How do I evaluate $int_{-infty}^{infty}int_{-infty}^{infty} e^{-(3x^2+2 sqrt 2 xy+3y^2)} mathrm dx,mathrm...












1












$begingroup$



Evaluate $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-3x^2-2 sqrt 2 xy - 3y^2right) , mathrm dx,mathrm dy$$




I first evaluate



$$int_{-infty}^{infty} int_{-infty}^{infty} expleft[-3bigl(x^2+ y^2bigr)right] ,mathrm dx,mathrm dy$$



using polar coordinates, which evaluates to $pi/3$. But I find difficulty to evaluate the double integral $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-2 sqrt 2 xyright) , mathrm dx,mathrm dy$$ Would anybody please help me finding it out?






real-analysis multivariable-calculus exponential-function multiple-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where are the limits of integration on $y$?
    $endgroup$
    – Shubham Johri
    Jan 12 at 18:17






  • 1




    $begingroup$
    How can you separate this integral into those two integrals? The integrand of the first integral is the product of the integrands of the latter two, not the sum.
    $endgroup$
    – Frpzzd
    Jan 12 at 18:19










  • $begingroup$
    Are you sure that what you want isn't $displaystyleint_{-infty}^inftyint_{-infty}^infty e^{-(3x^2+2sqrt2xy+3y^2)},mathrm dx,mathrm dy$?
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:23










  • $begingroup$
    Sorry I have corrected my problem.
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:23










  • $begingroup$
    See math.stackexchange.com/questions/653159/…, math.stackexchange.com/questions/384732/….
    $endgroup$
    – StubbornAtom
    Jan 12 at 20:04
















1












$begingroup$



Evaluate $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-3x^2-2 sqrt 2 xy - 3y^2right) , mathrm dx,mathrm dy$$




I first evaluate



$$int_{-infty}^{infty} int_{-infty}^{infty} expleft[-3bigl(x^2+ y^2bigr)right] ,mathrm dx,mathrm dy$$



using polar coordinates, which evaluates to $pi/3$. But I find difficulty to evaluate the double integral $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-2 sqrt 2 xyright) , mathrm dx,mathrm dy$$ Would anybody please help me finding it out?






real-analysis multivariable-calculus exponential-function multiple-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where are the limits of integration on $y$?
    $endgroup$
    – Shubham Johri
    Jan 12 at 18:17






  • 1




    $begingroup$
    How can you separate this integral into those two integrals? The integrand of the first integral is the product of the integrands of the latter two, not the sum.
    $endgroup$
    – Frpzzd
    Jan 12 at 18:19










  • $begingroup$
    Are you sure that what you want isn't $displaystyleint_{-infty}^inftyint_{-infty}^infty e^{-(3x^2+2sqrt2xy+3y^2)},mathrm dx,mathrm dy$?
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:23










  • $begingroup$
    Sorry I have corrected my problem.
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:23










  • $begingroup$
    See math.stackexchange.com/questions/653159/…, math.stackexchange.com/questions/384732/….
    $endgroup$
    – StubbornAtom
    Jan 12 at 20:04














1












1








1


1



$begingroup$



Evaluate $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-3x^2-2 sqrt 2 xy - 3y^2right) , mathrm dx,mathrm dy$$




I first evaluate



$$int_{-infty}^{infty} int_{-infty}^{infty} expleft[-3bigl(x^2+ y^2bigr)right] ,mathrm dx,mathrm dy$$



using polar coordinates, which evaluates to $pi/3$. But I find difficulty to evaluate the double integral $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-2 sqrt 2 xyright) , mathrm dx,mathrm dy$$ Would anybody please help me finding it out?






real-analysis multivariable-calculus exponential-function multiple-integral







share|cite|improve this question











$endgroup$





Evaluate $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-3x^2-2 sqrt 2 xy - 3y^2right) , mathrm dx,mathrm dy$$




I first evaluate



$$int_{-infty}^{infty} int_{-infty}^{infty} expleft[-3bigl(x^2+ y^2bigr)right] ,mathrm dx,mathrm dy$$



using polar coordinates, which evaluates to $pi/3$. But I find difficulty to evaluate the double integral $$int_{-infty}^{infty} int_{-infty}^{infty} expleft(-2 sqrt 2 xyright) , mathrm dx,mathrm dy$$ Would anybody please help me finding it out?






