Closed form for $intlimits_0^infty frac{x^alpha(1-x)^beta}{(x-c)^gamma(x-b)^gamma(x-bar{b})^gamma}...
$begingroup$
I would like to find a "closed form" for the integral
$$I(beta) = int_0^infty frac{x^alpha(1-x)^beta}{(x-c)^gamma(x-b)^gamma(x-bar{b})^gamma}mathrm{d}x$$
where $gamma=3/2$, $alpha,beta>0$ , ${b,bar{b}}$ is a pair of complex conjugates, and $c$ is real.
Defining $aequiv -alpha -beta +7/2$ and supposing $a>0$, we can write $I$ as $[1]$
$$ I(beta) = (-1)^beta B(a,alpha+1) Rleft(a;-beta,frac32,frac32,frac32;-1,-c,-b,-bar{b}right)$$
where $B$ is the Beta function, and $R$ is a generalized Carlson elliptic function. My question is, can this expression be simplified? Is there a way to write this R function in a "more closed form"? For instance, in terms of hypergeometric functions?
In particular, I am interested in taking the derivative with respect to $beta$ and then take the limit $beta rightarrow 0$ (the original integral has a log term). As noted here,
$$ int_0^{infty}!!!!frac{x^{alpha-1}ln^n(cx+d)}{(ax+b)^sigma(cx+d)^rho} ! mathrm dx=(-1)^nleft(!frac{d}{c}!right)^alpha!! b^{-sigma}!frac{partial^n}{partial rho^n}!!left[d^{-rho}B(alpha,sigma+rho-alpha) _2F_1!!left(!sigma,alpha;sigma+rho;1!-!frac{ad}{bc}!right)!right] $$
so maybe there is a similar formula for $I(beta)$ in terms of the Beta function and some hypergeometric function.
$[1]$:
By using equation $(text{T}.2)$ in page $224$.
integration definite-integrals hypergeometric-function elliptic-integrals beta-function
edited Jan 4 at 19:34
mrtaurho
5,68551540
asked Nov 14 '17 at 9:51
ASMASM
787
$endgroup$
add a comment |
$begingroup$
I would like to find a "closed form" for the integral
$$I(beta) = int_0^infty frac{x^alpha(1-x)^beta}{(x-c)^gamma(x-b)^gamma(x-bar{b})^gamma}mathrm{d}x$$
where $gamma=3/2$, $alpha,beta>0$ , ${b,bar{b}}$ is a pair of complex conjugates, and $c$ is real.
Defining $aequiv -alpha -beta +7/2$ and supposing $a>0$, we can write $I$ as $[1]$
$$ I(beta) = (-1)^beta B(a,alpha+1) Rleft(a;-beta,frac32,frac32,frac32;-1,-c,-b,-bar{b}right)$$
where $B$ is the Beta function, and $R$ is a generalized Carlson elliptic function. My question is, can this expression be simplified? Is there a way to write this R function in a "more closed form"? For instance, in terms of hypergeometric functions?
In particular, I am interested in taking the derivative with respect to $beta$ and then take the limit $beta rightarrow 0$ (the original integral has a log term). As noted here,
$$ int_0^{infty}!!!!frac{x^{alpha-1}ln^n(cx+d)}{(ax+b)^sigma(cx+d)^rho} ! mathrm dx=(-1)^nleft(!frac{d}{c}!right)^alpha!! b^{-sigma}!frac{partial^n}{partial rho^n}!!left[d^{-rho}B(alpha,sigma+rho-alpha) _2F_1!!left(!sigma,alpha;sigma+rho;1!-!frac{ad}{bc}!right)!right] $$
so maybe there is a similar formula for $I(beta)$ in terms of the Beta function and some hypergeometric function.
$[1]$:
By using equation $(text{T}.2)$ in page $224$.
integration definite-integrals hypergeometric-function elliptic-integrals beta-function
edited Jan 4 at 19:34
mrtaurho
5,68551540
asked Nov 14 '17 at 9:51
ASMASM
787
$endgroup$
add a comment |
$begingroup$
I would like to find a "closed form" for the integral
$$I(beta) = int_0^infty frac{x^alpha(1-x)^beta}{(x-c)^gamma(x-b)^gamma(x-bar{b})^gamma}mathrm{d}x$$
where $gamma=3/2$, $alpha,beta>0$ , ${b,bar{b}}$ is a pair of complex conjugates, and $c$ is real.
