$int_{mathbb R^n}e^{{-|x|}^n}dx$












4












$begingroup$


I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08
















4












$begingroup$


I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08














4












4








4





$begingroup$


I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration







share|cite|improve this question











$endgroup$




I'm trying to prove that $int_{mathbb R^n}e^{{-|x|}^n}dx = V(B_{n}(0,1))$ using induction and Cavalieri's principle.



Notice that $int_{mathbb R} e^{-|x|}dx = 2 = V(B_1(0, 1))$ so the basis of our induction is true.



Now suppose that for some $k-1$ we have $int_{mathbb R^{k-1}}e^{{-|x|}^{k-1}}dx = V(B_{k-1}(0,1))$



We want to calculate $int_{mathbb R^{k}}e^{-{|x|}^{k}}dx$, we can choose to extract it by spheres since this is a positive function, so this will be equal to $$lim_{r to infty}int_{x_1^2+dots x_k^2 leq r^2}e^{-|x|^{k}}dx.$$



Consider the intersection of this sphere with the plane $x_k = t$. From Cavalieri's principle:



$$
int_{x_1^2+dots x_k^2 leq r^2} e^{-|x|^{k}}dx
= int_{-r}^r
left( int_{x_1^2+dots+x_{k-1}^2 leq r^2-t^2}
e^{-(x_1^2+dots+x_{k-1}^2+t^2)^{k/2}}dx_1dx_2dots dx_{k-1}
right) dt$$



Not sure where to go from here....






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 16:00









gt6989b

34k22455




34k22455










asked Jan 16 at 15:57









Oria GruberOria Gruber

6,44332460




6,44332460












  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08


















  • $begingroup$
    Make rather a polar coordinate change?
    $endgroup$
    – Mindlack
    Jan 16 at 16:08










  • $begingroup$
    I don't think that's the intended way. I might be wrong.
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:08
















$begingroup$
Make rather a polar coordinate change?
$endgroup$
– Mindlack
Jan 16 at 16:08




$begingroup$
Make rather a polar coordinate change?
$endgroup$
– Mindlack
Jan 16 at 16:08












$begingroup$
I don't think that's the intended way. I might be wrong.
$endgroup$
– Oria Gruber
Jan 16 at 16:08




$begingroup$
I don't think that's the intended way. I might be wrong.
$endgroup$
– Oria Gruber
Jan 16 at 16:08










1 Answer
1






active

oldest

votes


















3












$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04
















3












$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04














3












3








3





$begingroup$

I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.






share|cite|improve this answer











$endgroup$



I think Cavalieri's principle means this way:
$$
int_{Bbb R^n}e^{-|x|^n}mathrm dx =int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt.
$$ (Explanation: imagine we are given $z=e^{-x^2-y^2}$. We can calculate the volume under the graph by cutting it along $z=tin (0,1)$, calculating the area of the cross-section and re-gathering it.) This gives
$$begin{eqnarray}
int_0^1 text{vol}{x:e^{-|x|^n}ge t}mathrm dt&=& int_0^1 text{vol}left[(-ln t)^{frac{1}{n}}cdot B_nright]mathrm dt\&=&text{vol}(B_n) int_0^1 (-ln t)mathrm dt\&=&text{vol}(B_n),
end{eqnarray}$$where $B_n$ is a $n$-dimensional unit ball and $ccdot B_n = {cx;|;xin B_n}=B_n(0,c)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 16:21

























answered Jan 16 at 16:09









SongSong

12.2k630




12.2k630












  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04


















  • $begingroup$
    Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 16:13










  • $begingroup$
    @OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
    $endgroup$
    – Song
    Jan 16 at 16:14












  • $begingroup$
    @OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
    $endgroup$
    – mathworker21
    Jan 16 at 16:19










  • $begingroup$
    @mathworker21 Yes! That is what exactly I did. Thanks for clarification.
    $endgroup$
    – Song
    Jan 16 at 16:20










  • $begingroup$
    May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
    $endgroup$
    – Oria Gruber
    Jan 16 at 17:04
















$begingroup$
Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
$endgroup$
– Oria Gruber
Jan 16 at 16:13




$begingroup$
Could you elaborate please? Aren't we supposed to look at cross sections? The valid "height" or $x_k$ coordinate for these cross sections is between $(-r,r)$ as $r$ tends to infinity? why $(0,1)$?
$endgroup$
– Oria Gruber
Jan 16 at 16:13












$begingroup$
@OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
$endgroup$
– Song
Jan 16 at 16:14






$begingroup$
@OriaGruber Yes, we do the cross section horizontally. Since $0< e^{-|x|^n}le 1$, $0<tle 1$ is sufficient. For example, imagine $z=e^{-x^2-y^2}$. We can calculate the volume under the graph of it by cutting it along $z=c in (0,1)$ and gathering it. I hope this makes it clear.
$endgroup$
– Song
Jan 16 at 16:14














$begingroup$
@OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
$endgroup$
– mathworker21
Jan 16 at 16:19




$begingroup$
@OriaGruber You can prove $int_X fdmu = int_0^infty mu({|f| > lambda}) dlambda$ for any measure space $(X,mu)$ and any $f in L^1$, by first proving it for simple functions and then doing linear combinations and taking limits.
$endgroup$
– mathworker21
Jan 16 at 16:19












$begingroup$
@mathworker21 Yes! That is what exactly I did. Thanks for clarification.
$endgroup$
– Song
Jan 16 at 16:20




$begingroup$
@mathworker21 Yes! That is what exactly I did. Thanks for clarification.
$endgroup$
– Song
Jan 16 at 16:20












$begingroup$
May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
$endgroup$
– Oria Gruber
Jan 16 at 17:04




$begingroup$
May I ask one more question @Song - Why $geq t$? Shouldn't it be equal $t$ since we are looking at a cross section with height $t$?
$endgroup$
– Oria Gruber
Jan 16 at 17:04


















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The Binding of Isaac: Rebirth/Afterbirth

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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