# Expressing the integral $\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{\left(1-x^3\right)\left(1-a^6x^3\right)}}$ in terms of elliptic integrals

Define the function $$\mathcal{I}:\left(0,1\right)\rightarrow\mathbb{R}$$ via the definite integral

$$\mathcal{I}{\left(p\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{4p}{\sqrt{\left(1-x^{3}\right)\left(1-p^{6}x^{3}\right)}}.\tag{1}$$

Question: Given $$p\in\left(0,1\right)$$, is there a (relatively) simple way to express $$\mathcal{I}{\left(p\right)}$$ in terms of elliptic integrals in Legendre normal form? Can $$\mathcal{I}{\left(p\right)}$$ possibly be written as a simple scalar multiple of a complete elliptic integral of the first kind?

Recall that for amplitude $$\theta\in\left(0,\frac{\pi}{2}\right)$$ and elliptic modulus $$\kappa\in\left(0,1\right)$$, the incomplete elliptic integral of the first kind $$F{\left(\theta,\kappa\right)}$$ is defined as

$$F{\left(\theta,\kappa\right)}:=\int_{0}^{\theta}\mathrm{d}\varphi\,\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}}.\tag{2}$$

Motivation:

I have noticed that of the questions on this site related to proving special closed form values of the Gauss hypergeometric function, $${_2F_1}{\left(a,b;c;z\right)}$$, an unusually large number of the more difficult ones can be shown equivalent to proving the particular hypergeometric function with parameters $$a=\frac12$$, $$b=\frac13$$, and $$c=\frac56$$, i.e., $${_2F_1}{\left(\frac12,\frac13;\frac56;z\right)}$$, takes some special (usually algebraic) value for some particular (usually rational) value of $$z$$, where $$0. So in the interest of generalization, I wanted to see how close one could get to obtaining a closed form for $$f(z):={_2F_1}{\left(\frac12,\frac13;\frac56;z\right)}$$ for arbitrary values of $$z\in(0,1)$$.

Using Euler's integral representation for the Gauss hypergeometric function with the assumption that $$0, we have:

\begin{align} f(z) &={_2F_1}{\left(\frac12,\frac13;\frac56;z\right)}\\ &=\frac{1}{\operatorname{B}{\left(\frac13,\frac12\right)}}\int_{0}^{1}t^{-\frac23}(1-t)^{-\frac12}(1-zt)^{-\frac12}\mathrm{d}t\\ &=\frac{3}{\operatorname{B}{\left(\frac13,\frac12\right)}}\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{(1-x^3)(1-zx^3)}},\tag{3}\\ \end{align}

where the beta function factor can alternatively be written in terms of the gamma function as,

$$\operatorname{B}{\left(\frac13,\frac12\right)}=\frac{\sqrt{3}}{2^{4/3}\pi}\left[\Gamma{\left(\frac13\right)}\right]^3.$$

The integral factor in the last line of $$(3)$$ is already superficially quite similar to the following integral representation of the complete elliptic integral of the first kind:

$$K(k)=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-k^2t^2)}}.$$

Moreover, the last line of $$(3)$$ implies that in order for $$f(z)$$ to take an algebraic value, the integral factor will be the product of an algebraic number $$A$$ with an elliptic integral singular value:

$$\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{(1-x^3)(1-zx^3)}}=A\,K{\left(k_3\right)}=A\frac{3^{1/4}}{2^{7/3}\pi}\left[\Gamma{\left(\frac13\right)}\right]^3.$$

This made me suspect that the integral $$\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{(1-x^3)(1-zx^3)}}=\mathcal{I}{(z^{1/6})}$$ might be capable of being written as a linear combination of complete elliptic integrals in general.

• Do you mind if I ask how you came up with the substitution $x=\frac{1}{a}(z-\sqrt{z^2-1})$ ? Thanks. – M.N.C.E. Oct 25 '14 at 9:18
• @M.N.C.E. See my question about integrals of reciprocal square-roots of 6th degree polynomials here. – David H Oct 25 '14 at 9:27
• A route I'll try to go after is that $_2 F_1(a,b;c;z)$ satisfies an ODE. So we can transfer the problem from integration to a boundary value problem. That's hardly a panacea, but it offers some possibilities. – Semiclassical Oct 27 '14 at 23:41
• For reference, and to to clarify my last comment: $y={_2F_1}(\frac12,\frac13;\frac56;z)$ satisfies the hypergeometric ODE $$6z(1-z)y''(z)+(5-11z)y'(z)-y(z)=0.$$ If we work in terms of $\alpha$ via $z=\alpha^6$, then the ODE becomes $$216(\alpha^6-1) \alpha \,y''(\alpha)+6(41\alpha^6-35)\,y'(\alpha)+\alpha^5\, y(\alpha)=0.$$ (This has the advantage that both solutions are analytic near 0, whereas the first had a second solution with leading term $z^{1/6}$.) – Semiclassical Oct 28 '14 at 3:18
• This is very closely related to this question. – Kirill Oct 28 '14 at 14:59

