Does this integral converge? $int frac{d^2x}{(2pi)^2} frac{d^2y}{(2pi)^2} [dots] exp [- frac{1}{4pi i}(x_1^2...
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Here I have a generalization of the Fresnel integral in two variables, and I'd like to check that it converges:
$$ frac{1}{2} int_{mathbb{R}^2 times mathbb{R}^2} frac{d^2x}{(2pi)^2} frac{d^2y}{(2pi)^2}
frac{big[2sinh big(frac{x_1 - x_2}{2}big)big]^2big[2sinh big(frac{y_1 - y_2}{2}big)big]^2}{prod_{i,j=1,2}big[2 cosh big(frac{x_i - y_j}{2}big)big]^2}
exp left[- frac{1}{4pi i}Big((x_1^2 - y_1^2)+(x_2^2- y_2^2)Big)right]$$
The domain of integration is that $(x_1, x_2)$ and $(y_1, y_2)$ vary all over $mathbb{R}^2$. This is from a physics computation.
Or prove that it diverges. This integral might give two different answers depending on which subspace we integrate first:
${x_1 = x_2} cap {y_1 = y_2}$ so that $sinh (x_1 - x_2) = sinh 0 = 0$.
${ x_1 = y_1} cap {x_2 = y_2}$ so that $e^{x_1^2 - y_1^2} = e^{x_2^2 - y_2^2} = e^0 = 1$
${ x_1 = y_2} cap {x_2 = y_1}$ so that $e^{x_1^2 - y_2^2} = e^{x_2^2 - y_1^2}= e^0 = 1$.
In the first case - without integrating over the remaining subspace $mathbb{R}^4 / big(mathbb{R}(1,-1) oplus mathbb{R}(1,-1)big) simeq mathbb{R}^2$ - we'd have that $int = 0$. My notes have $int = frac{1}{16}$ and I wonder what the other two iterated integrals have us.
This looks like it could be realted to the Fresnel integral.
$$ int_0^infty sin x^2 , dx = int_0^infty cos x^2 , dx = frac{sqrt{2pi}}{4} $$
improper-integrals multiple-integral
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add a comment |
$begingroup$
Here I have a generalization of the Fresnel integral in two variables, and I'd like to check that it converges:
$$ frac{1}{2} int_{mathbb{R}^2 times mathbb{R}^2} frac{d^2x}{(2pi)^2} frac{d^2y}{(2pi)^2}
frac{big[2sinh big(frac{x_1 - x_2}{2}big)big]^2big[2sinh big(frac{y_1 - y_2}{2}big)big]^2}{prod_{i,j=1,2}big[2 cosh big(frac{x_i - y_j}{2}big)big]^2}
exp left[- frac{1}{4pi i}Big((x_1^2 - y_1^2)+(x_2^2- y_2^2)Big)right]$$
The domain of integration is that $(x_1, x_2)$ and $(y_1, y_2)$ vary all over $mathbb{R}^2$. This is from a physics computation.
Or prove that it diverges. This integral might give two different answers depending on which subspace we integrate first:
${x_1 = x_2} cap {y_1 = y_2}$ so that $sinh (x_1 - x_2) = sinh 0 = 0$.
${ x_1 = y_1} cap {x_2 = y_2}$ so that $e^{x_1^2 - y_1^2} = e^{x_2^2 - y_2^2} = e^0 = 1$
${ x_1 = y_2} cap {x_2 = y_1}$ so that $e^{x_1^2 - y_2^2} = e^{x_2^2 - y_1^2}= e^0 = 1$.
In the first case - without integrating over the remaining subspace $mathbb{R}^4 / big(mathbb{R}(1,-1) oplus mathbb{R}(1,-1)big) simeq mathbb{R}^2$ - we'd have that $int = 0$. My notes have $int = frac{1}{16}$ and I wonder what the other two iterated integrals have us.
