Integral $int_{sqrt{33}}^inftyfrac{dx}{sqrt{x^3-11x^2+11x+121}}$
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How can we prove $$I:=int_{sqrt{33}}^inftyfrac{dx}{sqrt{x^3-11x^2+11x+121}}\=frac1{6sqrt2pi^2}Gamma(1/11)Gamma(3/11)Gamma(4/11)Gamma(5/11)Gamma(9/11)?$$
Thoughts of this integral
This integral is in the form $$intfrac{1}{sqrt{P(x)}}dx,$$where $deg P=3$. Therefore, this integral is an elliptic integral.
Also, I believe this integral is strongly related to Weierstrass elliptic function $wp(u)$. In order to find $g_2$ and $g_3$, substitute $x=t+11/3$ to get $$I=2int_{sqrt{33}-11/3}^inftyfrac{dt}{sqrt{4t^3-352/3t+6776/27}}$$
The question boils down to finding $wp(I;352/3,-6776/27)$ but I seem to be on the wrong track.
integration definite-integrals elliptic-integrals
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
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add a comment |
$begingroup$
How can we prove $$I:=int_{sqrt{33}}^inftyfrac{dx}{sqrt{x^3-11x^2+11x+121}}\=frac1{6sqrt2pi^2}Gamma(1/11)Gamma(3/11)Gamma(4/11)Gamma(5/11)Gamma(9/11)?$$
Thoughts of this integral
This integral is in the form $$intfrac{1}{sqrt{P(x)}}dx,$$where $deg P=3$. Therefore, this integral is an elliptic integral.
Also, I believe this integral is strongly related to Weierstrass elliptic function $wp(u)$. In order to find $g_2$ and $g_3$, substitute $x=t+11/3$ to get $$I=2int_{sqrt{33}-11/3}^inftyfrac{dt}{sqrt{4t^3-352/3t+6776/27}}$$
The question boils down to finding $wp(I;352/3,-6776/27)$ but I seem to be on the wrong track.
integration definite-integrals elliptic-integrals
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
$endgroup$
$begingroup$
I see a product of gamma values, probably the Beta function of Euler is used. Maybe it's better to prove the formula from right to left. (I don't have a solution)
$endgroup$
– FDP
Jan 18 at 15:12
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Possibly useful is a google search for "products of gamma functions" + "elliptic integrals".
$endgroup$
– Dave L. Renfro
Jan 18 at 15:30
$begingroup$
mathworld.wolfram.com/EllipticIntegralSingularValue.html
$endgroup$
– DavidP
Jan 18 at 16:33
$begingroup$
see something similar: math.stackexchange.com/q/2407578/515527
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– Zacky
Jan 18 at 19:36
$begingroup$
See related math.stackexchange.com/a/2391675/72031
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– Paramanand Singh
Feb 9 at 8:19
add a comment |
$begingroup$
How can we prove $$I:=int_{sqrt{33}}^inftyfrac{dx}{sqrt{x^3-11x^2+11x+121}}\=frac1{6sqrt2pi^2}Gamma(1/11)Gamma(3/11)Gamma(4/11)Gamma(5/11)Gamma(9/11)?$$
Thoughts of this integral
This integral is in the form $$intfrac{1}{sqrt{P(x)}}dx,$$where $deg P=3$. Therefore, this integral is an elliptic integral.
Also, I believe this integral is strongly related to Weierstrass elliptic function $wp(u)$. In order to find $g_2$ and $g_3$, substitute $x=t+11/3$ to get $$I=2int_{sqrt{33}-11/3}^inftyfrac{dt}{sqrt{4t^3-352/3t+6776/27}}$$
The question boils down to finding $wp(I;352/3,-6776/27)$ but I seem to be on the wrong track.
integration definite-integrals elliptic-integrals
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
$endgroup$
How can we prove $$I:=int_{sqrt{33}}^inftyfrac{dx}{sqrt{x^3-11x^2+11x+121}}\=frac1{6sqrt2pi^2}Gamma(1/11)Gamma(3/11)Gamma(4/11)Gamma(5/11)Gamma(9/11)?$$
Thoughts of this integral
This integral is in the form $$intfrac{1}{sqrt{P(x)}}dx,$$where $deg P=3$. Therefore, this integral is an elliptic integral.
Also, I believe this integral is strongly related to Weierstrass elliptic function $wp(u)$. In order to find $g_2$ and $g_3$, substitute $x=t+11/3$ to get $$I=2int_{sqrt{33}-11/3}^inftyfrac{dt}{sqrt{4t^3-352/3t+6776/27}}$$
The question boils down to finding $wp(I;352/3,-6776/27)$ but I seem to be on the wrong track.
integration definite-integrals elliptic-integrals
integration definite-integrals elliptic-integrals
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
asked Jan 18 at 13:41
Kemono ChenKemono Chen
3,3401844
3,3401844
$begingroup$
I see a product of gamma values, probably the Beta function of Euler is used. Maybe it's better to prove the formula from right to left. (I don't have a solution)
$endgroup$
– FDP
Jan 18 at 15:12
$begingroup$
Possibly useful is a google search for "products of gamma functions" + "elliptic integrals".
