Use of Tonelli's Theorem to calculate $int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x))...












0












$begingroup$


Define $f: [0,infty[times [0,infty[ to bar{mathbb R}$, $f(x,y):= xe^{-x^{2}(1+y^{2})}$ it is clear that $f(x,y)geq 0$ $lambda-$a.e. Therefore, Tonelli's theorem holds. This means:



$int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x)) dlambda (y)=int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)) dlambda (x)$



Looking exclusively at $int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)$, the only thing that comes to mind is "Riemann Integration", namely:



$int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)=lim_{nto infty}int_{[0,n]}xe^{-x^{2}(1+y^{2})}dy=lim_{nto infty}frac{1}{-xtimes2y}e^{-x^2(1+y^2)}vert_{0}^{n}$



Which gets me stuck as I have a $y$ in the denominator, however, I see no other way of doing it. Any other ideas?






real-analysis integration measure-theory multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you integrate over $y$ first, then it gives $frac{1}{2x}I$ where $I = int_{-infty}^infty e^{-t^2}dt$ is the Gaussian integral.
    $endgroup$
    – Song
    Dec 17 '18 at 20:51












  • $begingroup$
    Why $int^{infty}_{-infty}e^{-t^{2}}dt$ and not $int^{infty}_{0}e^{-t^{2}}dt$?
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:03












  • $begingroup$
    The factor $1/2$ is in front. Typically Gaussian integral refers to the former.
    $endgroup$
    – Song
    Dec 17 '18 at 21:07
















0












$begingroup$


Define $f: [0,infty[times [0,infty[ to bar{mathbb R}$, $f(x,y):= xe^{-x^{2}(1+y^{2})}$ it is clear that $f(x,y)geq 0$ $lambda-$a.e. Therefore, Tonelli's theorem holds. This means:



$int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x)) dlambda (y)=int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)) dlambda (x)$



Looking exclusively at $int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)$, the only thing that comes to mind is "Riemann Integration", namely:



$int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)=lim_{nto infty}int_{[0,n]}xe^{-x^{2}(1+y^{2})}dy=lim_{nto infty}frac{1}{-xtimes2y}e^{-x^2(1+y^2)}vert_{0}^{n}$



Which gets me stuck as I have a $y$ in the denominator, however, I see no other way of doing it. Any other ideas?






real-analysis integration measure-theory multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you integrate over $y$ first, then it gives $frac{1}{2x}I$ where $I = int_{-infty}^infty e^{-t^2}dt$ is the Gaussian integral.
    $endgroup$
    – Song
    Dec 17 '18 at 20:51












  • $begingroup$
    Why $int^{infty}_{-infty}e^{-t^{2}}dt$ and not $int^{infty}_{0}e^{-t^{2}}dt$?
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:03












  • $begingroup$
    The factor $1/2$ is in front. Typically Gaussian integral refers to the former.
    $endgroup$
    – Song
    Dec 17 '18 at 21:07














0












0








0





$begingroup$


Define $f: [0,infty[times [0,infty[ to bar{mathbb R}$, $f(x,y):= xe^{-x^{2}(1+y^{2})}$ it is clear that $f(x,y)geq 0$ $lambda-$a.e. Therefore, Tonelli's theorem holds. This means:



$int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x)) dlambda (y)=int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)) dlambda (x)$



Looking exclusively at $int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)$, the only thing that comes to mind is "Riemann Integration", namely:



$int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)=lim_{nto infty}int_{[0,n]}xe^{-x^{2}(1+y^{2})}dy=lim_{nto infty}frac{1}{-xtimes2y}e^{-x^2(1+y^2)}vert_{0}^{n}$



Which gets me stuck as I have a $y$ in the denominator, however, I see no other way of doing it. Any other ideas?






real-analysis integration measure-theory multivariable-calculus







share|cite|improve this question











$endgroup$




Define $f: [0,infty[times [0,infty[ to bar{mathbb R}$, $f(x,y):= xe^{-x^{2}(1+y^{2})}$ it is clear that $f(x,y)geq 0$ $lambda-$a.e. Therefore, Tonelli's theorem holds. This means:



