About the integrabity of $f(x) = |x|^{-(n+1)}$ and $f(x) = |x|^{-n}$ on $mathbb{R}^n$
$begingroup$
The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation
$$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
and let $epsilon to 0$
Please help me to verify my argument.
Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.
measure-theory proof-verification
asked Jan 14 at 9:14
ElementXElementX
404111
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add a comment |
$begingroup$
The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation
$$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
and let $epsilon to 0$
Please help me to verify my argument.
Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.
measure-theory proof-verification
asked Jan 14 at 9:14
ElementXElementX
404111
$endgroup$
add a comment |
$begingroup$
The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation
$$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
and let $epsilon to 0$
Please help me to verify my argument.
Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.
measure-theory proof-verification
asked Jan 14 at 9:14
ElementXElementX
404111
$endgroup$
The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation
$$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
and let $epsilon to 0$
Please help me to verify my argument.
Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.
measure-theory proof-verification
measure-theory proof-verification
asked Jan 14 at 9:14
ElementXElementX
404111
asked Jan 14 at 9:14
ElementXElementX
404111
asked Jan 14 at 9:14
ElementXElementX
404111
asked Jan 14 at 9:14
ElementXElementX
404111
asked Jan 14 at 9:14
ElementXElementX
404111
404111
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
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add a comment |
$begingroup$
I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
$$
When $k=1$, it becomes log-type and the integral also blows up.
So the result follows immediately. ;)
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
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add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
$begingroup$
Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
$begingroup$
Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
edited Jan 14 at 9:32
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 14 at 9:22
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
69.1k53169
add a comment |
add a comment |
$begingroup$
I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
$$
When $k=1$, it becomes log-type and the integral also blows up.
So the result follows immediately. ;)
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
$endgroup$
add a comment |
$begingroup$
I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
$$
When $k=1$, it becomes log-type and the integral also blows up.
So the result follows immediately. ;)
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
$endgroup$
add a comment |
$begingroup$
I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
$$
When $k=1$, it becomes log-type and the integral also blows up.
So the result follows immediately. ;)
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
$endgroup$
I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
$$
When $k=1$, it becomes log-type and the integral also blows up.
So the result follows immediately. ;)
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
answered Jan 14 at 9:33
Zixiao_LiuZixiao_Liu
938
938
add a comment |
add a comment |
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