General Procedure:

I will give an outline of a possible procedure (that I learned from a calculus book some time ago) to solve integrals of the form $$\mathcal{J}=\int\frac{{\rm d}x}{\sqrt{(x^2+ax+b)(x^2+cx+d)}}$$ First, let $x=t+\lambda$. \begin{align} \mathcal{J}=\int\frac{{\rm d}t}{\sqrt{(t^2+(2\lambda+a)t+\lambda^2+\lambda a+b)(t^2+(2\lambda+c)t+\lambda^2+\lambda c+d)}} \end{align} Next, let $\lambda=\dfrac{d-b}{a-c}$, then make the substitution $\displaystyle t=\mu v$ where $\displaystyle \mu=\sqrt{\lambda^2+\frac{ad-bc}{a-c}}$. $$\mathcal{J}=\frac{1}{\mu}\int\frac{{\rm d}v}{\sqrt{\left(v^2+Av+1\right)\left(v^2+Cv+1\right)}}$$ where $\displaystyle A=\frac{2\lambda+a}{\mu}$ and $\displaystyle C=\frac{2\lambda+c}{\mu}$. Follow this up with the substitution $\displaystyle v=\frac{1+w}{1-w}$. $$\mathcal{J}=\frac{2}{\mu}\int\frac{{\rm d}w}{\sqrt{((2-A)w^2+(2+A))((2-C)w^2+(2+C))}}$$ By applying the appropriate trigonometric substitution, it can be seen that $\mathcal{J}$ can be reduced to incomplete elliptic integrals.

Parameters

I will define the following parameters to save space and typing. \begin{align} \alpha_1=&\ \frac{1}{\sqrt{\left(a+1\right)^2\left(3a^2+2\sqrt{3}\sqrt{a^4+a^2+1}+3\right)}}\\ \alpha_2=&\ \frac{1}{\sqrt{\left(a-1\right)^2\left(3a^2+2\sqrt{3}\sqrt{a^4+a^2+1}+3\right)}}\\ \beta_1=&\ \frac{\sqrt{3}a^2-\sqrt{a^4+a^2+1}+\sqrt{3}a+\sqrt{3}}{\sqrt{3}a^2+\sqrt{a^4+a^2+1}+\sqrt{3}a+\sqrt{3}}\\ \beta_2=&\ \frac{\sqrt{3}a^2-\sqrt{a^4+a^2+1}-\sqrt{3}a+\sqrt{3}}{\sqrt{3}a^2+\sqrt{a^4+a^2+1}-\sqrt{3}a+\sqrt{3}}\\ \xi_1=&\ \frac{\sqrt{a^4+a^2+1}-\sqrt{3}a}{\sqrt{a^4+a^2+1}+\sqrt{3}a}\\ \xi_2=&\ \frac{\sqrt{a^4+a^2+1}+\sqrt{3}a}{\sqrt{a^4+a^2+1}-\sqrt{3}a} \end{align}