$int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x)) dlambda (y)=int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)) dlambda (x)$



Looking exclusively at $int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)$, the only thing that comes to mind is "Riemann Integration", namely:



$int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)=lim_{nto infty}int_{[0,n]}xe^{-x^{2}(1+y^{2})}dy=lim_{nto infty}frac{1}{-xtimes2y}e^{-x^2(1+y^2)}vert_{0}^{n}$



Which gets me stuck as I have a $y$ in the denominator, however, I see no other way of doing it. Any other ideas?






real-analysis integration measure-theory multivariable-calculus




real-analysis integration measure-theory multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 20:38







SABOY

















asked Dec 17 '18 at 20:21









SABOYSABOY

656311




656311












  • $begingroup$
    If you integrate over $y$ first, then it gives $frac{1}{2x}I$ where $I = int_{-infty}^infty e^{-t^2}dt$ is the Gaussian integral.
    $endgroup$
    – Song
    Dec 17 '18 at 20:51












  • $begingroup$
    Why $int^{infty}_{-infty}e^{-t^{2}}dt$ and not $int^{infty}_{0}e^{-t^{2}}dt$?
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:03












  • $begingroup$
    The factor $1/2$ is in front. Typically Gaussian integral refers to the former.
    $endgroup$
    – Song
    Dec 17 '18 at 21:07


















  • $begingroup$
    If you integrate over $y$ first, then it gives $frac{1}{2x}I$ where $I = int_{-infty}^infty e^{-t^2}dt$ is the Gaussian integral.
    $endgroup$
    – Song
    Dec 17 '18 at 20:51












  • $begingroup$
    Why $int^{infty}_{-infty}e^{-t^{2}}dt$ and not $int^{infty}_{0}e^{-t^{2}}dt$?
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:03












  • $begingroup$
    The factor $1/2$ is in front. Typically Gaussian integral refers to the former.
    $endgroup$
    – Song
    Dec 17 '18 at 21:07
















$begingroup$
If you integrate over $y$ first, then it gives $frac{1}{2x}I$ where $I = int_{-infty}^infty e^{-t^2}dt$ is the Gaussian integral.
$endgroup$
– Song
Dec 17 '18 at 20:51






$begingroup$
If you integrate over $y$ first, then it gives $frac{1}{2x}I$ where $I = int_{-infty}^infty e^{-t^2}dt$ is the Gaussian integral.
$endgroup$
– Song
Dec 17 '18 at 20:51














$begingroup$
Why $int^{infty}_{-infty}e^{-t^{2}}dt$ and not $int^{infty}_{0}e^{-t^{2}}dt$?
$endgroup$
– SABOY
Dec 17 '18 at 21:03






$begingroup$
Why $int^{infty}_{-infty}e^{-t^{2}}dt$ and not $int^{infty}_{0}e^{-t^{2}}dt$?
$endgroup$
– SABOY
Dec 17 '18 at 21:03














$begingroup$
The factor $1/2$ is in front. Typically Gaussian integral refers to the former.
$endgroup$
– Song
Dec 17 '18 at 21:07




$begingroup$
The factor $1/2$ is in front. Typically Gaussian integral refers to the former.
$endgroup$
– Song
Dec 17 '18 at 21:07










1 Answer
1






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2












$begingroup$

Since the Lebesgue integral agrees with the Riemann integral if we have absolutely Riemann-integrable functions we can first integrate with respect to $x$ to get:



$$int_{0}^{infty}xe^{-x^2(1+y^2)}dlambda(x) = frac{-1}{2}frac{e^{-x^2(1+y^2)}}{1+y^2}Bigvert_{0}^{infty} = frac{1}{2}frac{1}{1+y^2}.$$



