Challenging definite integral of hypergeometric functions The Next CEO of Stack OverflowSymbolic Integration involving hypergeometric functionsClosed form for integral of inverse hyperbolic function in terms of $_4F_3$Definite integral of arcsine over square-root of quadraticSimplifying real part of hypergeometric function with complex parametersEvaluating a certain integral which generalizes the $_3F_2$ hypergeometric functionWhat is $_3F_2left(1,frac32,2; frac43,frac53; frac427right)$ as an integral?Definite integral of Hypergeometric function 2F1Integral involving the Gauss Hypergeometric functions with quadratic argumentRewriting Appell's Hypergeometric Function $F_1$ in terms of Gauss' Hypergeometric Function $_2F_1$Definite integral of a hypergeometric function
Strange use of "whether ... than ..." in official text
Aggressive Under-Indexing and no data for missing index
Point distance program written without a framework
Is it OK to decorate a log book cover?
Yu-Gi-Oh cards in Python 3
Is there a difference between "Fahrstuhl" and "Aufzug"?
Expressing the idea of having a very busy time
How do I fit a non linear curve?
Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?
How to avoid supervisors with prejudiced views?
Cannot shrink btrfs filesystem although there is still data and metadata space left : ERROR: unable to resize '/home': No space left on device
Ising model simulation
Are the names of these months realistic?
Vector calculus integration identity problem
Why is information "lost" when it got into a black hole?
What is the process for purifying your home if you believe it may have been previously used for pagan worship?
How to use ReplaceAll on an expression that contains a rule
Free fall ellipse or parabola?
Is it ever safe to open a suspicious HTML file (e.g. email attachment)?
Calculate the Mean mean of two numbers
Do scriptures give a method to recognize a truly self-realized person/jivanmukta?
IC has pull-down resistors on SMBus lines?
If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?
Help! I cannot understand this game’s notations!
Challenging definite integral of hypergeometric functions
The Next CEO of Stack OverflowSymbolic Integration involving hypergeometric functionsClosed form for integral of inverse hyperbolic function in terms of $_4F_3$Definite integral of arcsine over square-root of quadraticSimplifying real part of hypergeometric function with complex parametersEvaluating a certain integral which generalizes the $_3F_2$ hypergeometric functionWhat is $_3F_2left(1,frac32,2; frac43,frac53; frac427right)$ as an integral?Definite integral of Hypergeometric function 2F1Integral involving the Gauss Hypergeometric functions with quadratic argumentRewriting Appell's Hypergeometric Function $F_1$ in terms of Gauss' Hypergeometric Function $_2F_1$Definite integral of a hypergeometric function
$begingroup$
Given $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$, define the function $mathcalJ_ell,m,nleft(a,b,cright)$ via the definite integral
$$smallmathcalJ_ell,m,nleft(a,b,cright):=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright),_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right).$$
The Gauss hypergeometric function $_2F_1$ may be defined via the infinite series
$$_2F_1left(alpha,beta;gamma;zright):=sum_n=0^inftyfracleft(alpharight)_n,left(betaright)_nleft(gammaright)_ncdotfracz^nn!;~~~smallleft.$$
Question: Does the integral above possess a closed form representation in terms of hypergeometric (or simpler) functions?
Possible starting point: Lacking any obvious way to evaluate this integral, one strategy that came to mind was to expand the integral as a multiple infinite series, which could then be converted into some hypergeometric representation (assuming such exists). As you can see below, things get messy, and it's hard to tell if any headway is being made.
Let $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$. We find
$$beginalign
mathcalJ_ell,m,nleft(a,b,cright)
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~times,_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right)\
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~timessum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!left(1-tright)^k\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n+k-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,sum_j=0^inftybinomj+ell-1ja^jt^m+j-frac12left(1-tright)^ell+n+k-frac12\
&~~~~~times,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!sum_j=0^inftybinomj+ell-1ja^j\
&~~~~~timesint_0^1mathrmdt,t^m+j-frac12left(1-tright)^ell+n+k-frac12,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~times,_3F_2left(frac12,frac12+m,frac12+m+j;frac32+m,1+ell+m+n+j+k;bright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~timessum_p=0^inftyfracleft(frac12right)_p,left(frac12+mright)_p,left(frac12+m+jright)_p,b^pleft(frac32+mright)_p,left(1+ell+m+n+j+kright)_p,p!.\
endalign$$
Does anybody have any ideas for finishing the derivation, or another approach entirely? This is a difficult and time-consuming problem, but also one that I'm quite eager to crack. Any help would be greatly appreciated!
integration definite-integrals special-functions closed-form hypergeometric-function
edited Mar 28 at 3:40
clathratus
5,0361438
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
$endgroup$
add a comment |
$begingroup$
Given $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$, define the function $mathcalJ_ell,m,nleft(a,b,cright)$ via the definite integral
$$smallmathcalJ_ell,m,nleft(a,b,cright):=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright),_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right).$$
The Gauss hypergeometric function $_2F_1$ may be defined via the infinite series
$$_2F_1left(alpha,beta;gamma;zright):=sum_n=0^inftyfracleft(alpharight)_n,left(betaright)_nleft(gammaright)_ncdotfracz^nn!;~~~smallleft.$$
Question: Does the integral above possess a closed form representation in terms of hypergeometric (or simpler) functions?
