Definite integral of a hypergeometric function Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Bounding a polynomial from belowSimplifying real part of hypergeometric function with complex parametersQuestion about Meijer-G definition and identityIntegral with incomplete gamma functionProof of Integral (Gradshteyn & Ryzhik 3.462.1)Challenging definite integral of hypergeometric functionsDefinite integral of exponential, power and Bessel function : a hypergeometric function?Double Gaussian definite integral with one variable limitQuestion over limit of integrand for definite integral.Sum involving hypergeometric 2F2 functionBounding a polynomial from below
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Definite integral of a hypergeometric function
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Bounding a polynomial from belowSimplifying real part of hypergeometric function with complex parametersQuestion about Meijer-G definition and identityIntegral with incomplete gamma functionProof of Integral (Gradshteyn & Ryzhik 3.462.1)Challenging definite integral of hypergeometric functionsDefinite integral of exponential, power and Bessel function : a hypergeometric function?Double Gaussian definite integral with one variable limitQuestion over limit of integrand for definite integral.Sum involving hypergeometric 2F2 functionBounding a polynomial from below
$begingroup$
This question is related to Bounding a polynomial from below but perhaps more straightforward.
Let $sigma >0$ be fixed. For even $k in mathbbN cup 0$, we consider the hypergeometric function
beginequation
varphi_k(x) = _2F_1(-k,k+sigma+frac12;frac12;x^2), quad x in (-1,1).
endequation
My question is the following:
Can we find a "closed form" for the definite integral
beginequation
int_-1^1 (1-x^2)^sigma varphi_k(x)^2 dx
endequation
for all even $kin mathbbN cup 0$?
Some remarks:
Since $k$ is an integer, $varphi_k$ terminates and is thus a polynomial of the form
beginequation
varphi_k(x) = sum_j=0^k (-1)^j k choose j b_j , x^2j quad x in (-1,1),
endequation
where
$b_j = fracBig(k+sigma+frac12Big)_jBig(frac12 Big)_j$
and for $s in mathbbR$, $(s)_j$ denotes the Pochhammer symbol
beginequation
(s)_j=begincases1&j=0\s(s+1)cdots (s+j-1)&j>0.endcases
endequation
In particular, $varphi_0(x) =1$.
Writing the square of the sum as the double sum of the product of the coefficients, and using the linearity of the integral, I obtain something of the form
beginequation
sum_j=0^k sum_ell = 0^k (-1)^j+ell k choose ell k choose j fracb_j b_ell2(j+ell)+1 int_-1^1 (1-x^2)^sigma x^2(ell+j) dx.
endequation
Moreover, according to Wolfram Alpha,
beginequation
int_-1^1 (1-x^2)^sigma x^2(ell+j) dx = fracGamma(j+ell+frac12)Gamma(sigma+1)Gamma(sigma+ell+j+frac32).
endequation
I am interested however in knowing if this can be reduced to an expression which does not contain a double sum and is more "compact".
I made this remark in the previous question, and perhaps they might be useful.
polynomials definite-integrals special-functions
asked Mar 27 at 20:01
bgskbgsk
190111
$endgroup$
add a comment |
$begingroup$
This question is related to Bounding a polynomial from below but perhaps more straightforward.
Let $sigma >0$ be fixed. For even $k in mathbbN cup 0$, we consider the hypergeometric function
beginequation
varphi_k(x) = _2F_1(-k,k+sigma+frac12;frac12;x^2), quad x in (-1,1).
endequation
My question is the following:
Can we find a "closed form" for the definite integral
beginequation
int_-1^1 (1-x^2)^sigma varphi_k(x)^2 dx
endequation
for all even $kin mathbbN cup 0$?
Some remarks:
Since $k$ is an integer, $varphi_k$ terminates and is thus a polynomial of the form
beginequation
varphi_k(x) = sum_j=0^k (-1)^j k choose j b_j , x^2j quad x in (-1,1),
endequation
where
$b_j = fracBig(k+sigma+frac12Big)_jBig(frac12 Big)_j$
and for $s in mathbbR$, $(s)_j$ denotes the Pochhammer symbol
beginequation
(s)_j=begincases1&j=0\s(s+1)cdots (s+j-1)&j>0.endcases
endequation
In particular, $varphi_0(x) =1$.
Writing the square of the sum as the double sum of the product of the coefficients, and using the linearity of the integral, I obtain something of the form
beginequation
sum_j=0^k sum_ell = 0^k (-1)^j+ell k choose ell k choose j fracb_j b_ell2(j+ell)+1 int_-1^1 (1-x^2)^sigma x^2(ell+j) dx.
endequation
Moreover, according to Wolfram Alpha,
beginequation
int_-1^1 (1-x^2)^sigma x^2(ell+j) dx = fracGamma(j+ell+frac12)Gamma(sigma+1)Gamma(sigma+ell+j+frac32).
endequation
I am interested however in knowing if this can be reduced to an expression which does not contain a double sum and is more "compact".
I made this remark in the previous question, and perhaps they might be useful.
polynomials definite-integrals special-functions
asked Mar 27 at 20:01
bgskbgsk
190111
$endgroup$
1
$begingroup$
I don't have a solution, but I'd note that $$ (1-x^2)^sigma phi_k(x) = ,_2F_1(-k-sigma,k+frac12;frac12;x^2)$$ That may be useful.
$endgroup$
– Adam Latosiński
Mar 28 at 0:21
add a comment |
$begingroup$
This question is related to Bounding a polynomial from below but perhaps more straightforward.
