Closed form for $intlimits_0^infty frac{x^alpha(1-x)^beta}{(x-c)^gamma(x-b)^gamma(x-bar{b})^gamma}...
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
4,04721234
asked Nov 14 '17 at 9:51
ASMASM
787
add a comment |
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
4,04721234
asked Nov 14 '17 at 9:51
ASMASM
787
add a comment |
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
4,04721234
asked Nov 14 '17 at 9:51
ASMASM
787
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
4,04721234
asked Nov 14 '17 at 9:51
ASMASM
787
edited Jan 4 at 19:34
mrtaurho
4,04721234
asked Nov 14 '17 at 9:51
ASMASM
787
edited Jan 4 at 19:34
mrtaurho
4,04721234
edited Jan 4 at 19:34
mrtaurho
4,04721234
edited Jan 4 at 19:34
mrtaurho
4,04721234
4,04721234
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 |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2519650%2fclosed-form-for-int-limits-0-infty-fracx-alpha1-x-betax-c-gammax%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2519650%2fclosed-form-for-int-limits-0-infty-fracx-alpha1-x-betax-c-gammax%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