Possible starting point: Lacking any obvious way to evaluate this integral, one strategy that came to mind was to expand the integral as a multiple infinite series, which could then be converted into some hypergeometric representation (assuming such exists). As you can see below, things get messy, and it's hard to tell if any headway is being made.
Let $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$. We find
$$beginalign
mathcalJ_ell,m,nleft(a,b,cright)
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~times,_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right)\
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~timessum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!left(1-tright)^k\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n+k-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,sum_j=0^inftybinomj+ell-1ja^jt^m+j-frac12left(1-tright)^ell+n+k-frac12\
&~~~~~times,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!sum_j=0^inftybinomj+ell-1ja^j\
&~~~~~timesint_0^1mathrmdt,t^m+j-frac12left(1-tright)^ell+n+k-frac12,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~times,_3F_2left(frac12,frac12+m,frac12+m+j;frac32+m,1+ell+m+n+j+k;bright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~timessum_p=0^inftyfracleft(frac12right)_p,left(frac12+mright)_p,left(frac12+m+jright)_p,b^pleft(frac32+mright)_p,left(1+ell+m+n+j+kright)_p,p!.\
endalign$$
Does anybody have any ideas for finishing the derivation, or another approach entirely? This is a difficult and time-consuming problem, but also one that I'm quite eager to crack. Any help would be greatly appreciated!
integration definite-integrals special-functions closed-form hypergeometric-function
edited Mar 28 at 3:40
clathratus
5,0361438
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
$endgroup$
add a comment |
$begingroup$
Given $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$, define the function $mathcalJ_ell,m,nleft(a,b,cright)$ via the definite integral
$$smallmathcalJ_ell,m,nleft(a,b,cright):=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright),_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right).$$
The Gauss hypergeometric function $_2F_1$ may be defined via the infinite series
$$_2F_1left(alpha,beta;gamma;zright):=sum_n=0^inftyfracleft(alpharight)_n,left(betaright)_nleft(gammaright)_ncdotfracz^nn!;~~~smallleft.$$
Question: Does the integral above possess a closed form representation in terms of hypergeometric (or simpler) functions?
Possible starting point: Lacking any obvious way to evaluate this integral, one strategy that came to mind was to expand the integral as a multiple infinite series, which could then be converted into some hypergeometric representation (assuming such exists). As you can see below, things get messy, and it's hard to tell if any headway is being made.
Let $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$. We find
$$beginalign
mathcalJ_ell,m,nleft(a,b,cright)
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~times,_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right)\
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~timessum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!left(1-tright)^k\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n+k-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,sum_j=0^inftybinomj+ell-1ja^jt^m+j-frac12left(1-tright)^ell+n+k-frac12\
&~~~~~times,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!sum_j=0^inftybinomj+ell-1ja^j\
&~~~~~timesint_0^1mathrmdt,t^m+j-frac12left(1-tright)^ell+n+k-frac12,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~times,_3F_2left(frac12,frac12+m,frac12+m+j;frac32+m,1+ell+m+n+j+k;bright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~timessum_p=0^inftyfracleft(frac12right)_p,left(frac12+mright)_p,left(frac12+m+jright)_p,b^pleft(frac32+mright)_p,left(1+ell+m+n+j+kright)_p,p!.\
endalign$$
Does anybody have any ideas for finishing the derivation, or another approach entirely? This is a difficult and time-consuming problem, but also one that I'm quite eager to crack. Any help would be greatly appreciated!
integration definite-integrals special-functions closed-form hypergeometric-function
edited Mar 28 at 3:40
clathratus
5,0361438
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
$endgroup$
Given $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$, define the function $mathcalJ_ell,m,nleft(a,b,cright)$ via the definite integral
$$smallmathcalJ_ell,m,nleft(a,b,cright):=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright),_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right).$$
The Gauss hypergeometric function $_2F_1$ may be defined via the infinite series
$$_2F_1left(alpha,beta;gamma;zright):=sum_n=0^inftyfracleft(alpharight)_n,left(betaright)_nleft(gammaright)_ncdotfracz^nn!;~~~smallleft.$$
Question: Does the integral above possess a closed form representation in terms of hypergeometric (or simpler) functions?
Possible starting point: Lacking any obvious way to evaluate this integral, one strategy that came to mind was to expand the integral as a multiple infinite series, which could then be converted into some hypergeometric representation (assuming such exists). As you can see below, things get messy, and it's hard to tell if any headway is being made.
