroup$
    Yeah, that's true.
    $endgroup$
    – Asit Srivastava
    Jan 5 at 17:37






  • 1




    $begingroup$
    $displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 5 at 19:47


















0












$begingroup$


enter image description here



I have solved the question. But I have problem with the condition on n.



This is what I think.



RHS converges if $n>1$(because of Zeta function).



For LHS to converge, the given function must tend to zero as $x$ tends to infinity. This is true for all $nin R$. Also, the given function must converge to some value when $x$ tends to $0$. When $n=2$ the function on LHS tends to $1$ when $x$ tends to $0$. When $n>2$ it tends to zero. For $n<2$ it tends to infinity. So the basic conditions for convergence of integral on LHS are $ngeq 2$ (I have not checked convergence with any test since only this condition gives me the answer).



Why do the conditions on $n$ do not match on both sides of the equation?






convergence riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
    $endgroup$
    – b00n heT
    Jan 5 at 17:36










  • $begingroup$
    Yeah, that's true.
    $endgroup$
    – Asit Srivastava
    Jan 5 at 17:37






  • 1




    $begingroup$
    $displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 5 at 19:47
















0












0








0





$begingroup$


enter image description here



I have solved the question. But I have problem with the condition on n.



This is what I think.



RHS converges if $n>1$(because of Zeta function).



For LHS to converge, the given function must tend to zero as $x$ tends to infinity. This is true for all $nin R$. Also, the given function must converge to some value when $x$ tends to $0$. When $n=2$ the function on LHS tends to $1$ when $x$ tends to $0$. When $n>2$ it tends to zero. For $n<2$ it tends to infinity. So the basic conditions for convergence of integral on LHS are $ngeq 2$ (I have not checked convergence with any test since only this condition gives me the answer).



Why do the conditions on $n$ do not match on both sides of the equation?






convergence riemann-zeta







share|cite|improve this question











$endgroup$




enter image description here



I have solved the question. But I have problem with the condition on n.



This is what I think.



RHS converges if $n>1$(because of Zeta function).



For LHS to converge, the given function must tend to zero as $x$ tends to infinity. This is true for all $nin R$. Also, the given function must converge to some value when $x$ tends to $0$. When $n=2$ the function on LHS tends to $1$ when $x$ tends to $0$. When $n>2$ it tends to zero. For $n<2$ it tends to infinity. So the basic conditions for convergence of integral on LHS are $ngeq 2$ (I have not checked convergence with any test since only this condition gives me the answer).



Why do the conditions on $n$ do not match on both sides of the equation?






convergence riemann-zeta




convergence riemann-zeta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 17:37







Asit Srivastava

















asked Jan 5 at 17:31









Asit SrivastavaAsit Srivastava

257




257












  • $begingroup$
    The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
    $endgroup$
    – b00n heT
    Jan 5 at 17:36










  • $begingroup$
    Yeah, that's true.
    $endgroup$
    – Asit Srivastava
    Jan 5 at 17:37






  • 1




    $begingroup$
    $displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 5 at 19:47




















  • $begingroup$
    The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
    $endgroup$
    – b00n heT
    Jan 5 at 17:36










  • $begingroup$
    Yeah, that's true.
    $endgroup$
    – Asit Srivastava
    Jan 5 at 17:37






  • 1




    $begingroup$
    $displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
    $endgroup$
    – Felix Marin
    Jan 5 at 19:47


















$begingroup$
The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
$endgroup$
– b00n heT
Jan 5 at 17:36




$begingroup$
The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
$endgroup$
– b00n heT
Jan 5 at 17:36












$begingroup$
Yeah, that's true.
$endgroup$
– Asit Srivastava
Jan 5 at 17:37




$begingroup$
Yeah, that's true.
$endgroup$
– Asit Srivastava
Jan 5 at 17:37




1




1




$begingroup$
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
$endgroup$
– Felix Marin
Jan 5 at 19:47






$begingroup$
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
$endgroup$
– Felix Marin
Jan 5 at 19:47












1 Answer
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$begingroup$

In fact, the LHS integral is finite for $n=1+epsilon$, $forallepsilon>0$. To see this, note that the integrand behaves like $frac{1}{x^{1-epsilon}}$ near $x=0$. Since
$$
int_0^1 frac{1}{x^{1-epsilon}}dx = frac{1}{epsilon}x^{epsilon}Big|^1_0=frac{1}{epsilon}<infty,
$$ by comparison test, we have $$int_0^1 frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty.$$ And since the integrand decays exponentially, we have
$$
int_0^infty frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty,quadforallepsilon>0.
$$ This shows that the both sides converge for $n>1$.






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    $begingroup$

    In fact, the LHS integral is finite for $n=1+epsilon$, $forallepsilon>0$. To see this, note that the integrand behaves like $frac{1}{x^{1-epsilon}}$ near $x=0$. Since
    $$
    int_0^1 frac{1}{x^{1-epsilon}}dx = frac{1}{epsilon}x^{epsilon}Big|^1_0=frac{1}{epsilon}<infty,
    $$     by comparison test, we have $$int_0^1 frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty.$$ And since the integrand decays exponentially, we have
    $$
    int_0^infty frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty,quadforallepsilon>0.
    $$     This shows that the both sides converge for $n>1$.






    share|cite|improve this answer









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      1












      $begingroup$

      In fact, the LHS integral is finite for $n=1+epsilon$, $forallepsilon>0$. To see this, note that the integrand behaves like $frac{1}{x^{1-epsilon}}$ near $x=0$. Since
      $$
      int_0^1 frac{1}{x^{1-epsilon}}dx = frac{1}{epsilon}x^{epsilon}Big|^1_0=frac{1}{epsilon}<infty,
      $$       by comparison test, we have $$int_0^1 frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty.$$ And since the integrand decays exponentially, we have
      $$
      int_0^infty frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty,quadforallepsilon>0.
      $$       This shows that the both sides converge for $n>1$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        In fact, the LHS integral is finite for $n=1+epsilon$, $forallepsilon>0$. To see this, note that the integrand behaves like $frac{1}{x^{1-epsilon}}$ near $x=0$. Since
        $$
        int_0^1 frac{1}{x^{1-epsilon}}dx = frac{1}{epsilon}x^{epsilon}Big|^1_0=frac{1}{epsilon}<infty,
        $$         by comparison test, we have $$int_0^1 frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty.$$ And since the integrand decays exponentially, we have
        $$
        int_0^infty frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty,quadforallepsilon>0.
        $$         This shows that the both sides converge for $n>1$.






        share|cite|improve this answer









        $endgroup$



        In fact, the LHS integral is finite for $n=1+epsilon$, $forallepsilon>0$. To see this, note that the integrand behaves like $frac{1}{x^{1-epsilon}}$ near $x=0$. Since
        $$
        int_0^1 frac{1}{x^{1-epsilon}}dx = frac{1}{epsilon}x^{epsilon}Big|^1_0=frac{1}{epsilon}<infty,
        $$         by comparison test, we have $$int_0^1 frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty.$$ And since the integrand decays exponentially, we have
        $$
        int_0^infty frac{x^{1+epsilon}e^x}{(e^x -1)^2}dx<infty,quadforallepsilon>0.
        $$         This shows that the both sides converge for $n>1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 19:07









        SongSong

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