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I found myself with the following integral

$$ \int_{b_1}^{b_2} \sqrt{\frac{(b-b_1)(b_2-b)(b_3-b)}{(b_4-b)}} \ db $$

with $ b_1 < b_2 < b_3 < b_4 $. I know that

$$ \int_{b_1}^{b_2} \frac{db}{\sqrt{(b-b_1)(b_2-b)(b_3-b)(b_4-b)}} $$

is equal to

$$ \frac{2}{(b_4-b_2)(b_3-b_1)} K(k) $$

where $K(k)$ is the complete elliptic integral of first kind, so I suspect that this integral is somehow reducible to a linear combination of elliptic integrals, but I can't find the right way.

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  • $\begingroup$ Mathematica produces some indefinite integrals, but I don't trust them, since they use expressions that are either invalid or complex (and therefore possibly invalid through poor choices of $i$ vs $-i$). $\endgroup$
    – user44143
    Commented May 12, 2022 at 17:08
  • $\begingroup$ Unlike the integral which "you know", this one is not an elliptic integral (the integrand has 5 singularities), so it in probably not expressed in terms of the ordinary special functions. $\endgroup$
    – Alexandre Eremenko
    Commented May 12, 2022 at 18:40
  • $\begingroup$ The integral can be evaluated using elliptic functions, see my answer below. $\endgroup$
    – Fred Hucht
    Commented Jan 9, 2023 at 10:10

2 Answers 2

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The expression obtained by @Robert can be simplified with the Imaginary-Argument Transformation from DLMF 19.7.7. Then, the limit $t\to 1^-$ can be performed. The result can be further simplified and written in many ways due to the large number of elliptic functions relations. The simplest expression I found reads \begin{align} &\int_0^1 \mathrm db \,\sqrt{\frac{b(1-b)(b_3-b)}{b_4-b}}\\ &=\frac{1}{4 \sqrt{b_4}}\left[ b_4 (3 b_4-b_3-1) E(k) - b_3 (b_4+b_3-1) K(k)\\ \quad+ \frac{b_3}{\sqrt{b_3-1}} \left[(b_3-1)^2 + 2(b_3+1)b_4 - 3 b_4^2\right] \Pi \left(\tfrac{1}{1-b_3},k\right) \right]\tag{1} \end{align} with parameter and elliptic modulus $$ m=k^2=\frac{b_4-b_3}{\left(1-b_3\right) b_4}.\tag{2} $$ One can rewrite this expression with the Imaginary-Modulus Transformation DLMF 19.7.5 to get $0<\tilde m<1$ and real $0<\tilde k<1$, i.e., ${\tilde k}{}^2=\frac{b_4-b_3}{b_3(b_4-1)}$, however the expressions become more complicated. For $b_3=2, b_4=3$ this last formulation gives the representation obtained in the comment by @Robert.

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We may as well rescale and translate so $b_1 = 0$ and $b_2 = 1$, leaving just two parameters instead of four. Maple produces a result involving a limit of elliptic integrals: $$ \frac{\underset{t \rightarrow 1-}{\mathrm{lim}}\frac{i \left(b_{3}^{2}+\left(2 b_{4}-2\right) b_{3}-3 b_{4}^{2}+2 b_{4}+1\right) \left(\sqrt{t}-t^{\frac{3}{2}}\right) \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \sqrt{b_{3}-1}\, \Pi \left(\frac{i \sqrt{b_{3}-1}\, \sqrt{t}}{\sqrt{b_{3}}\, \sqrt{1-t}}, \frac{b_{3}}{b_{3}-1}, \frac{\sqrt{b_{4}-1}\, \sqrt{b_{3}}}{\sqrt{b_{4}}\, \sqrt{b_{3}-1}}\right)+\left(i \left(b_{3}-3 b_{4}+1\right) \left(\sqrt{t}-t^{\frac{3}{2}}\right) b_{4} \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \sqrt{b_{3}-1}\, E\left(\frac{i \sqrt{b_{3}-1}\, \sqrt{t}}{\sqrt{b_{3}}\, \sqrt{1-t}}, \frac{\sqrt{b_{4}-1}\, \sqrt{b_{3}}}{\sqrt{b_{4}}\, \sqrt{b_{3}-1}}\right)-i \left(b_{3}-b_{4}-1\right) \left(\sqrt{t}-t^{\frac{3}{2}}\right) \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \sqrt{b_{3}-1}\, F\left(\frac{i \sqrt{b_{3}-1}\, \sqrt{t}}{\sqrt{b_{3}}\, \sqrt{1-t}}, \frac{\sqrt{b_{4}-1}\, \sqrt{b_{3}}}{\sqrt{b_{4}}\, \sqrt{b_{3}-1}}\right)+t \left(-2 \sqrt{b_{4}}\, \sqrt{b_{3}-t}\, \sqrt{b_{4}-t}\, \left(t -1\right) \sqrt{-\left(t -1\right) \left(-b_{4}+t \right) \left(-b_{3}+t \right)}+\left(-b_{3}+t \right) \sqrt{1-t}\, \left(\left(-3 t -b_{3}-1\right) b_{4}^{\frac{3}{2}}+3 b_{4}^{\frac{5}{2}}+\left(t b_{3}+t \right) \sqrt{b_{4}}\right)\right)\right) \left(b_{3}-1\right)}{\sqrt{b_{3}-t}\, \sqrt{t}\, \sqrt{b_{4}-t}\, \left(t -1\right)}}{\sqrt{b_{4}}\, \left(4 b_{3}-4\right)} $$

Let's try a special case: $b_3 = 2$, $b_4 = 3$. Numerical evaluation of the above result and numerical integration, both using 20 decimal digits, agree on the value $0.30296476900449078284$.

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  • $\begingroup$ I would like this answer better with a formula for the limit; it feels incomplete to me now. $\endgroup$
    – user44143
    Commented May 13, 2022 at 1:48
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    $\begingroup$ I notice Maple can give explicit formulas for particular integer $b_3,b_4$ after this scaling. $\endgroup$
    – Brendan McKay
    Commented Oct 10, 2022 at 1:04
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    $\begingroup$ @BrendanMcKay Good point. For example, the case $b_3=2$ and $b_4 = 3$ is $$-3 \boldsymbol{EllipticPi}\! \left(-\frac{1}{2}, \frac{1}{2}\right)+3 \boldsymbol{\mathit{EllipticE}}\! \left(\frac{1}{2}\right)$$ However, I don't see a simple pattern for other values of $b_3$ and $b_4$. $\endgroup$
    – Robert Israel
    Commented Oct 12, 2022 at 0:55

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