Let $ellinmathbbNlandleft(m,nright)inmathbbZ_ge0^2landleft(a,b,cright)inmathbbR^3land0<a<b<1land0<c<1$. We find
$$beginalign
mathcalJ_ell,m,nleft(a,b,cright)
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~times,_2F_1left(frac12,n+frac12;n+frac32;cleft(1-tright)right)\
&=int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&~~~~~timessum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!left(1-tright)^k\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,fract^m-frac12left(1-tright)^ell+n+k-frac12left(1-atright)^ell,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=smallsum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!int_0^1mathrmdt,sum_j=0^inftybinomj+ell-1ja^jt^m+j-frac12left(1-tright)^ell+n+k-frac12\
&~~~~~times,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_k=0^inftyfracleft(frac12right)_k,left(n+frac12right)_k,c^kleft(n+frac32right)_k,k!sum_j=0^inftybinomj+ell-1ja^j\
&~~~~~timesint_0^1mathrmdt,t^m+j-frac12left(1-tright)^ell+n+k-frac12,_2F_1left(frac12,m+frac12;m+frac32;btright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~times,_3F_2left(frac12,frac12+m,frac12+m+j;frac32+m,1+ell+m+n+j+k;bright)\
&=sum_j=0^inftysum_k=0^inftyfracbinomj+ell-1jleft(frac12right)_k,left(n+frac12right)_k,a^j,c^kleft(n+frac32right)_k,k!,operatornameBleft(frac12+m+j,frac12+ell+n+kright)\
&~~~~~timessum_p=0^inftyfracleft(frac12right)_p,left(frac12+mright)_p,left(frac12+m+jright)_p,b^pleft(frac32+mright)_p,left(1+ell+m+n+j+kright)_p,p!.\
endalign$$
Does anybody have any ideas for finishing the derivation, or another approach entirely? This is a difficult and time-consuming problem, but also one that I'm quite eager to crack. Any help would be greatly appreciated!
integration definite-integrals special-functions closed-form hypergeometric-function
integration definite-integrals special-functions closed-form hypergeometric-function
edited Mar 28 at 3:40
clathratus
5,0361438
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
edited Mar 28 at 3:40
clathratus
5,0361438
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
edited Mar 28 at 3:40
clathratus
5,0361438
edited Mar 28 at 3:40
clathratus
5,0361438
edited Mar 28 at 3:40
clathratus
5,0361438
5,0361438
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
asked Aug 12 '16 at 17:20
David HDavid H
21.7k24693
21.7k24693
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $m,ninmathbb Z_ge0$ the integrand can be rewritten in terms of elementary functions using that
$$z^m-frac12_2F_1left(frac12,m+frac12;m+frac32;zright)=fracleft(frac12right)_m+1m!left[frac2arcsinsqrt zz-sqrt1-zcdotsum_k=1^mfrac(k-1)!,z^k-frac32left(frac12right)_kright].$$
The initial integral therefore breaks down into finite number of pieces of 3 different types - however, neither of them looks computable.
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
add a comment |
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%2f1890376%2fchallenging-definite-integral-of-hypergeometric-functions%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $m,ninmathbb Z_ge0$ the integrand can be rewritten in terms of elementary functions using that
$$z^m-frac12_2F_1left(frac12,m+frac12;m+frac32;zright)=fracleft(frac12right)_m+1m!left[frac2arcsinsqrt zz-sqrt1-zcdotsum_k=1^mfrac(k-1)!,z^k-frac32left(frac12right)_kright].$$
The initial integral therefore breaks down into finite number of pieces of 3 different types - however, neither of them looks computable.
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
add a comment |
$begingroup$
For $m,ninmathbb Z_ge0$ the integrand can be rewritten in terms of elementary functions using that
$$z^m-frac12_2F_1left(frac12,m+frac12;m+frac32;zright)=fracleft(frac12right)_m+1m!left[frac2arcsinsqrt zz-sqrt1-zcdotsum_k=1^mfrac(k-1)!,z^k-frac32left(frac12right)_kright].$$
The initial integral therefore breaks down into finite number of pieces of 3 different types - however, neither of them looks computable.
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
add a comment |
$begingroup$
For $m,ninmathbb Z_ge0$ the integrand can be rewritten in terms of elementary functions using that
$$z^m-frac12_2F_1left(frac12,m+frac12;m+frac32;zright)=fracleft(frac12right)_m+1m!left[frac2arcsinsqrt zz-sqrt1-zcdotsum_k=1^mfrac(k-1)!,z^k-frac32left(frac12right)_kright].$$
The initial integral therefore breaks down into finite number of pieces of 3 different types - however, neither of them looks computable.
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
For $m,ninmathbb Z_ge0$ the integrand can be rewritten in terms of elementary functions using that
$$z^m-frac12_2F_1left(frac12,m+frac12;m+frac32;zright)=fracleft(frac12right)_m+1m!left[frac2arcsinsqrt zz-sqrt1-zcdotsum_k=1^mfrac(k-1)!,z^k-frac32left(frac12right)_kright].$$
The initial integral therefore breaks down into finite number of pieces of 3 different types - however, neither of them looks computable.
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
answered Aug 17 '16 at 11:41
Start wearing purpleStart wearing purple
47.3k12135192
47.3k12135192
add a comment |
add a comment |
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%2f1890376%2fchallenging-definite-integral-of-hypergeometric-functions%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
