# To evaluate $\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$

$$f(a,b)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt{x^2+a^2}\sqrt{x^2+b^2}}$$

To use Landen’s transformation

$$f(a,b)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt{x^2+(\frac{a+b}{2})^2}\sqrt{x^2+ab}}$$

$$f(a,b)=f\left(\frac{a+b}{2},\sqrt{ab}\right)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt{x^2+c^2}\sqrt{x^2+c^2}}=\int_0^{+\infty} \frac{\;\mathrm dx}{x^2+c^2}=\frac{\pi}{2c}$$

$$c=\operatorname{AGM}(a,b)$$

$$\operatorname{AGM}$$ is Arithmetic–geometric mean

I would like to find similar transform for:

$$a,b,c>0$$

$$g(a,b,c)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$$

Do you know similar kind of transform method for that integral?

Or which other methods can be used to evaluate that improper integral?

Thanks a lot for answers and helps.

UPDATE:

I would like to share my results. Maybe someone can give some ideas to go forward.

$$g(a,b,c)=\frac{1}{F(a,b,c)}=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$$
$$\frac{1}{F(a.k,b.k,c.k)}=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3k^3}\sqrt[3]{x^3+b^3k^3}\sqrt[3]{x^3+c^3k^3}}$$

$$\frac{1}{F(a.k,b.k,c.k)}=\int_0^{+\infty} \frac{\;\mathrm dx}{k^3\sqrt[3]{\frac{x^3}{k^3}+a^3}\sqrt[3]{\frac{x^3}{k^3}+b^3}\sqrt[3]{\frac{x^3}{k^3}+c^3}}$$
$$ku=x$$
$$\frac{1}{F(a.k,b.k,c.k)}=\int_0^{+\infty} \frac{\;k du}{k^3\sqrt[3]{u^3+a^3}\sqrt[3]{u^3+b^3}\sqrt[3]{u^3+c^3}}$$

$$F(a.k,b.k,c.k)=k^2F(a,b,c)$$

$$k=\frac{1}{a}$$

$$F\left(1,\frac{b}{a},\frac{c}{a}\right)=\frac{1}{a^2}F(a,b,c)$$

$$F\left(a,b,c\right)=a^2F\left(1,\frac{b}{a},\frac{c}{a}\right)$$

$$\frac{b}{a}=t$$

$$\frac{c}{a}=z$$

$$F(a,at,az)=a^2F(1,t,z)=a^2H(t,z)$$

$$\frac{1}{F(a,at,az)}=\frac{1}{a^2F(1,t,z)}=\frac{1}{a^2H(t,z)}=\frac{1}{a^2}\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+1}\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$

Now the problem is for 2 variables.
$$\frac{1}{H(t,z)}=V(t,z)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+1}\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$

We can get a partial differential equation from here

$$\frac{\partial V(t,z)}{\partial t}=-\int_0^{+\infty} \frac{t^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+t^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$

$$\frac{\partial V(t,z)}{\partial z}=-\int_0^{+\infty} \frac{z^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+z^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$

$$\frac{\partial^2 V(t,z)}{\partial t \partial z }=\int_0^{+\infty} \frac{t^2z^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+t^3)(x^3+z^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}=$$

$$=\frac{1 }{z^3-t^3}\int_0^{+\infty} \frac{t^2z^2\;\mathrm dx}{\sqrt[3]{x^3+1}(x^3+t^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}+\frac{1 }{t^3-z^3}\int_0^{+\infty} \frac{t^2z^2\; \mathrm dx}{\sqrt[3]{x^3+1}(x^3+z^3)\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$

$$(t^3 -z^3 )\frac{\partial^2 V(t,z)}{\partial t \partial z }=z^2\frac{\partial V(t,z)}{\partial t }- t^2\frac{\partial V(t,z)}{\partial z }$$

Now I am looking for a transform that

$$t=k(s,y)$$
$$z=l(s,y)$$
To get same equation

$$(s^3 -y^3 )\frac{\partial^2 V(s,y)}{\partial s \partial y }=y^2\frac{\partial V(s,y)}{\partial s }- s^2\frac{\partial V(s,y)}{\partial y }$$

UPDATE 2: I tried to use a variable change .