real-analysis multivariable-calculus exponential-function multiple-integral




real-analysis multivariable-calculus exponential-function multiple-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 20:01









StubbornAtom

5,72111138




5,72111138










asked Jan 12 at 18:16









Dbchatto67Dbchatto67

575116




575116












  • $begingroup$
    Where are the limits of integration on $y$?
    $endgroup$
    – Shubham Johri
    Jan 12 at 18:17






  • 1




    $begingroup$
    How can you separate this integral into those two integrals? The integrand of the first integral is the product of the integrands of the latter two, not the sum.
    $endgroup$
    – Frpzzd
    Jan 12 at 18:19










  • $begingroup$
    Are you sure that what you want isn't $displaystyleint_{-infty}^inftyint_{-infty}^infty e^{-(3x^2+2sqrt2xy+3y^2)},mathrm dx,mathrm dy$?
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:23










  • $begingroup$
    Sorry I have corrected my problem.
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:23










  • $begingroup$
    See math.stackexchange.com/questions/653159/…, math.stackexchange.com/questions/384732/….
    $endgroup$
    – StubbornAtom
    Jan 12 at 20:04


















  • $begingroup$
    Where are the limits of integration on $y$?
    $endgroup$
    – Shubham Johri
    Jan 12 at 18:17






  • 1




    $begingroup$
    How can you separate this integral into those two integrals? The integrand of the first integral is the product of the integrands of the latter two, not the sum.
    $endgroup$
    – Frpzzd
    Jan 12 at 18:19










  • $begingroup$
    Are you sure that what you want isn't $displaystyleint_{-infty}^inftyint_{-infty}^infty e^{-(3x^2+2sqrt2xy+3y^2)},mathrm dx,mathrm dy$?
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:23










  • $begingroup$
    Sorry I have corrected my problem.
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:23










  • $begingroup$
    See math.stackexchange.com/questions/653159/…, math.stackexchange.com/questions/384732/….
    $endgroup$
    – StubbornAtom
    Jan 12 at 20:04
















$begingroup$
Where are the limits of integration on $y$?
$endgroup$
– Shubham Johri
Jan 12 at 18:17




$begingroup$
Where are the limits of integration on $y$?
$endgroup$
– Shubham Johri
Jan 12 at 18:17




1




1




$begingroup$
How can you separate this integral into those two integrals? The integrand of the first integral is the product of the integrands of the latter two, not the sum.
$endgroup$
– Frpzzd
Jan 12 at 18:19




$begingroup$
How can you separate this integral into those two integrals? The integrand of the first integral is the product of the integrands of the latter two, not the sum.
$endgroup$
– Frpzzd
Jan 12 at 18:19












$begingroup$
Are you sure that what you want isn't $displaystyleint_{-infty}^inftyint_{-infty}^infty e^{-(3x^2+2sqrt2xy+3y^2)},mathrm dx,mathrm dy$?
$endgroup$
– José Carlos Santos
Jan 12 at 18:23




$begingroup$
Are you sure that what you want isn't $displaystyleint_{-infty}^inftyint_{-infty}^infty e^{-(3x^2+2sqrt2xy+3y^2)},mathrm dx,mathrm dy$?
$endgroup$
– José Carlos Santos
Jan 12 at 18:23












$begingroup$
Sorry I have corrected my problem.
$endgroup$
– Dbchatto67
Jan 12 at 18:23




$begingroup$
Sorry I have corrected my problem.
$endgroup$
– Dbchatto67
Jan 12 at 18:23












$begingroup$
See math.stackexchange.com/questions/653159/…, math.stackexchange.com/questions/384732/….
$endgroup$
– StubbornAtom
Jan 12 at 20:04




$begingroup$
See math.stackexchange.com/questions/653159/…, math.stackexchange.com/questions/384732/….
$endgroup$
– StubbornAtom
Jan 12 at 20:04










2 Answers
2






active

oldest

votes


















3












$begingroup$

Hint: Doing the change of variables $x=X+Y$ and $y=X-Y$, your integral becomes$$2int_{-infty}^inftyint_{-infty}^infty expleft[-left(2sqrt2+6right)X^2-left(-2sqrt2+6right)Y^2right],mathrm dX,mathrm dY.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I find the integrand to be $2 e^{-((6 + 2 sqrt 2) x^2 + (6 - 2 sqrt 2)y^2)}$. Would you please check your calculation?
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:37












  • $begingroup$
    @Dbchatto67 Right! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:45










  • $begingroup$
    @ChaseRyanTaylor No, I don't mind.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 20:51



















4












$begingroup$

$$3x^2+2sqrt{2} xy + 3y^2
=begin{bmatrix}x & y end{bmatrix}
begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}
begin{bmatrix}x \ y end{bmatrix}$$
so the integrand is
$$exp(- v^top Omega v/2)$$
where $v = begin{bmatrix}x \ y end{bmatrix}$ and $Omega = 2begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}$.