Evaluation of $\ \mathcal{I}(a)$

It was derived by the OP that \begin{align} \mathcal{I}(a) =&\underbrace{\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a-1)^2}{2a}x-\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}}_{\mathcal{I}_1}\\ &+\underbrace{\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a+1)^2}{2a}x+\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}}_{\mathcal{I}_2} \end{align} Applying the aforementioned procedure and using Wolfram Alpha to do all the algebra, we get \begin{align} \mathcal{I}_1 =&\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a-1)^2}{2a}x-\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}\\ =&\frac{1}{\sqrt{3}\sqrt{a^4+a^2+1}}\int^\infty_\frac{\sqrt{3}(a^2+a+1)}{\sqrt{a^4+a^2+1}}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{2(2a^2+a+2)}{\sqrt{3}\sqrt{a^4+a^2+1}}x+1\right)\left(x^2-\frac{2\sqrt{3}\ a}{\sqrt{a^4+a^2+1}}+1\right)}}\\ =&\alpha_1\int^1_{\beta_1}\frac{{\rm d}x}{\sqrt{\left(x^2-\beta_1^2\right)\left(x^2+\xi_1\right)}} =\alpha_1\int^{\arccos{\beta_1}}_{0}\frac{\sec{\varphi}\tan{\varphi}}{\sqrt{(\sec^2{\varphi}-1)(\beta_1^2\sec^2{\varphi}+\xi_1)}}{\rm d}\varphi\\ =&\alpha_1\int^{\arccos{\beta_1}}_{0}\frac{{\rm d}\varphi}{\sqrt{\beta_1^2+\xi_1\cos^2{\varphi}}} =\alpha_1\int^{\arccos{\beta_1}}_{0}\frac{{\rm d}\varphi}{\sqrt{(\beta_1^2+\xi_1)-\xi_1\sin^2{\varphi}}}\\ =&\frac{\alpha_1}{\sqrt{\beta_1^2+\xi_1}}\int^{\arccos{\beta_1}}_{0}\frac{{\rm d}\varphi}{\sqrt{1-\frac{\xi_1}{\beta_1^2+\xi_1}\sin^2{\varphi}}} =\color{red}{\frac{\alpha_1}{\sqrt{\beta_1^2+\xi_1}}\mathbf{F}\left(\arccos{\beta_1},\sqrt{\frac{\xi_1}{\beta_1^2+\xi_1}}\right)} \end{align} for $\mathcal{I}_1$, and we obtain (I omit the steps of trigonometric manipulations since they are the same) \begin{align} \mathcal{I}_2 =&\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a+1)^2}{2a}x+\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}\\ =&\frac{1}{\sqrt{3}\sqrt{a^4+a^2+1}}\int^\infty_{\frac{\sqrt{3}(a^2-a+1)}{\sqrt{a^4+a^2+1}}}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{2(2a^2-a+2)}{\sqrt{3}\sqrt{a^4+a^2+1}}x+1\right)\left(x^2+\frac{2\sqrt{3}\ a}{\sqrt{a^4+a^2+1}}x+1\right)}}\\ =&\alpha_2\int^1_{\beta_2}\frac{{\rm d}x}{\sqrt{\left(x^2-\beta_2^2\right)\left(x^2+\xi_2\right)}}=\color{blue}{\frac{\alpha_2}{\sqrt{\beta_2^2+\xi_2}}\mathbf{F}\left(\arccos{\beta_2},\sqrt{\frac{\xi_2}{\beta_2^2+\xi_2}}\right)} \end{align} Therefore $$\mathcal{I}(a)=\frac{\alpha_1}{\sqrt{\beta_1^2+\xi_1}}\mathbf{F}\left(\arccos{\beta_1},\sqrt{\frac{\xi_1}{\beta_1^2+\xi_1}}\right)+\frac{\alpha_2}{\sqrt{\beta_2^2+\xi_2}}\mathbf{F}\left(\arccos{\beta_2},\sqrt{\frac{\xi_2}{\beta_2^2+\xi_2}}\right)$$ I have substituted in some random values of $a$ into $\mathcal{I}_1$ and it seems to match numerically. However, I have not verified $\mathcal{I}_2$ yet so there might be a mistake.

• I find your answer useful because it explains how to prove Prop.3.145(1) of Gradshteyn, which I didn't previously know how to do, so +1). However, I don't think it's particularly relevant to the problem at hand. Do you have any thoughts about the conjecture I made at the bottom of my post? – David H Oct 25 '14 at 13:10
• @DavidH As of now, I do not have an approach in mind to verify your conjecture, sorry for that. However, in the event that I do find a viable method to prove or disprove your conjecture, I would post it (that is assuming that someone else doesn't post their approach before I do). – M.N.C.E. Oct 25 '14 at 13:18
• FYI, the conjecture appears to be false. Apparently I should have tested more values. =p – David H Oct 25 '14 at 18:03
• @DavidH It is quite a pity that your conjecture is false, it would have saved me alot of work if it were true :P. In any case, I have edited my answer slightly and may I ask if it suits the question requirements better this time round? I will probably check my answer for errors only tomorrow. – M.N.C.E. Oct 28 '14 at 12:57