Defining $aequiv -alpha -beta +7/2$ and supposing $a>0$, we can write $I$ as $[1]$
$$ I(beta) = (-1)^beta B(a,alpha+1) Rleft(a;-beta,frac32,frac32,frac32;-1,-c,-b,-bar{b}right)$$
where $B$ is the Beta function, and $R$ is a generalized Carlson elliptic function. My question is, can this expression be simplified? Is there a way to write this R function in a "more closed form"? For instance, in terms of hypergeometric functions?
In particular, I am interested in taking the derivative with respect to $beta$ and then take the limit $beta rightarrow 0$ (the original integral has a log term). As noted here,
$$ int_0^{infty}!!!!frac{x^{alpha-1}ln^n(cx+d)}{(ax+b)^sigma(cx+d)^rho} ! mathrm dx=(-1)^nleft(!frac{d}{c}!right)^alpha!! b^{-sigma}!frac{partial^n}{partial rho^n}!!left[d^{-rho}B(alpha,sigma+rho-alpha) _2F_1!!left(!sigma,alpha;sigma+rho;1!-!frac{ad}{bc}!right)!right] $$
so maybe there is a similar formula for $I(beta)$ in terms of the Beta function and some hypergeometric function.
$[1]$:
By using equation $(text{T}.2)$ in page $224$.
integration definite-integrals hypergeometric-function elliptic-integrals beta-function
edited Jan 4 at 19:34
mrtaurho
5,68551540
asked Nov 14 '17 at 9:51
ASMASM
787
$endgroup$
I would like to find a "closed form" for the integral
$$I(beta) = int_0^infty frac{x^alpha(1-x)^beta}{(x-c)^gamma(x-b)^gamma(x-bar{b})^gamma}mathrm{d}x$$
where $gamma=3/2$, $alpha,beta>0$ , ${b,bar{b}}$ is a pair of complex conjugates, and $c$ is real.
Defining $aequiv -alpha -beta +7/2$ and supposing $a>0$, we can write $I$ as $[1]$
$$ I(beta) = (-1)^beta B(a,alpha+1) Rleft(a;-beta,frac32,frac32,frac32;-1,-c,-b,-bar{b}right)$$
where $B$ is the Beta function, and $R$ is a generalized Carlson elliptic function. My question is, can this expression be simplified? Is there a way to write this R function in a "more closed form"? For instance, in terms of hypergeometric functions?
In particular, I am interested in taking the derivative with respect to $beta$ and then take the limit $beta rightarrow 0$ (the original integral has a log term). As noted here,
$$ int_0^{infty}!!!!frac{x^{alpha-1}ln^n(cx+d)}{(ax+b)^sigma(cx+d)^rho} ! mathrm dx=(-1)^nleft(!frac{d}{c}!right)^alpha!! b^{-sigma}!frac{partial^n}{partial rho^n}!!left[d^{-rho}B(alpha,sigma+rho-alpha) _2F_1!!left(!sigma,alpha;sigma+rho;1!-!frac{ad}{bc}!right)!right] $$
so maybe there is a similar formula for $I(beta)$ in terms of the Beta function and some hypergeometric function.
$[1]$:
By using equation $(text{T}.2)$ in page $224$.
integration definite-integrals hypergeometric-function elliptic-integrals beta-function
integration definite-integrals hypergeometric-function elliptic-integrals beta-function
edited Jan 4 at 19:34
mrtaurho
5,68551540
asked Nov 14 '17 at 9:51
ASMASM
787
edited Jan 4 at 19:34
mrtaurho
5,68551540
asked Nov 14 '17 at 9:51
ASMASM
787
edited Jan 4 at 19:34
mrtaurho
5,68551540
edited Jan 4 at 19:34
mrtaurho
5,68551540
edited Jan 4 at 19:34
mrtaurho
5,68551540
5,68551540
asked Nov 14 '17 at 9:51
ASMASM
787
asked Nov 14 '17 at 9:51
ASMASM
787
asked Nov 14 '17 at 9:51
ASMASM
787
787
add a comment |
add a comment |
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