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












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

35.5k22557




35.5k22557










asked Jan 16 at 15:57









Oria GruberOria Gruber

6,53732462




6,53732462












  • $begingroup$
    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

18.5k21651




18.5k21651












  • $begingroup$
    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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Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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