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Evaluating $int_0^infty x^2alpha-1(e^-x^alpha-e^-r^alpha x^alpha)^n-2e^-x^alpha(1+r^alpha),dx$
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I am looking for a tractable closed form of the following integral where $alpha>0,r>1$:
$$I=int_0^infty x^2alpha-1(e^-x^alpha-e^-r^alpha x^alpha)^n-2e^-x^alpha(1+r^alpha),dxtag1$$
It can be assumed that $nge2$ is an integer.
Plugging $I$ directly into Mathematica, I get a sort of ugly answer involving non-elementary functions. Not to mention there is some confusion over the convergence of the integral.
Is there a way to tackle the integral by hand and arrive at a form involving elementary functions?
The context is in finding the density of $(max_1le ile nX_i/min_1le ile n X_i)$ where $X_1,X_2,cdots,X_n$ are independent random variables having the common Weibull density
$$f(x)=alpha x^alpha-1e^-x^alphamathbf1_x>0,,quadalpha>0$$
I had used a change of variables $(max X_i,min X_i)toleft(fracmax X_imin X_i,min X_iright)$ to be faced with $(1)$ trying to find the marginal density $fracmax X_imin X_i$ from the joint density of $left(fracmax X_imin X_i,min X_iright)$.
integration probability-distributions improper-integrals closed-form
asked Jul 18 at 16:04
StubbornAtom
3,79311134
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up vote
1
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I am looking for a tractable closed form of the following integral where $alpha>0,r>1$:
$$I=int_0^infty x^2alpha-1(e^-x^alpha-e^-r^alpha x^alpha)^n-2e^-x^alpha(1+r^alpha),dxtag1$$
It can be assumed that $nge2$ is an integer.
Plugging $I$ directly into Mathematica, I get a sort of ugly answer involving non-elementary functions. Not to mention there is some confusion over the convergence of the integral.
Is there a way to tackle the integral by hand and arrive at a form involving elementary functions?
The context is in finding the density of $(max_1le ile nX_i/min_1le ile n X_i)$ where $X_1,X_2,cdots,X_n$ are independent random variables having the common Weibull density
$$f(x)=alpha x^alpha-1e^-x^alphamathbf1_x>0,,quadalpha>0$$
I had used a change of variables $(max X_i,min X_i)toleft(fracmax X_imin X_i,min X_iright)$ to be faced with $(1)$ trying to find the marginal density $fracmax X_imin X_i$ from the joint density of $left(fracmax X_imin X_i,min X_iright)$.
integration probability-distributions improper-integrals closed-form
asked Jul 18 at 16:04
StubbornAtom
3,79311134
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am looking for a tractable closed form of the following integral where $alpha>0,r>1$:
$$I=int_0^infty x^2alpha-1(e^-x^alpha-e^-r^alpha x^alpha)^n-2e^-x^alpha(1+r^alpha),dxtag1$$
It can be assumed that $nge2$ is an integer.
Plugging $I$ directly into Mathematica, I get a sort of ugly answer involving non-elementary functions. Not to mention there is some confusion over the convergence of the integral.
Is there a way to tackle the integral by hand and arrive at a form involving elementary functions?
The context is in finding the density of $(max_1le ile nX_i/min_1le ile n X_i)$ where $X_1,X_2,cdots,X_n$ are independent random variables having the common Weibull density
$$f(x)=alpha x^alpha-1e^-x^alphamathbf1_x>0,,quadalpha>0$$
I had used a change of variables $(max X_i,min X_i)toleft(fracmax X_imin X_i,min X_iright)$ to be faced with $(1)$ trying to find the marginal density $fracmax X_imin X_i$ from the joint density of $left(fracmax X_imin X_i,min X_iright)$.
integration probability-distributions improper-integrals closed-form
asked Jul 18 at 16:04
StubbornAtom
3,79311134
I am looking for a tractable closed form of the following integral where $alpha>0,r>1$:
$$I=int_0^infty x^2alpha-1(e^-x^alpha-e^-r^alpha x^alpha)^n-2e^-x^alpha(1+r^alpha),dxtag1$$
It can be assumed that $nge2$ is an integer.
Plugging $I$ directly into Mathematica, I get a sort of ugly answer involving non-elementary functions. Not to mention there is some confusion over the convergence of the integral.
Is there a way to tackle the integral by hand and arrive at a form involving elementary functions?
