Simplify $int_a^b mathrm{d}x_1 …mathrm{d}x_m int_c^d mathrm{d}y_1 …mathrm{d}y_n prod_{i=1}^m prod_{j=1}^n...
$begingroup$
Let $a le b$ and $cle d$ be real numbers and $f$ a real function. I am struggling with an integral of the form:
$$I =
int_a^b mathrm{d}x_1 ...mathrm{d}x_m
int_c^d mathrm{d}y_1 ...mathrm{d}y_n
prod_{i=1}^m prod_{j=1}^n f(x_i y_j)$$
Can it be reduced to a single or double integration somehow?
I am interested in the large $m,n$ behavior of $I$. Specifically, suppose that the ratio $alpha = m/n$ is fixed. Can we evaluate in general the limit:
$$lim_{nrightarrowinfty} frac{1}{n} ln I$$
So far, these are the only reductions I've been able to do:
$$begin{aligned}
I & = int_a^b mathrm d x_1 dots mathrm d x_m left( int_c^d mathrm d y
prod_{i = 1}^m f (x_i y) right)^n\
& = int_c^d mathrm d y_1 ldots mathrm d y_m left( int_a^b mathrm d x
prod_{j = 1}^n f (x y_j) right)^m
end{aligned}$$
integration definite-integrals multiple-integral large-deviation-theory
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
$endgroup$
add a comment |
$begingroup$
Let $a le b$ and $cle d$ be real numbers and $f$ a real function. I am struggling with an integral of the form:
$$I =
int_a^b mathrm{d}x_1 ...mathrm{d}x_m
int_c^d mathrm{d}y_1 ...mathrm{d}y_n
prod_{i=1}^m prod_{j=1}^n f(x_i y_j)$$
Can it be reduced to a single or double integration somehow?
I am interested in the large $m,n$ behavior of $I$. Specifically, suppose that the ratio $alpha = m/n$ is fixed. Can we evaluate in general the limit:
$$lim_{nrightarrowinfty} frac{1}{n} ln I$$
So far, these are the only reductions I've been able to do:
$$begin{aligned}
I & = int_a^b mathrm d x_1 dots mathrm d x_m left( int_c^d mathrm d y
prod_{i = 1}^m f (x_i y) right)^n\
& = int_c^d mathrm d y_1 ldots mathrm d y_m left( int_a^b mathrm d x
prod_{j = 1}^n f (x y_j) right)^m
end{aligned}$$
integration definite-integrals multiple-integral large-deviation-theory
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
$endgroup$
$begingroup$
Even in the simplest case $[a,b]=[c,d]=[0,1]$ with $f(x)=x^p$ $(p>0)$, we have $$I=int_{[0,1]^m}dmathrm{x}int_{[0,1]^n}dmathrm{y},left(x_1^n cdots x_m^n y_1^m cdots y_n^m right)=frac{1}{(pn+1)^m(pm+1)^n}.$$ In this case, along $m/n sim alpha in (0, 1)$, we have $log I sim -(alpha + 1)nlog n$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Nov 30 '18 at 23:24
$begingroup$
@SangchulLee Lovely. Can this be generalized somehow?
$endgroup$
– becko
Dec 3 '18 at 16:32
add a comment |
$begingroup$
Let $a le b$ and $cle d$ be real numbers and $f$ a real function. I am struggling with an integral of the form:
$$I =
int_a^b mathrm{d}x_1 ...mathrm{d}x_m
int_c^d mathrm{d}y_1 ...mathrm{d}y_n
prod_{i=1}^m prod_{j=1}^n f(x_i y_j)$$
Can it be reduced to a single or double integration somehow?
I am interested in the large $m,n$ behavior of $I$. Specifically, suppose that the ratio $alpha = m/n$ is fixed. Can we evaluate in general the limit:
$$lim_{nrightarrowinfty} frac{1}{n} ln I$$
So far, these are the only reductions I've been able to do:
$$begin{aligned}
I & = int_a^b mathrm d x_1 dots mathrm d x_m left( int_c^d mathrm d y
prod_{i = 1}^m f (x_i y) right)^n\
& = int_c^d mathrm d y_1 ldots mathrm d y_m left( int_a^b mathrm d x
prod_{j = 1}^n f (x y_j) right)^m
end{aligned}$$
integration definite-integrals multiple-integral large-deviation-theory
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
$endgroup$
Let $a le b$ and $cle d$ be real numbers and $f$ a real function. I am struggling with an integral of the form:
$$I =
int_a^b mathrm{d}x_1 ...mathrm{d}x_m
int_c^d mathrm{d}y_1 ...mathrm{d}y_n
prod_{i=1}^m prod_{j=1}^n f(x_i y_j)$$
Can it be reduced to a single or double integration somehow?
