Condition on convergence of an integral
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
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
add a comment |
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
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
– b00n heT
2 days ago
Yeah, that's true.
– Asit Srivastava
2 days ago
1
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
– Felix Marin
2 days ago
add a comment |
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
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
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
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
edited 2 days ago
Asit Srivastava
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
asked 2 days ago
Asit SrivastavaAsit Srivastava
257
257
The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
– b00n heT
2 days ago
Yeah, that's true.
– Asit Srivastava
2 days ago
1
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
– Felix Marin
2 days ago
add a comment |
The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
– b00n heT
2 days ago
Yeah, that's true.
– Asit Srivastava
2 days ago
1
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
– Felix Marin
2 days ago
The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
– b00n heT
2 days ago
The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
– b00n heT
2 days ago
Yeah, that's true.
– Asit Srivastava
2 days ago
Yeah, that's true.
– Asit Srivastava
2 days ago
1
1
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
– Felix Marin
2 days ago
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
– Felix Marin
2 days ago
add a comment |
1 Answer
1
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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 $n>1$.
answered 2 days ago
SongSong
6,115318
add a comment |
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1 Answer
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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 $n>1$.
answered 2 days ago
SongSong
6,115318
add a comment |
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 $n>1$.
answered 2 days ago
SongSong
6,115318
add a comment |
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 $n>1$.
answered 2 days ago
SongSong
6,115318
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 $n>1$.
answered 2 days ago
SongSong
6,115318
answered 2 days ago
SongSong
6,115318
answered 2 days ago
SongSong
6,115318
answered 2 days ago
SongSong
6,115318
6,115318
add a comment |
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The integrand actually converges to zero as $xtoinfty$ for any $n$ Due to the square in the denominator!
– b00n heT
2 days ago
Yeah, that's true.
– Asit Srivastava
2 days ago
1
$displaystyle{x^{n},mathrm{e}^{x} over left({mathrm{e}^{x} - 1}right)^{2}} sim x^{n - 2}$ as $displaystyle x to 0^{+}$. So, in order to converge it must be $displaystyle Releft({n - 2}right) > - 1 implies color{red}{Releft({n }right) > 1}$ !!!.
– Felix Marin
2 days ago