Find $lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$
$begingroup$
$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$
This seems like an apt situation to utilize dominating convergence.
$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.
$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$
Any hints?
real-analysis integration measure-theory convergence
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
$endgroup$
|
show 5 more comments
$begingroup$
$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$
This seems like an apt situation to utilize dominating convergence.
$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.
$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$
Any hints?
real-analysis integration measure-theory convergence
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
$endgroup$
1
$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27
$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46
1
$begingroup$
I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
$endgroup$
– SABOY
Dec 4 '18 at 20:49
1
$begingroup$
maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
$endgroup$
– Masacroso
Dec 4 '18 at 23:01
1
$begingroup$
@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
$endgroup$
– Masacroso
Dec 4 '18 at 23:07
|
show 5 more comments
$begingroup$
$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$
This seems like an apt situation to utilize dominating convergence.
$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.
$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$
Any hints?
real-analysis integration measure-theory convergence
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
$endgroup$
$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$
This seems like an apt situation to utilize dominating convergence.
$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.
$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$
Any hints?
real-analysis integration measure-theory convergence
real-analysis integration measure-theory convergence
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
asked Dec 4 '18 at 20:20
SABOYSABOY
588311
588311
1
$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27
$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46
1
$begingroup$
I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
$endgroup$
– SABOY
Dec 4 '18 at 20:49
1
$begingroup$
maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
$endgroup$
– Masacroso
Dec 4 '18 at 23:01
1
$begingroup$
@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
$endgroup$
– Masacroso
Dec 4 '18 at 23:07
|
show 5 more comments
1
$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27
$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46
1
$begingroup$
I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
$endgroup$
– SABOY
Dec 4 '18 at 20:49
1
$begingroup$
maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
$endgroup$
– Masacroso
Dec 4 '18 at 23:01
1
$begingroup$
@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
$endgroup$
– Masacroso
Dec 4 '18 at 23:07
1
1
$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27
$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27
$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46
$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46
1
1
$begingroup$
I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
$endgroup$
– SABOY
Dec 4 '18 at 20:49
$begingroup$
I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
$endgroup$
– SABOY
Dec 4 '18 at 20:49
1
1
$begingroup$
maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
$endgroup$
– Masacroso
Dec 4 '18 at 23:01
$begingroup$
maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
$endgroup$
– Masacroso
Dec 4 '18 at 23:01
1
1
$begingroup$
@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
$endgroup$
– Masacroso
Dec 4 '18 at 23:07
$begingroup$
@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
$endgroup$
– Masacroso
Dec 4 '18 at 23:07
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Note that for $x > 0$ we have
$$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
Thus for all $x >0$
$$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
$$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
and we cannot apply the dominated convergence theorem.
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
$endgroup$
add a comment |
$begingroup$
To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.
You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
Note that for $x > 0$ we have
$$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
Thus for all $x >0$
$$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
$$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
and we cannot apply the dominated convergence theorem.
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
$endgroup$
add a comment |
$begingroup$
Note that for $x > 0$ we have
$$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
Thus for all $x >0$
$$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
$$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
and we cannot apply the dominated convergence theorem.
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
$endgroup$
add a comment |
$begingroup$
Note that for $x > 0$ we have
$$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
Thus for all $x >0$
$$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
$$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
and we cannot apply the dominated convergence theorem.
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
$endgroup$
Note that for $x > 0$ we have
$$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
Thus for all $x >0$
$$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
$$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
and we cannot apply the dominated convergence theorem.
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
answered Dec 4 '18 at 20:27
p4schp4sch
4,995217
4,995217
add a comment |
add a comment |
$begingroup$
To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.
You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
$endgroup$
add a comment |
$begingroup$
To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.
You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
$endgroup$
add a comment |
$begingroup$
To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.
You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
$endgroup$
To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.
You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
answered Dec 4 '18 at 20:27
Umberto P.Umberto P.
38.8k13064
38.8k13064
add a comment |
add a comment |
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$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27
$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46
1
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I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
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– SABOY
Dec 4 '18 at 20:49
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maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
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– Masacroso
Dec 4 '18 at 23:01
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@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
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– Masacroso
Dec 4 '18 at 23:07