To find more special cases of the Incomplete Beta function $\text B_z(a,b)$ in terms of Elliptic $\text E(x,k)$ using Mathematica notation:

The goal is to find values of:

$$\text B_z(a,b)=\int z^{a-1}(1-z)^{b-1}dz\mathop=^\text{set}\int\sqrt{1-k\sin^2(x)}dx=\text E(x,k);a,b\in\Bbb R,k\ne 0,1$$

The only values for $k\ne 0,1$ found is:

$$\frac14 \text B_{\sin^2(2x)}\left(\frac12,\frac34\right)=\text E(x,2) $$


$$\frac58 \text B_{\sin^2(2x)}\left(\frac32,\frac34\right)=\text E(x,2)-\frac12\sin(2x)\cos^\frac32(2x) $$

which may or may not be simplified with parameter transformation formulas


The following formulas use Elliptic $\text F(x,k)$ and contain $k=-1,\frac12,2,\sqrt[\pm3]{-1}$ with the principal root:

$$\frac14\text B_{\sin^2(2x)}\left(\frac12,\frac14\right)=\text F(x,2)$$

$$\frac14\text B_{\sin^4(x)}\left(\frac14,\frac12\right)=\text F(x,-1)$$

$$\frac{i}{\sqrt 2}\left(\frac12\text B_{\sec^4(x)}\left(\frac14,\frac12\right)-\text L\right)=\text F\left(x,\frac12\right)$$

$$\frac{\sqrt[12]{-1}}{2\cdot3^\frac34}\text B_\frac{3\sqrt3i}{\left(1+\sqrt[3]{-1}-3\csc^2(x)\right)^3}\left(\frac16,\frac12\right)=\text F\left(x,\sqrt[3]{-1}\right)$$

which work for the first positive period of the elliptic integrals where $\text L=$the Lemniscate Constant

but are there any other formulas for EllipticE? Please correct me and give me feedback!

More Motivation:

It looks like using $a,b\in\{\frac n2,\frac n3,\frac n4,\frac n6\},n\in\Bbb N$ gives elliptic integrals. Here are some identities involving Inverse Weierstrass P $\wp^{-1}(z;a,b)$ which unfortunately only are the Elliptic F function from above, not Elliptic E:

$$-\frac{\text B_\frac{a}{4z^2}\left(\frac14,\frac12\right)}{2\sqrt2\sqrt[4]a}=\wp^{-1}(z;a,0)$$

$$-\frac{(-1)^\frac23\text B_{-\frac{4z^3}b}\left(\frac13,\frac12\right)}{3\cdot2^\frac23\sqrt[6]b}-\frac{2\omega_2}{\sqrt 3\sqrt[6]b}=\wp^{-1}(z;0,-b)$$

$$-\frac{\text B_\frac b{4z^3}\left(\frac16,\frac12\right)}{3\cdot2^\frac23\sqrt[6]b}=\wp^{-1}(z;0,b)$$

with $\omega_2=$Omega-$2$ Constant

Reason: The final destination is to invert any other special cases to help find the:

Inverse of elliptic integral of second kind

using Inverse Beta Regularized, but it is simpler to use the Incomplete Beta function because it is more familiar. Your answer will get credit of course.

If there are no other special cases for $$\text B_z(a,b)=\text E(x,k) $$

then please prove it.


1 Answer 1


We can use the reciprocal modulus transformation to characterize $E\left(x,\frac{1}{2}\right)$ as a sum of two incomplete beta functions.

