1
$\begingroup$

In [1] (refered to as "the handbook"), it is said that

... the more general integral (Eq 575.16)

$$ \int R(\tau)\sqrt{(\tau-r_1)(\tau-r_2)(\tau-r_3)(\tau-r_4)(\tau-r_5)(\tau-r_6)}\,d\tau $$ can be reduced to elliptic integrals if the six branch-points $r_1,...,r_6$ form three pairs of points of an involution.

First of all, I think there is a typo; it should be $R(\tau)/\sqrt{\cdots}$. Then, the handbook cites [2] for reference. The relavant context seems to be from the bottom of page 289 until the top of page 291. However, my poor German knowledge does not find there a satisfying discussion about roots of polynomials of degree six.

So, where can I find a proof of this statement? How does this reduction looks like?

For reference, the integral that I'm looking at looks like $$ \int \frac{R(\tau)}{\sqrt{a_0\tau^6+a_1\tau^5+a_2\tau^4+a_3\tau^3-a_2\tau^2+a_1\tau^1-a_0}}\,d\tau.$$ Similar to Eqs. 579.00 and 581.00 of the handbook, but not exactly the same. In particular, the polynomial in the square root has three pais of roots, each pair of the form $rr'=-1$.

[1] Byrd, P.F.; Friedman, M.D., Handbook of elliptic integrals for engineers and scientists. 2nd ed., revised, Die Grundlehren der mathematischen Wissenschaften, 67. Berlin-Heidelberg-New York: Springer-Verlag. XVI, 358 p. with 22 fig. (1971). ZBL0213.16602.

[2] Königsberger, L., On the reduction of hyperelliptic integrals to elliptic integrals., Borchardt J. LXXXV, 273-294 (1878). ZBL10.0323.01.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.