By using the density of a $N(0, Sigma)$ distribution we have
$$frac{1}{sqrt{(2 pi)^2 det (Omega^{-1})}} int_{-infty}^infty int_{-infty}^infty exp(-v^top Omega v / 2) , dx , dy = 1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    fascinating . . . definitely learned something new
    $endgroup$
    – Chase Ryan Taylor
    Jan 12 at 18:37










  • $begingroup$
    I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $expbigl(-v^topOmega v/2bigr)$, then what allows $dv=dbegin{bmatrix}x \ y end{bmatrix}$ to stand for $dx,dy$, from the original differential?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:46












  • $begingroup$
    @ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $int_{-infty}^infty int_{-infty}^infty cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral.
    $endgroup$
    – angryavian
    Jan 13 at 6:48










  • $begingroup$
    If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:49











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071214%2fhow-do-i-evaluate-int-infty-infty-int-infty-infty-e-3x22-s%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Hint: Doing the change of variables $x=X+Y$ and $y=X-Y$, your integral becomes$$2int_{-infty}^inftyint_{-infty}^infty expleft[-left(2sqrt2+6right)X^2-left(-2sqrt2+6right)Y^2right],mathrm dX,mathrm dY.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I find the integrand to be $2 e^{-((6 + 2 sqrt 2) x^2 + (6 - 2 sqrt 2)y^2)}$. Would you please check your calculation?
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:37












  • $begingroup$
    @Dbchatto67 Right! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:45










  • $begingroup$
    @ChaseRyanTaylor No, I don't mind.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 20:51
















3












$begingroup$

Hint: Doing the change of variables $x=X+Y$ and $y=X-Y$, your integral becomes$$2int_{-infty}^inftyint_{-infty}^infty expleft[-left(2sqrt2+6right)X^2-left(-2sqrt2+6right)Y^2right],mathrm dX,mathrm dY.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I find the integrand to be $2 e^{-((6 + 2 sqrt 2) x^2 + (6 - 2 sqrt 2)y^2)}$. Would you please check your calculation?
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:37












  • $begingroup$
    @Dbchatto67 Right! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:45










  • $begingroup$
    @ChaseRyanTaylor No, I don't mind.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 20:51














3












3








3





$begingroup$

Hint: Doing the change of variables $x=X+Y$ and $y=X-Y$, your integral becomes$$2int_{-infty}^inftyint_{-infty}^infty expleft[-left(2sqrt2+6right)X^2-left(-2sqrt2+6right)Y^2right],mathrm dX,mathrm dY.$$






share|cite|improve this answer











$endgroup$



Hint: Doing the change of variables $x=X+Y$ and $y=X-Y$, your integral becomes$$2int_{-infty}^inftyint_{-infty}^infty expleft[-left(2sqrt2+6right)X^2-left(-2sqrt2+6right)Y^2right],mathrm dX,mathrm dY.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 12 at 20:46









Chase Ryan Taylor

4,38021530




4,38021530










answered Jan 12 at 18:26









José Carlos SantosJosé Carlos Santos

157k22126227




157k22126227












  • $begingroup$
    I find the integrand to be $2 e^{-((6 + 2 sqrt 2) x^2 + (6 - 2 sqrt 2)y^2)}$. Would you please check your calculation?
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:37












  • $begingroup$
    @Dbchatto67 Right! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:45










  • $begingroup$
    @ChaseRyanTaylor No, I don't mind.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 20:51


















  • $begingroup$
    I find the integrand to be $2 e^{-((6 + 2 sqrt 2) x^2 + (6 - 2 sqrt 2)y^2)}$. Would you please check your calculation?
    $endgroup$
    – Dbchatto67
    Jan 12 at 18:37












  • $begingroup$
    @Dbchatto67 Right! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:45