The context is in finding the density of $(max_1le ile nX_i/min_1le ile n X_i)$ where $X_1,X_2,cdots,X_n$ are independent random variables having the common Weibull density
$$f(x)=alpha x^alpha-1e^-x^alphamathbf1_x>0,,quadalpha>0$$
I had used a change of variables $(max X_i,min X_i)toleft(fracmax X_imin X_i,min X_iright)$ to be faced with $(1)$ trying to find the marginal density $fracmax X_imin X_i$ from the joint density of $left(fracmax X_imin X_i,min X_iright)$.
integration probability-distributions improper-integrals closed-form
asked Jul 18 at 16:04
StubbornAtom
3,79311134
asked Jul 18 at 16:04
StubbornAtom
3,79311134
asked Jul 18 at 16:04
StubbornAtom
3,79311134
asked Jul 18 at 16:04
StubbornAtom
3,79311134
3,79311134
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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up vote
2
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You can start with the substitution $x^alpha = y$ . If we let $R = r^alpha$ , it leads to
$$ I = frac1alpha int limits_0^infty y left(mathrme^-y - mathrme^-R yright)^n-2 mathrme^-(1+R) y , mathrmd y , .$$
In this form it is even more evident that there are no convergence issues for the given ranges of the parameters.
If you want to avoid special functions, you should probably expand the integrand using the binomial theorem:
beginalign
I &= frac1alpha sum limits_k=0^n-2 n-2choose k (-1)^k int limits_0^infty y mathrme^-[R(k+1) + n-2-k+1] y , mathrmd y \
&= frac1alpha sum limits_k=0^n-2 n-2choose k frac(-1)^k[n + (r^alpha-1)(k+1)]^2 , .
endalign
I do not see further simplifications for general values of the parameters, but it is a finite sum of nice functions.
In order to get a solution in terms of special functions, the substitution $mathrme^-(R-1) y = t$ can be used to find
beginalign
I &= - frac1alpha(R-1)^2 int limits_0^1 ln(t) t^fracnR-1 (1-t)^n-2 , mathrmd t \
&= - frac1alpha (r^alpha -1)^2 operatornamepartial_1 operatornameB left(fracnr^alpha - 1 + 1 , n-1right)
endalign
with the beta function $operatornameB$, which eventually reduces to the result given by Mathematica.
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
Thanks, the expression with the binomial coefficient was all I was looking for.
– StubbornAtom
Jul 18 at 17:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You can start with the substitution $x^alpha = y$ . If we let $R = r^alpha$ , it leads to
$$ I = frac1alpha int limits_0^infty y left(mathrme^-y - mathrme^-R yright)^n-2 mathrme^-(1+R) y , mathrmd y , .$$
In this form it is even more evident that there are no convergence issues for the given ranges of the parameters.
If you want to avoid special functions, you should probably expand the integrand using the binomial theorem:
beginalign
I &= frac1alpha sum limits_k=0^n-2 n-2choose k (-1)^k int limits_0^infty y mathrme^-[R(k+1) + n-2-k+1] y , mathrmd y \
&= frac1alpha sum limits_k=0^n-2 n-2choose k frac(-1)^k[n + (r^alpha-1)(k+1)]^2 , .
endalign
I do not see further simplifications for general values of the parameters, but it is a finite sum of nice functions.
In order to get a solution in terms of special functions, the substitution $mathrme^-(R-1) y = t$ can be used to find
beginalign
I &= - frac1alpha(R-1)^2 int limits_0^1 ln(t) t^fracnR-1 (1-t)^n-2 , mathrmd t \
&= - frac1alpha (r^alpha -1)^2 operatornamepartial_1 operatornameB left(fracnr^alpha - 1 + 1 , n-1right)
endalign
with the beta function $operatornameB$, which eventually reduces to the result given by Mathematica.
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
Thanks, the expression with the binomial coefficient was all I was looking for.
– StubbornAtom
Jul 18 at 17:53
add a comment |Â
up vote
2
down vote
accepted
You can start with the substitution $x^alpha = y$ . If we let $R = r^alpha$ , it leads to
$$ I = frac1alpha int limits_0^infty y left(mathrme^-y - mathrme^-R yright)^n-2 mathrme^-(1+R) y , mathrmd y , .$$
In this form it is even more evident that there are no convergence issues for the given ranges of the parameters.