This looks like it could be realted to the Fresnel integral.
$$ int_0^infty sin x^2 , dx = int_0^infty cos x^2 , dx = frac{sqrt{2pi}}{4} $$
improper-integrals multiple-integral
$endgroup$
add a comment |
$begingroup$
Here I have a generalization of the Fresnel integral in two variables, and I'd like to check that it converges:
$$ frac{1}{2} int_{mathbb{R}^2 times mathbb{R}^2} frac{d^2x}{(2pi)^2} frac{d^2y}{(2pi)^2}
frac{big[2sinh big(frac{x_1 - x_2}{2}big)big]^2big[2sinh big(frac{y_1 - y_2}{2}big)big]^2}{prod_{i,j=1,2}big[2 cosh big(frac{x_i - y_j}{2}big)big]^2}
exp left[- frac{1}{4pi i}Big((x_1^2 - y_1^2)+(x_2^2- y_2^2)Big)right]$$
The domain of integration is that $(x_1, x_2)$ and $(y_1, y_2)$ vary all over $mathbb{R}^2$. This is from a physics computation.
Or prove that it diverges. This integral might give two different answers depending on which subspace we integrate first:
${x_1 = x_2} cap {y_1 = y_2}$ so that $sinh (x_1 - x_2) = sinh 0 = 0$.
${ x_1 = y_1} cap {x_2 = y_2}$ so that $e^{x_1^2 - y_1^2} = e^{x_2^2 - y_2^2} = e^0 = 1$
${ x_1 = y_2} cap {x_2 = y_1}$ so that $e^{x_1^2 - y_2^2} = e^{x_2^2 - y_1^2}= e^0 = 1$.
In the first case - without integrating over the remaining subspace $mathbb{R}^4 / big(mathbb{R}(1,-1) oplus mathbb{R}(1,-1)big) simeq mathbb{R}^2$ - we'd have that $int = 0$. My notes have $int = frac{1}{16}$ and I wonder what the other two iterated integrals have us.
This looks like it could be realted to the Fresnel integral.
$$ int_0^infty sin x^2 , dx = int_0^infty cos x^2 , dx = frac{sqrt{2pi}}{4} $$
improper-integrals multiple-integral
$endgroup$
Here I have a generalization of the Fresnel integral in two variables, and I'd like to check that it converges:
$$ frac{1}{2} int_{mathbb{R}^2 times mathbb{R}^2} frac{d^2x}{(2pi)^2} frac{d^2y}{(2pi)^2}
frac{big[2sinh big(frac{x_1 - x_2}{2}big)big]^2big[2sinh big(frac{y_1 - y_2}{2}big)big]^2}{prod_{i,j=1,2}big[2 cosh big(frac{x_i - y_j}{2}big)big]^2}
exp left[- frac{1}{4pi i}Big((x_1^2 - y_1^2)+(x_2^2- y_2^2)Big)right]$$
The domain of integration is that $(x_1, x_2)$ and $(y_1, y_2)$ vary all over $mathbb{R}^2$. This is from a physics computation.
Or prove that it diverges. This integral might give two different answers depending on which subspace we integrate first:
${x_1 = x_2} cap {y_1 = y_2}$ so that $sinh (x_1 - x_2) = sinh 0 = 0$.
${ x_1 = y_1} cap {x_2 = y_2}$ so that $e^{x_1^2 - y_1^2} = e^{x_2^2 - y_2^2} = e^0 = 1$
${ x_1 = y_2} cap {x_2 = y_1}$ so that $e^{x_1^2 - y_2^2} = e^{x_2^2 - y_1^2}= e^0 = 1$.
In the first case - without integrating over the remaining subspace $mathbb{R}^4 / big(mathbb{R}(1,-1) oplus mathbb{R}(1,-1)big) simeq mathbb{R}^2$ - we'd have that $int = 0$. My notes have $int = frac{1}{16}$ and I wonder what the other two iterated integrals have us.
This looks like it could be realted to the Fresnel integral.
$$ int_0^infty sin x^2 , dx = int_0^infty cos x^2 , dx = frac{sqrt{2pi}}{4} $$
improper-integrals multiple-integral
improper-integrals multiple-integral
edited Dec 16 '18 at 21:46
cactus314
asked Dec 16 '18 at 21:01
cactus314cactus314
15.5k42269
15.5k42269
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