Integrating Inverse square root of a polynomial
I'm Looking for integrals of the following kind
$intlimits_{0}^{M} left( -x^3 +bx^2 -omega right)^{-frac{1}{2}} ,dx , ,$
for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.
I haven't found any good way of calculating this integral.
definite-integrals cubic-equations elliptic-integrals
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
add a comment |
I'm Looking for integrals of the following kind
$intlimits_{0}^{M} left( -x^3 +bx^2 -omega right)^{-frac{1}{2}} ,dx , ,$
for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.
I haven't found any good way of calculating this integral.
definite-integrals cubic-equations elliptic-integrals
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^{-frac12}mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13
Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31
add a comment |
I'm Looking for integrals of the following kind
$intlimits_{0}^{M} left( -x^3 +bx^2 -omega right)^{-frac{1}{2}} ,dx , ,$
for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.
I haven't found any good way of calculating this integral.
definite-integrals cubic-equations elliptic-integrals
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
I'm Looking for integrals of the following kind
$intlimits_{0}^{M} left( -x^3 +bx^2 -omega right)^{-frac{1}{2}} ,dx , ,$
for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.
I haven't found any good way of calculating this integral.
definite-integrals cubic-equations elliptic-integrals
definite-integrals cubic-equations elliptic-integrals
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
edited Jul 17 '16 at 7:54
Amir Sagiv
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
asked Jun 15 '16 at 17:45
Amir SagivAmir Sagiv
1708
1708
The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^{-frac12}mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13
Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31
add a comment |
The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^{-frac12}mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13
Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31
The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^{-frac12}mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13
The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^{-frac12}mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13
Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31
Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31
add a comment |
1 Answer
1
active
oldest
votes
I think you'll find everything you need and more here:
http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf
You're welcome.
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
1
I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think you'll find everything you need and more here:
http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf
You're welcome.
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
1
I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53
add a comment |
I think you'll find everything you need and more here:
http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf
You're welcome.
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
1
I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53
add a comment |
I think you'll find everything you need and more here:
http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf
You're welcome.
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
I think you'll find everything you need and more here:
http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf
You're welcome.
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
answered Jun 15 '16 at 17:54
Mathemagician1234Mathemagician1234
13.9k24058
13.9k24058
1
I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53
add a comment |
1
I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53
1
1
I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53
I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53
add a comment |
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The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^{-frac12}mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13
Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31