Computing the Lebesgue integral over a ball
$begingroup$
Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$
For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$
and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?
Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?
I tried:
$f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.
So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.
For computing the integrals, I thought about using the transformation formula:
$int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$
Here I don't know how to continue. Is this method correct or is there another way to do this?
measure-theory lebesgue-integral lebesgue-measure
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
$endgroup$
add a comment |
$begingroup$
Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$
For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$
and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?
Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?
I tried:
$f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.
So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.
For computing the integrals, I thought about using the transformation formula:
$int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$
Here I don't know how to continue. Is this method correct or is there another way to do this?
measure-theory lebesgue-integral lebesgue-measure
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
$endgroup$
add a comment |
$begingroup$
Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$
For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$
and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?
Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?
I tried:
$f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.
So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.
For computing the integrals, I thought about using the transformation formula:
$int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$
Here I don't know how to continue. Is this method correct or is there another way to do this?
measure-theory lebesgue-integral lebesgue-measure
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
$endgroup$
Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.
Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$
For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$
and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?
Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?
I tried:
$f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.
So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.
For computing the integrals, I thought about using the transformation formula:
$int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$
Here I don't know how to continue. Is this method correct or is there another way to do this?
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
edited Dec 1 '18 at 16:52
Olsgur
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
asked Dec 1 '18 at 13:48
OlsgurOlsgur
564
564
add a comment |
add a comment |
1 Answer
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$begingroup$
Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
$$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
$$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
$$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
$$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
$endgroup$
Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
$$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
answered Dec 1 '18 at 17:03
Christian BlatterChristian Blatter
173k7113326
173k7113326
add a comment |
add a comment |
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