I am interested in the large $m,n$ behavior of $I$. Specifically, suppose that the ratio $alpha = m/n$ is fixed. Can we evaluate in general the limit:
$$lim_{nrightarrowinfty} frac{1}{n} ln I$$
So far, these are the only reductions I've been able to do:
$$begin{aligned}
I & = int_a^b mathrm d x_1 dots mathrm d x_m left( int_c^d mathrm d y
prod_{i = 1}^m f (x_i y) right)^n\
& = int_c^d mathrm d y_1 ldots mathrm d y_m left( int_a^b mathrm d x
prod_{j = 1}^n f (x y_j) right)^m
end{aligned}$$
integration definite-integrals multiple-integral large-deviation-theory
integration definite-integrals multiple-integral large-deviation-theory
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
edited Nov 30 '18 at 22:26
becko
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
asked Nov 29 '18 at 17:48
beckobecko
2,34931942
2,34931942
$begingroup$
Even in the simplest case $[a,b]=[c,d]=[0,1]$ with $f(x)=x^p$ $(p>0)$, we have $$I=int_{[0,1]^m}dmathrm{x}int_{[0,1]^n}dmathrm{y},left(x_1^n cdots x_m^n y_1^m cdots y_n^m right)=frac{1}{(pn+1)^m(pm+1)^n}.$$ In this case, along $m/n sim alpha in (0, 1)$, we have $log I sim -(alpha + 1)nlog n$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Nov 30 '18 at 23:24
$begingroup$
@SangchulLee Lovely. Can this be generalized somehow?
$endgroup$
– becko
Dec 3 '18 at 16:32
add a comment |
$begingroup$
Even in the simplest case $[a,b]=[c,d]=[0,1]$ with $f(x)=x^p$ $(p>0)$, we have $$I=int_{[0,1]^m}dmathrm{x}int_{[0,1]^n}dmathrm{y},left(x_1^n cdots x_m^n y_1^m cdots y_n^m right)=frac{1}{(pn+1)^m(pm+1)^n}.$$ In this case, along $m/n sim alpha in (0, 1)$, we have $log I sim -(alpha + 1)nlog n$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Nov 30 '18 at 23:24
$begingroup$
@SangchulLee Lovely. Can this be generalized somehow?
$endgroup$
– becko
Dec 3 '18 at 16:32
$begingroup$
Even in the simplest case $[a,b]=[c,d]=[0,1]$ with $f(x)=x^p$ $(p>0)$, we have $$I=int_{[0,1]^m}dmathrm{x}int_{[0,1]^n}dmathrm{y},left(x_1^n cdots x_m^n y_1^m cdots y_n^m right)=frac{1}{(pn+1)^m(pm+1)^n}.$$ In this case, along $m/n sim alpha in (0, 1)$, we have $log I sim -(alpha + 1)nlog n$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Nov 30 '18 at 23:24
$begingroup$
Even in the simplest case $[a,b]=[c,d]=[0,1]$ with $f(x)=x^p$ $(p>0)$, we have $$I=int_{[0,1]^m}dmathrm{x}int_{[0,1]^n}dmathrm{y},left(x_1^n cdots x_m^n y_1^m cdots y_n^m right)=frac{1}{(pn+1)^m(pm+1)^n}.$$ In this case, along $m/n sim alpha in (0, 1)$, we have $log I sim -(alpha + 1)nlog n$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Nov 30 '18 at 23:24
$begingroup$
@SangchulLee Lovely. Can this be generalized somehow?
$endgroup$
– becko
Dec 3 '18 at 16:32
$begingroup$
@SangchulLee Lovely. Can this be generalized somehow?
$endgroup$
– becko
Dec 3 '18 at 16:32
add a comment |
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$begingroup$
Even in the simplest case $[a,b]=[c,d]=[0,1]$ with $f(x)=x^p$ $(p>0)$, we have $$I=int_{[0,1]^m}dmathrm{x}int_{[0,1]^n}dmathrm{y},left(x_1^n cdots x_m^n y_1^m cdots y_n^m right)=frac{1}{(pn+1)^m(pm+1)^n}.$$ In this case, along $m/n sim alpha in (0, 1)$, we have $log I sim -(alpha + 1)nlog n$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Nov 30 '18 at 23:24
$begingroup$
@SangchulLee Lovely. Can this be generalized somehow?
$endgroup$
– becko
Dec 3 '18 at 16:32