$endgroup$
– Dave L. Renfro
Jan 18 at 15:30
$begingroup$
mathworld.wolfram.com/EllipticIntegralSingularValue.html
$endgroup$
– DavidP
Jan 18 at 16:33
$begingroup$
see something similar: math.stackexchange.com/q/2407578/515527
$endgroup$
– Zacky
Jan 18 at 19:36
$begingroup$
See related math.stackexchange.com/a/2391675/72031
$endgroup$
– Paramanand Singh
Feb 9 at 8:19
add a comment |
$begingroup$
I see a product of gamma values, probably the Beta function of Euler is used. Maybe it's better to prove the formula from right to left. (I don't have a solution)
$endgroup$
– FDP
Jan 18 at 15:12
$begingroup$
Possibly useful is a google search for "products of gamma functions" + "elliptic integrals".
$endgroup$
– Dave L. Renfro
Jan 18 at 15:30
$begingroup$
mathworld.wolfram.com/EllipticIntegralSingularValue.html
$endgroup$
– DavidP
Jan 18 at 16:33
$begingroup$
see something similar: math.stackexchange.com/q/2407578/515527
$endgroup$
– Zacky
Jan 18 at 19:36
$begingroup$
See related math.stackexchange.com/a/2391675/72031
$endgroup$
– Paramanand Singh
Feb 9 at 8:19
$begingroup$
I see a product of gamma values, probably the Beta function of Euler is used. Maybe it's better to prove the formula from right to left. (I don't have a solution)
$endgroup$
– FDP
Jan 18 at 15:12
$begingroup$
I see a product of gamma values, probably the Beta function of Euler is used. Maybe it's better to prove the formula from right to left. (I don't have a solution)
$endgroup$
– FDP
Jan 18 at 15:12
$begingroup$
Possibly useful is a google search for "products of gamma functions" + "elliptic integrals".
$endgroup$
– Dave L. Renfro
Jan 18 at 15:30
$begingroup$
Possibly useful is a google search for "products of gamma functions" + "elliptic integrals".
$endgroup$
– Dave L. Renfro
Jan 18 at 15:30
$begingroup$
mathworld.wolfram.com/EllipticIntegralSingularValue.html
$endgroup$
– DavidP
Jan 18 at 16:33
$begingroup$
mathworld.wolfram.com/EllipticIntegralSingularValue.html
$endgroup$
– DavidP
Jan 18 at 16:33
$begingroup$
see something similar: math.stackexchange.com/q/2407578/515527
$endgroup$
– Zacky
Jan 18 at 19:36
$begingroup$
see something similar: math.stackexchange.com/q/2407578/515527
$endgroup$
– Zacky
Jan 18 at 19:36
$begingroup$
See related math.stackexchange.com/a/2391675/72031
$endgroup$
– Paramanand Singh
Feb 9 at 8:19
$begingroup$
See related math.stackexchange.com/a/2391675/72031
$endgroup$
– Paramanand Singh
Feb 9 at 8:19
add a comment |
1 Answer
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Referring to Zacky’s comment it suffices to turn the cubic denominator into a quadratic one, then perform the Landen transform to come to an elliptic integral of the first kind equivalent to K(k11)_
fjaclot
answered Jan 21 at 10:56
fjaclotfjaclot
312
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add a comment |
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1 Answer
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Referring to Zacky’s comment it suffices to turn the cubic denominator into a quadratic one, then perform the Landen transform to come to an elliptic integral of the first kind equivalent to K(k11)_
fjaclot
answered Jan 21 at 10:56
fjaclotfjaclot
312
$endgroup$
add a comment |
$begingroup$
Referring to Zacky’s comment it suffices to turn the cubic denominator into a quadratic one, then perform the Landen transform to come to an elliptic integral of the first kind equivalent to K(k11)_
fjaclot
answered Jan 21 at 10:56
fjaclotfjaclot
312
$endgroup$
add a comment |
$begingroup$
Referring to Zacky’s comment it suffices to turn the cubic denominator into a quadratic one, then perform the Landen transform to come to an elliptic integral of the first kind equivalent to K(k11)_
fjaclot
answered Jan 21 at 10:56
fjaclotfjaclot
312
$endgroup$
Referring to Zacky’s comment it suffices to turn the cubic denominator into a quadratic one, then perform the Landen transform to come to an elliptic integral of the first kind equivalent to K(k11)_
fjaclot
answered Jan 21 at 10:56
fjaclotfjaclot
312
answered Jan 21 at 10:56
fjaclotfjaclot
312
answered Jan 21 at 10:56
fjaclotfjaclot
312
answered Jan 21 at 10:56
fjaclotfjaclot
312
312
add a comment |
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$begingroup$
I see a product of gamma values, probably the Beta function of Euler is used. Maybe it's better to prove the formula from right to left. (I don't have a solution)
$endgroup$
– FDP
Jan 18 at 15:12
$begingroup$
Possibly useful is a google search for "products of gamma functions" + "elliptic integrals".
$endgroup$
– Dave L. Renfro
Jan 18 at 15:30
$begingroup$
mathworld.wolfram.com/EllipticIntegralSingularValue.html
$endgroup$
– DavidP
Jan 18 at 16:33
$begingroup$
see something similar: math.stackexchange.com/q/2407578/515527
$endgroup$
– Zacky
Jan 18 at 19:36
$begingroup$
See related math.stackexchange.com/a/2391675/72031
$endgroup$
– Paramanand Singh
Feb 9 at 8:19