  • $begingroup$
    @ChaseRyanTaylor No, I don't mind.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 20:51
















$begingroup$
I find the integrand to be $2 e^{-((6 + 2 sqrt 2) x^2 + (6 - 2 sqrt 2)y^2)}$. Would you please check your calculation?
$endgroup$
– Dbchatto67
Jan 12 at 18:37






$begingroup$
I find the integrand to be $2 e^{-((6 + 2 sqrt 2) x^2 + (6 - 2 sqrt 2)y^2)}$. Would you please check your calculation?
$endgroup$
– Dbchatto67
Jan 12 at 18:37














$begingroup$
@Dbchatto67 Right! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 12 at 18:45




$begingroup$
@Dbchatto67 Right! I've edited my answer. Thank you.
$endgroup$
– José Carlos Santos
Jan 12 at 18:45












$begingroup$
@ChaseRyanTaylor No, I don't mind.
$endgroup$
– José Carlos Santos
Jan 12 at 20:51




$begingroup$
@ChaseRyanTaylor No, I don't mind.
$endgroup$
– José Carlos Santos
Jan 12 at 20:51











4












$begingroup$

$$3x^2+2sqrt{2} xy + 3y^2
=begin{bmatrix}x & y end{bmatrix}
begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}
begin{bmatrix}x \ y end{bmatrix}$$
so the integrand is
$$exp(- v^top Omega v/2)$$
where $v = begin{bmatrix}x \ y end{bmatrix}$ and $Omega = 2begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}$.



By using the density of a $N(0, Sigma)$ distribution we have
$$frac{1}{sqrt{(2 pi)^2 det (Omega^{-1})}} int_{-infty}^infty int_{-infty}^infty exp(-v^top Omega v / 2) , dx , dy = 1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    fascinating . . . definitely learned something new
    $endgroup$
    – Chase Ryan Taylor
    Jan 12 at 18:37










  • $begingroup$
    I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $expbigl(-v^topOmega v/2bigr)$, then what allows $dv=dbegin{bmatrix}x \ y end{bmatrix}$ to stand for $dx,dy$, from the original differential?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:46












  • $begingroup$
    @ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $int_{-infty}^infty int_{-infty}^infty cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral.
    $endgroup$
    – angryavian
    Jan 13 at 6:48










  • $begingroup$
    If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:49
















4












$begingroup$

$$3x^2+2sqrt{2} xy + 3y^2
=begin{bmatrix}x & y end{bmatrix}
begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}
begin{bmatrix}x \ y end{bmatrix}$$
so the integrand is
$$exp(- v^top Omega v/2)$$
where $v = begin{bmatrix}x \ y end{bmatrix}$ and $Omega = 2begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}$.



By using the density of a $N(0, Sigma)$ distribution we have
$$frac{1}{sqrt{(2 pi)^2 det (Omega^{-1})}} int_{-infty}^infty int_{-infty}^infty exp(-v^top Omega v / 2) , dx , dy = 1.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    fascinating . . . definitely learned something new
    $endgroup$
    – Chase Ryan Taylor
    Jan 12 at 18:37










  • $begingroup$
    I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $expbigl(-v^topOmega v/2bigr)$, then what allows $dv=dbegin{bmatrix}x \ y end{bmatrix}$ to stand for $dx,dy$, from the original differential?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:46












  • $begingroup$
    @ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $int_{-infty}^infty int_{-infty}^infty cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral.
    $endgroup$
    – angryavian
    Jan 13 at 6:48










  • $begingroup$
    If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:49














4












4








4





$begingroup$

$$3x^2+2sqrt{2} xy + 3y^2
=begin{bmatrix}x & y end{bmatrix}
begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}
begin{bmatrix}x \ y end{bmatrix}$$
so the integrand is
$$exp(- v^top Omega v/2)$$
where $v = begin{bmatrix}x \ y end{bmatrix}$ and $Omega = 2begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}$.



By using the density of a $N(0, Sigma)$ distribution we have
$$frac{1}{sqrt{(2 pi)^2 det (Omega^{-1})}} int_{-infty}^infty int_{-infty}^infty exp(-v^top Omega v / 2) , dx , dy = 1.$$






share|cite|improve this answer











$endgroup$



$$3x^2+2sqrt{2} xy + 3y^2
=begin{bmatrix}x & y end{bmatrix}
begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}
begin{bmatrix}x \ y end{bmatrix}$$
so the integrand is
$$exp(- v^top Omega v/2)$$
where $v = begin{bmatrix}x \ y end{bmatrix}$ and $Omega = 2begin{bmatrix} 3 & sqrt{2} \ sqrt{2} & 3 end{bmatrix}$.