Now integrate



$$frac{1}{2}int_{0}^{infty}frac{1}{1+y^2}dlambda(y)$$



NOTE: A primitive for $xe^{-x^2(1+y^2)}$ with respect to $y$ is not given by $frac{1}{-x^2cdot 2y}e^{-xy^2}$! Rather



$$frac{1}{x}frac{d}{dy}frac{1}{2y}e^{-x^2y^2} = frac{1}{x}cdotfrac{-2yx^2e^{-x^2y^2}(2y)-2e^{-x^2y^2}}{4y^2}=-xe^{-x^2y^2}-frac{e^{-x^2y^2}}{2xy^2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can this help me find $int_{[0,infty[}e^{-t^{2}}d lambda (t)$ ?
    $endgroup$
    – SABOY
    Dec 17 '18 at 20:43










  • $begingroup$
    Is that what you are trying to do?
    $endgroup$
    – Olof Rubin
    Dec 17 '18 at 20:44










  • $begingroup$
    Yes, that is why I needed to use Tonelli, to switch integrals, and integrate with respect to $y$
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:06













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Since the Lebesgue integral agrees with the Riemann integral if we have absolutely Riemann-integrable functions we can first integrate with respect to $x$ to get:



$$int_{0}^{infty}xe^{-x^2(1+y^2)}dlambda(x) = frac{-1}{2}frac{e^{-x^2(1+y^2)}}{1+y^2}Bigvert_{0}^{infty} = frac{1}{2}frac{1}{1+y^2}.$$



Now integrate



$$frac{1}{2}int_{0}^{infty}frac{1}{1+y^2}dlambda(y)$$



NOTE: A primitive for $xe^{-x^2(1+y^2)}$ with respect to $y$ is not given by $frac{1}{-x^2cdot 2y}e^{-xy^2}$! Rather



$$frac{1}{x}frac{d}{dy}frac{1}{2y}e^{-x^2y^2} = frac{1}{x}cdotfrac{-2yx^2e^{-x^2y^2}(2y)-2e^{-x^2y^2}}{4y^2}=-xe^{-x^2y^2}-frac{e^{-x^2y^2}}{2xy^2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can this help me find $int_{[0,infty[}e^{-t^{2}}d lambda (t)$ ?
    $endgroup$
    – SABOY
    Dec 17 '18 at 20:43










  • $begingroup$
    Is that what you are trying to do?
    $endgroup$
    – Olof Rubin
    Dec 17 '18 at 20:44










  • $begingroup$
    Yes, that is why I needed to use Tonelli, to switch integrals, and integrate with respect to $y$
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:06


















2












$begingroup$

Since the Lebesgue integral agrees with the Riemann integral if we have absolutely Riemann-integrable functions we can first integrate with respect to $x$ to get:



$$int_{0}^{infty}xe^{-x^2(1+y^2)}dlambda(x) = frac{-1}{2}frac{e^{-x^2(1+y^2)}}{1+y^2}Bigvert_{0}^{infty} = frac{1}{2}frac{1}{1+y^2}.$$



Now integrate



$$frac{1}{2}int_{0}^{infty}frac{1}{1+y^2}dlambda(y)$$



NOTE: A primitive for $xe^{-x^2(1+y^2)}$ with respect to $y$ is not given by $frac{1}{-x^2cdot 2y}e^{-xy^2}$! Rather



$$frac{1}{x}frac{d}{dy}frac{1}{2y}e^{-x^2y^2} = frac{1}{x}cdotfrac{-2yx^2e^{-x^2y^2}(2y)-2e^{-x^2y^2}}{4y^2}=-xe^{-x^2y^2}-frac{e^{-x^2y^2}}{2xy^2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How can this help me find $int_{[0,infty[}e^{-t^{2}}d lambda (t)$ ?
    $endgroup$
    – SABOY
    Dec 17 '18 at 20:43










  • $begingroup$
    Is that what you are trying to do?
    $endgroup$
    – Olof Rubin
    Dec 17 '18 at 20:44










  • $begingroup$
    Yes, that is why I needed to use Tonelli, to switch integrals, and integrate with respect to $y$
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:06
















