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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.

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There's no typo in 575.16, for $R(t)\sqrt{P(t)}=R(t)P(t)/\sqrt{P(t)}=S(t)/\sqrt{P(t)}$. Regarding the main question, suppose the $r_i$'s form three pairs of points of an involution (i.e. three points, and the images of those three points under a self-inverse Möbius transformation). Then there is a linear fractional transformation mapping those $r_i$'s onto the points $$z,\frac1z,ze^{2\pi i/3},\frac1ze^{2\pi i/3},ze^{-2\pi i/3},\frac1ze^{-2\pi i/3}$$ where $z$ is some complex number. When this transformation is applied to the integral, it takes the form $\int R(t)/\sqrt{t^6+at^3+1}\,dt$. The polynomial under root being a palindromic sextic, it can be reduced to elliptic integrals by substituting $u=t+\frac1t$ (B&F 579.01, 581.02): $$2\int_0^x\frac{R(t)}{\sqrt{P(t)}}\,dt=\int_{1+1/x^2}^{\infty}\frac{R_1(t)+(t+2)R_2(t)}{\sqrt{(t+2)Q(t)}}\,dt+\int_{1+1/x^2}^{\infty}\frac{R_1(t)+(t-2)R_2(t)}{\sqrt{(t-2)Q(t)}}\,dt$$ $$2\int_1^x\frac{R(t)}{\sqrt{P(t)}}\,dt=\int_2^{1+1/x^2}\frac{R_1(t)+(t-2)R_2(t)}{\sqrt{(t-2)Q(t)}}\,dt-\int_2^{1+1/x^2}\frac{R_1(t)+(t+2)R_2(t)}{\sqrt{(t+2)Q(t)}}\,dt$$ where $P(t)=t^3Q(t+1/t)$ and

  • in the first case, $x\le1$ and $R((t-\sqrt{t^2-4})/2)=R_1(t)+R_2(t)\sqrt{t^2-4}$
  • in the second case, $x\ge1$ and $R((t+\sqrt{t^2-4})/2)=R_1(t)-R_2(t)\sqrt{t^2-4}$

Since the polynomials under root are now quartics ($Q$ is cubic, the "folding" of palindromic $P$ onto itself), elliptic integrals can be used. I reduced the sextic further than what B&F requires for convenience's sake of the desired proof.

For your specific integral: $$\int\frac{R(t)}{\sqrt{a_0t^6+a_1t^5+a_2t^4+a_3t^3-a_2t^2+a_1t-a_0}}\,dt$$ the reduction to elliptic integrals requires a complex substitution $u=-it$: $$=\int\frac{iR(iu)}{\sqrt{-a_0u^6+(a_1i)u^5+a_2u^4-(a_3i)u^3+a_2u^2+(a_1i)u-a_0}}\,du$$ You will have to consider branches carefully, but the reduction is still possible. (Elliptic integrals accept arbitrary complex numbers for their arguments.)

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  • $\begingroup$ Thank you for the answer! I was afraid of complex argument at the time of asking this question. I'm now more used to it. $\endgroup$
    – Hao Chen
    Dec 15, 2020 at 11:16

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