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











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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


















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