Recall the reciprocal modulus transformation (proof in the appendix):

$$E\left(\phi,\frac{1}{p} \right) = \frac{1}{\sqrt{p}}E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)-\frac{1-\sqrt{p}^2}{\sqrt{p}}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)$$

If we put $p=2$ and using

$$\frac14 \text B_{\sin^2(2x)}\left(\frac12,\frac34\right)=\text E(x,2) $$


$$\frac14\text B_{\sin^2(2x)}\left(\frac12,\frac14\right)=\text F(x,2)$$

\begin{align*}E\left(\phi,\frac{1}{2} \right) = &\frac{1}{\sqrt{2}}E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right),2\right)+\frac{1}{\sqrt{2}}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right),2\right)\\ =& \frac{1}{4\sqrt{2}}B_{\sin^2\left(2\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right)\right)}\left(\frac12,\frac34\right)+ \frac{1}{4\sqrt{2}} B_{\sin^2\left(2\arcsin\left(\frac{\sin \phi}{\sqrt{2}}\right)\right)}\left(\frac12,\frac14\right)\\ =& \frac{1}{4\sqrt{2}}B_{1-\cos^4(\phi)}\left(\frac12,\frac34\right)+ \frac{1}{4\sqrt{2}} B_{1-\cos^4(\phi)}\left(\frac12,\frac14\right) \end{align*}

So the incomplete elliptic integral of the second kind with $\displaystyle k=\frac{1}{2}$ is also related to the incomplete beta function through the following sum:

$$\boxed{E\left(\phi, \frac{1}{2} \right) = \frac{1}{4\sqrt{2}}B_{1-\cos^4(\phi)}\left(\frac12,\frac34\right)+ \frac{1}{4\sqrt{2}} B_{1-\cos^4(\phi)}\left(\frac12,\frac14\right)} $$

Edit (Appendix): Proof of the reciprocal modulus transformation.

We will use the notation from Wolfram:

$$F(\alpha,m ) = \int_{0}^{\alpha} \frac{1}{\sqrt{1-m\sin^2\alpha}}$$ Typically, the reciprocal modulus transformation is used to transform modulus exceeding the unity to the standard modulus form $0<p\leq 1$. However, nothing prevent us to use it the other way around:

\begin{align*} E\left(\phi,\frac{1}{p} \right) = &\int_{0}^{\phi} \frac{1}{\sqrt{1-\frac{1}{p}\sin^2\theta }} d\theta \\ =& \int_{0}^{\arcsin\left(\frac{\sin\phi}{\sqrt{p}}\right)} \frac{\sqrt{p}\cos^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi \quad \left( \sin\varphi \mapsto \frac{\sin\theta}{\sqrt{p}}\right)\\ =& \sqrt{p}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right) - \sqrt{p}\int_{0}^{\arcsin\left(\frac{\sin\phi}{\sqrt{p}}\right)} \frac{\sin^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi \end{align*} The last integral is known as $D(\alpha,p)$ and has the solution.

$$ D(\alpha,p) = \int_{0}^{\alpha} \frac{\sin^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi = \frac{F(\alpha,p)-E(\alpha,p)}{p}$$

(To prove this just apply the standard definition to the right hand side.)


\begin{align*} E\left(\phi,\frac{1}{p} \right) =& \sqrt{p}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right) - \sqrt{p}\int_{0}^{\arcsin\left(\frac{\sin\phi}{\sqrt{p}}\right)} \frac{\sin^2\varphi }{\sqrt{1-p\sin^2\varphi }} d\varphi\\ =& \sqrt{p}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right) - \frac{F\left(\arcsin\left(\frac{\phi}{\sqrt{p}}\right),p\right)-E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)}{\sqrt{p}}\\ =& \frac{1}{\sqrt{p}} E\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right)-\frac{1-p}{\sqrt{p}}F\left(\arcsin\left(\frac{\sin \phi}{\sqrt{p}}\right),p\right) \end{align*}

  • 1
    $\begingroup$ @TymaGaidash it seems that k=2,1/2 (using Wolfram notation) are the only values that works but I'm continue looking for another transformation $\endgroup$
    – Bertrand87
    May 11, 2022 at 5:39
  • $\begingroup$ Where did you find this specific reciprocal modulus transformation? $\endgroup$ Mar 17, 2023 at 17:25
  • 1
    $\begingroup$ Hi @TymaGaidash. It is a relatively well known result for elliptic integrals. One reference is the proposition 114.01 in Friedman and Byrd (1971) Handbook of Elliptic Integrals for Engineers and Scientists. I added the proof in the appendix. $\endgroup$
    – Bertrand87
    Mar 17, 2023 at 20:23

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