By using the density of a $N(0, Sigma)$ distribution we have
$$frac{1}{sqrt{(2 pi)^2 det (Omega^{-1})}} int_{-infty}^infty int_{-infty}^infty exp(-v^top Omega v / 2) , dx , dy = 1.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 15:43

























answered Jan 12 at 18:27









angryavianangryavian

40.6k23280




40.6k23280












  • $begingroup$
    fascinating . . . definitely learned something new
    $endgroup$
    – Chase Ryan Taylor
    Jan 12 at 18:37










  • $begingroup$
    I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $expbigl(-v^topOmega v/2bigr)$, then what allows $dv=dbegin{bmatrix}x \ y end{bmatrix}$ to stand for $dx,dy$, from the original differential?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:46












  • $begingroup$
    @ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $int_{-infty}^infty int_{-infty}^infty cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral.
    $endgroup$
    – angryavian
    Jan 13 at 6:48










  • $begingroup$
    If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:49


















  • $begingroup$
    fascinating . . . definitely learned something new
    $endgroup$
    – Chase Ryan Taylor
    Jan 12 at 18:37










  • $begingroup$
    I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $expbigl(-v^topOmega v/2bigr)$, then what allows $dv=dbegin{bmatrix}x \ y end{bmatrix}$ to stand for $dx,dy$, from the original differential?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:46












  • $begingroup$
    @ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $int_{-infty}^infty int_{-infty}^infty cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral.
    $endgroup$
    – angryavian
    Jan 13 at 6:48










  • $begingroup$
    If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different?
    $endgroup$
    – Chase Ryan Taylor
    Jan 13 at 6:49
















$begingroup$
fascinating . . . definitely learned something new
$endgroup$
– Chase Ryan Taylor
Jan 12 at 18:37




$begingroup$
fascinating . . . definitely learned something new
$endgroup$
– Chase Ryan Taylor
Jan 12 at 18:37












$begingroup$
I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $expbigl(-v^topOmega v/2bigr)$, then what allows $dv=dbegin{bmatrix}x \ y end{bmatrix}$ to stand for $dx,dy$, from the original differential?
$endgroup$
– Chase Ryan Taylor
Jan 13 at 6:46






$begingroup$
I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $expbigl(-v^topOmega v/2bigr)$, then what allows $dv=dbegin{bmatrix}x \ y end{bmatrix}$ to stand for $dx,dy$, from the original differential?
$endgroup$
– Chase Ryan Taylor
Jan 13 at 6:46














$begingroup$
@ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $int_{-infty}^infty int_{-infty}^infty cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral.
$endgroup$
– angryavian
Jan 13 at 6:48




$begingroup$
@ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $int_{-infty}^infty int_{-infty}^infty cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral.
$endgroup$
– angryavian
Jan 13 at 6:48












$begingroup$
If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different?
$endgroup$
– Chase Ryan Taylor
Jan 13 at 6:49




$begingroup$
If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different?
$endgroup$
– Chase Ryan Taylor
Jan 13 at 6:49


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071214%2fhow-do-i-evaluate-int-infty-infty-int-infty-infty-e-3x22-s%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
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
Read more

File:Tiny Toon Adventures Wacky Sports JP Title.png

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! File:Tiny Toon Adventures Wacky Sports JP Title.png From The Cutting Room Floor Jump to: navigation, search File File history File usage No higher resolution available. Tiny_Toon_Adventures_Wacky_Sports_JP_Title.png ‎ (160 × 144 pixels, file size: 3 KB, MIME type: image/png ) This image is a screenshot of a(n) Game Boy title screen. File history Click on a date/time to view the file as it appeared at that time. Date/Time Thumbnail Dimensions User Comment current 21:31, 20 August 2016 160 × 144 (3 KB) TheBluegeek (Talk | contribs) 17:25, 29 August 2013 160 × 144 (3 KB) Luigi-San (Talk | contribs) {{titlescreen|Game Boy}} You cannot overwrite this file. File usage The following page links to this file: Tiny Toon Advent
Read more