Let $sigma >0$ be fixed. For even $k in mathbbN cup 0$, we consider the hypergeometric function
beginequation
varphi_k(x) = _2F_1(-k,k+sigma+frac12;frac12;x^2), quad x in (-1,1).
endequation
My question is the following:
Can we find a "closed form" for the definite integral
beginequation
int_-1^1 (1-x^2)^sigma varphi_k(x)^2 dx
endequation
for all even $kin mathbbN cup 0$?
Some remarks:
Since $k$ is an integer, $varphi_k$ terminates and is thus a polynomial of the form
beginequation
varphi_k(x) = sum_j=0^k (-1)^j k choose j b_j , x^2j quad x in (-1,1),
endequation
where
$b_j = fracBig(k+sigma+frac12Big)_jBig(frac12 Big)_j$
and for $s in mathbbR$, $(s)_j$ denotes the Pochhammer symbol
beginequation
(s)_j=begincases1&j=0\s(s+1)cdots (s+j-1)&j>0.endcases
endequation
In particular, $varphi_0(x) =1$.
Writing the square of the sum as the double sum of the product of the coefficients, and using the linearity of the integral, I obtain something of the form
beginequation
sum_j=0^k sum_ell = 0^k (-1)^j+ell k choose ell k choose j fracb_j b_ell2(j+ell)+1 int_-1^1 (1-x^2)^sigma x^2(ell+j) dx.
endequation
Moreover, according to Wolfram Alpha,
beginequation
int_-1^1 (1-x^2)^sigma x^2(ell+j) dx = fracGamma(j+ell+frac12)Gamma(sigma+1)Gamma(sigma+ell+j+frac32).
endequation
I am interested however in knowing if this can be reduced to an expression which does not contain a double sum and is more "compact".
I made this remark in the previous question, and perhaps they might be useful.
polynomials definite-integrals special-functions
asked Mar 27 at 20:01
bgskbgsk
190111
$endgroup$
This question is related to Bounding a polynomial from below but perhaps more straightforward.
Let $sigma >0$ be fixed. For even $k in mathbbN cup 0$, we consider the hypergeometric function
beginequation
varphi_k(x) = _2F_1(-k,k+sigma+frac12;frac12;x^2), quad x in (-1,1).
endequation
My question is the following:
Can we find a "closed form" for the definite integral
beginequation
int_-1^1 (1-x^2)^sigma varphi_k(x)^2 dx
endequation
for all even $kin mathbbN cup 0$?
Some remarks:
Since $k$ is an integer, $varphi_k$ terminates and is thus a polynomial of the form
beginequation
varphi_k(x) = sum_j=0^k (-1)^j k choose j b_j , x^2j quad x in (-1,1),
endequation
where
$b_j = fracBig(k+sigma+frac12Big)_jBig(frac12 Big)_j$
and for $s in mathbbR$, $(s)_j$ denotes the Pochhammer symbol
beginequation
(s)_j=begincases1&j=0\s(s+1)cdots (s+j-1)&j>0.endcases
endequation
In particular, $varphi_0(x) =1$.
Writing the square of the sum as the double sum of the product of the coefficients, and using the linearity of the integral, I obtain something of the form
beginequation
sum_j=0^k sum_ell = 0^k (-1)^j+ell k choose ell k choose j fracb_j b_ell2(j+ell)+1 int_-1^1 (1-x^2)^sigma x^2(ell+j) dx.
endequation
Moreover, according to Wolfram Alpha,
beginequation
int_-1^1 (1-x^2)^sigma x^2(ell+j) dx = fracGamma(j+ell+frac12)Gamma(sigma+1)Gamma(sigma+ell+j+frac32).
endequation
I am interested however in knowing if this can be reduced to an expression which does not contain a double sum and is more "compact".
I made this remark in the previous question, and perhaps they might be useful.
polynomials definite-integrals special-functions
polynomials definite-integrals special-functions
asked Mar 27 at 20:01
bgskbgsk
190111
asked Mar 27 at 20:01
bgskbgsk
190111
edited Mar 27 at 20:29
bgsk
asked Mar 27 at 20:01
bgskbgsk
190111
asked Mar 27 at 20:01
bgskbgsk
190111
asked Mar 27 at 20:01
bgskbgsk
190111
190111
1
$begingroup$
I don't have a solution, but I'd note that $$ (1-x^2)^sigma phi_k(x) = ,_2F_1(-k-sigma,k+frac12;frac12;x^2)$$ That may be useful.
$endgroup$
– Adam Latosiński
Mar 28 at 0:21
add a comment |
1
$begingroup$
I don't have a solution, but I'd note that $$ (1-x^2)^sigma phi_k(x) = ,_2F_1(-k-sigma,k+frac12;frac12;x^2)$$ That may be useful.
$endgroup$
– Adam Latosiński
Mar 28 at 0:21
1
1
$begingroup$
I don't have a solution, but I'd note that $$ (1-x^2)^sigma phi_k(x) = ,_2F_1(-k-sigma,k+frac12;frac12;x^2)$$ That may be useful.
$endgroup$
– Adam Latosiński
Mar 28 at 0:21
$begingroup$
I don't have a solution, but I'd note that $$ (1-x^2)^sigma phi_k(x) = ,_2F_1(-k-sigma,k+frac12;frac12;x^2)$$ That may be useful.
$endgroup$
– Adam Latosiński
Mar 28 at 0:21
add a comment |
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$begingroup$
I don't have a solution, but I'd note that $$ (1-x^2)^sigma phi_k(x) = ,_2F_1(-k-sigma,k+frac12;frac12;x^2)$$ That may be useful.
$endgroup$
– Adam Latosiński
Mar 28 at 0:21