If you want to avoid special functions, you should probably expand the integrand using the binomial theorem:
beginalign
I &= frac1alpha sum limits_k=0^n-2 n-2choose k (-1)^k int limits_0^infty y mathrme^-[R(k+1) + n-2-k+1] y , mathrmd y \
&= frac1alpha sum limits_k=0^n-2 n-2choose k frac(-1)^k[n + (r^alpha-1)(k+1)]^2 , .
endalign
I do not see further simplifications for general values of the parameters, but it is a finite sum of nice functions.
In order to get a solution in terms of special functions, the substitution $mathrme^-(R-1) y = t$ can be used to find
beginalign
I &= - frac1alpha(R-1)^2 int limits_0^1 ln(t) t^fracnR-1 (1-t)^n-2 , mathrmd t \
&= - frac1alpha (r^alpha -1)^2 operatornamepartial_1 operatornameB left(fracnr^alpha - 1 + 1 , n-1right)
endalign
with the beta function $operatornameB$, which eventually reduces to the result given by Mathematica.
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
Thanks, the expression with the binomial coefficient was all I was looking for.
– StubbornAtom
Jul 18 at 17:53
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You can start with the substitution $x^alpha = y$ . If we let $R = r^alpha$ , it leads to
$$ I = frac1alpha int limits_0^infty y left(mathrme^-y - mathrme^-R yright)^n-2 mathrme^-(1+R) y , mathrmd y , .$$
In this form it is even more evident that there are no convergence issues for the given ranges of the parameters.
If you want to avoid special functions, you should probably expand the integrand using the binomial theorem:
beginalign
I &= frac1alpha sum limits_k=0^n-2 n-2choose k (-1)^k int limits_0^infty y mathrme^-[R(k+1) + n-2-k+1] y , mathrmd y \
&= frac1alpha sum limits_k=0^n-2 n-2choose k frac(-1)^k[n + (r^alpha-1)(k+1)]^2 , .
endalign
I do not see further simplifications for general values of the parameters, but it is a finite sum of nice functions.
In order to get a solution in terms of special functions, the substitution $mathrme^-(R-1) y = t$ can be used to find
beginalign
I &= - frac1alpha(R-1)^2 int limits_0^1 ln(t) t^fracnR-1 (1-t)^n-2 , mathrmd t \
&= - frac1alpha (r^alpha -1)^2 operatornamepartial_1 operatornameB left(fracnr^alpha - 1 + 1 , n-1right)
endalign
with the beta function $operatornameB$, which eventually reduces to the result given by Mathematica.
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
You can start with the substitution $x^alpha = y$ . If we let $R = r^alpha$ , it leads to
$$ I = frac1alpha int limits_0^infty y left(mathrme^-y - mathrme^-R yright)^n-2 mathrme^-(1+R) y , mathrmd y , .$$
In this form it is even more evident that there are no convergence issues for the given ranges of the parameters.
If you want to avoid special functions, you should probably expand the integrand using the binomial theorem:
beginalign
I &= frac1alpha sum limits_k=0^n-2 n-2choose k (-1)^k int limits_0^infty y mathrme^-[R(k+1) + n-2-k+1] y , mathrmd y \
&= frac1alpha sum limits_k=0^n-2 n-2choose k frac(-1)^k[n + (r^alpha-1)(k+1)]^2 , .
endalign
I do not see further simplifications for general values of the parameters, but it is a finite sum of nice functions.
In order to get a solution in terms of special functions, the substitution $mathrme^-(R-1) y = t$ can be used to find
beginalign
I &= - frac1alpha(R-1)^2 int limits_0^1 ln(t) t^fracnR-1 (1-t)^n-2 , mathrmd t \
&= - frac1alpha (r^alpha -1)^2 operatornamepartial_1 operatornameB left(fracnr^alpha - 1 + 1 , n-1right)
endalign
with the beta function $operatornameB$, which eventually reduces to the result given by Mathematica.
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
edited Jul 18 at 17:36
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
answered Jul 18 at 17:27
ComplexYetTrivial
2,607624
2,607624
Thanks, the expression with the binomial coefficient was all I was looking for.
– StubbornAtom
Jul 18 at 17:53
add a comment |Â
Thanks, the expression with the binomial coefficient was all I was looking for.
– StubbornAtom
Jul 18 at 17:53
Thanks, the expression with the binomial coefficient was all I was looking for.
– StubbornAtom
Jul 18 at 17:53
Thanks, the expression with the binomial coefficient was all I was looking for.
– StubbornAtom
Jul 18 at 17:53
add a comment |Â
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