2












2








2





$begingroup$

Since the Lebesgue integral agrees with the Riemann integral if we have absolutely Riemann-integrable functions we can first integrate with respect to $x$ to get:



$$int_{0}^{infty}xe^{-x^2(1+y^2)}dlambda(x) = frac{-1}{2}frac{e^{-x^2(1+y^2)}}{1+y^2}Bigvert_{0}^{infty} = frac{1}{2}frac{1}{1+y^2}.$$



Now integrate



$$frac{1}{2}int_{0}^{infty}frac{1}{1+y^2}dlambda(y)$$



NOTE: A primitive for $xe^{-x^2(1+y^2)}$ with respect to $y$ is not given by $frac{1}{-x^2cdot 2y}e^{-xy^2}$! Rather



$$frac{1}{x}frac{d}{dy}frac{1}{2y}e^{-x^2y^2} = frac{1}{x}cdotfrac{-2yx^2e^{-x^2y^2}(2y)-2e^{-x^2y^2}}{4y^2}=-xe^{-x^2y^2}-frac{e^{-x^2y^2}}{2xy^2}$$






share|cite|improve this answer











$endgroup$



Since the Lebesgue integral agrees with the Riemann integral if we have absolutely Riemann-integrable functions we can first integrate with respect to $x$ to get:



$$int_{0}^{infty}xe^{-x^2(1+y^2)}dlambda(x) = frac{-1}{2}frac{e^{-x^2(1+y^2)}}{1+y^2}Bigvert_{0}^{infty} = frac{1}{2}frac{1}{1+y^2}.$$



Now integrate



$$frac{1}{2}int_{0}^{infty}frac{1}{1+y^2}dlambda(y)$$



NOTE: A primitive for $xe^{-x^2(1+y^2)}$ with respect to $y$ is not given by $frac{1}{-x^2cdot 2y}e^{-xy^2}$! Rather



$$frac{1}{x}frac{d}{dy}frac{1}{2y}e^{-x^2y^2} = frac{1}{x}cdotfrac{-2yx^2e^{-x^2y^2}(2y)-2e^{-x^2y^2}}{4y^2}=-xe^{-x^2y^2}-frac{e^{-x^2y^2}}{2xy^2}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 20:45

























answered Dec 17 '18 at 20:29









Olof RubinOlof Rubin

872317




872317












  • $begingroup$
    How can this help me find $int_{[0,infty[}e^{-t^{2}}d lambda (t)$ ?
    $endgroup$
    – SABOY
    Dec 17 '18 at 20:43










  • $begingroup$
    Is that what you are trying to do?
    $endgroup$
    – Olof Rubin
    Dec 17 '18 at 20:44










  • $begingroup$
    Yes, that is why I needed to use Tonelli, to switch integrals, and integrate with respect to $y$
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:06




















  • $begingroup$
    How can this help me find $int_{[0,infty[}e^{-t^{2}}d lambda (t)$ ?
    $endgroup$
    – SABOY
    Dec 17 '18 at 20:43










  • $begingroup$
    Is that what you are trying to do?
    $endgroup$
    – Olof Rubin
    Dec 17 '18 at 20:44










  • $begingroup$
    Yes, that is why I needed to use Tonelli, to switch integrals, and integrate with respect to $y$
    $endgroup$
    – SABOY
    Dec 17 '18 at 21:06


















$begingroup$
How can this help me find $int_{[0,infty[}e^{-t^{2}}d lambda (t)$ ?
$endgroup$
– SABOY
Dec 17 '18 at 20:43




$begingroup$
How can this help me find $int_{[0,infty[}e^{-t^{2}}d lambda (t)$ ?
$endgroup$
– SABOY
Dec 17 '18 at 20:43












$begingroup$
Is that what you are trying to do?
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:44




$begingroup$
Is that what you are trying to do?
$endgroup$
– Olof Rubin
Dec 17 '18 at 20:44












$begingroup$
Yes, that is why I needed to use Tonelli, to switch integrals, and integrate with respect to $y$
$endgroup$
– SABOY
Dec 17 '18 at 21:06






$begingroup$
Yes, that is why I needed to use Tonelli, to switch integrals, and integrate with respect to $y$
$endgroup$
– SABOY
Dec 17 '18 at 21:06




















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