$$\frac{1}{H(t,z)}=V(t,z)=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+1}\sqrt[3]{x^3+t^3}\sqrt[3]{x^3+z^3}}$$

$$u=\frac{x}{\sqrt[3]{x^3+1}}$$

$$\frac{du}{1-u^3}=\frac{dx}{\sqrt[3]{x^3+1}}$$

$$\frac{1}{H(t,z)}=V(t,z)=\int_0^{1} \frac{\;du}{(1-u^3)\sqrt[3]{\frac{u^3}{1-u^3}+t^3}\sqrt[3]{\frac{u^3}{1-u^3}+z^3}}=$$

$$=\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{t^3+(1-t^3)u^3}\sqrt[3]{z^3+(1-z^3)u^3}}$$

$$=\frac{1}{\sqrt[3]{(1-t^3)}\sqrt[3]{(1-z^3)}}\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{(\cfrac{t}{\sqrt[3]{(1-t^3)}})^3+u^3}\sqrt[3]{(\cfrac{z}{\sqrt[3]{(1-z^3)}})^3+u^3}}$$

If we combine the results above, we can get

$$g(a,b,c)=\frac{1}{F(a,b,c)}=\int_0^{+\infty} \frac{\;\mathrm dx}{\sqrt[3]{x^3+a^3}\sqrt[3]{x^3+b^3}\sqrt[3]{x^3+c^3}}$$

$$u=\frac{x}{\sqrt[3]{x^3+a^3}}$$

$$\frac{du}{1-u^3}=\frac{dx}{\sqrt[3]{x^3+a^3}}$$

$$g(a,b,c)=\frac{1}{F(a,b,c)}=\int_0^{1} \frac{\;du}{(1-u^3)\sqrt[3]{\frac{a^3u^3}{1-u^3}+b^3}\sqrt[3]{\frac{a^3u^3}{1-u^3}+c^3}}=\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{b^3-(b^3-a^3)u^3}\sqrt[3]{c^3-(c^3-a^3)u^3}}$$

$$g(a,b,c)=\frac{1}{F(a,b,c)}=\frac{1}{bc}\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{1-(1-(a/b)^3)u^3}\sqrt[3]{1-(1-(a/c)^3)u^3}}$$

$$\frac{1}{T(x,y)}=\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{1-x^3u^3}\sqrt[3]{1-y^3u^3}}$$

$$F(a,b,c)=bc .T(\sqrt[3]{1-(a/b)^3},\sqrt[3]{1-(a/c)^3})=ac.T(\sqrt[3]{1-(b/a)^3},\sqrt[3]{1-(b/c)^3})=ab.T(\sqrt[3]{1-(c/a)^3},\sqrt[3]{1-(c/b)^3})$$

It seems If we solve $$\int_0^{1} \frac{\;du}{\sqrt[3]{1-u^3}\sqrt[3]{1-x^3u^3}\sqrt[3]{1-y^3u^3}}$$, It will be enough.

Maybe It can be other group of integrals or can be found a relation with elliptic integrals
such as $$\int_0^{1} \frac{\;du}{\sqrt{1-u^2}\sqrt{1-x^2u^2}}$$. I do not know yet.

There is a known AGM-like transformation for your integral
$$g(a,b,c) = \int_0^\infty \frac{dx} { \root 3 \of {x^3+a^3} \root 3 \of {x^3+b^3} \root 3 \of {x^3+c^3} }.$$
It was obtained only a few years ago, and is considerably more complicated
and harder to prove than Landen’s transformation;
it’s inspired by the familiar AGM, but does not reduce to it,
and must be developed on its own terms.

Define
$$I(a,b,c) = \int_0^\infty \frac{dt}{\root 3 \of {t(t+a^3)(t+b^3)(t+c^3)}}.$$
The change of variables $x^3 = 1/t$, $dx = -t^{-4/3} dt/3$ gives
$$g(a,b,c) = \frac13\int_0^\infty \frac{dt} {\root 3 \of {t(1+a^3t)(1+b^3t)(1+c^3t)}} = \frac1{3abc} I(a^{-1},b^{-1},c^{-1}).$$

Now the AGM-like identity is
$$I(a,b,c) = I(a’,b’,c’)$$
where the transformation $\psi: (a,b,c) \mapsto (a’,b’,c’)$ is given by
$$a’ = \frac13(a+b+c), \quad b’, c’ = \left[\frac12 \left( \frac{a^2 (b+c) + b^2(c+a) + c^2(a+b)}{3} \pm \frac{(a-b)(b-c)(c-a)}{\sqrt{-27}} \right)\right]^{1/3}.$$
Note the $\sqrt{-27}$ in the denominator: if $a,b,c$ are distinct
positive reals then $b’,c’$ are complex conjugates (chosen to be
the principal cube roots), but then another application of $\psi$
returns positive $a”,b”,c”$ that are much closer than $a,b,c$.
Repeated application of $\psi$ yields a sequence that converges cubically
to $(M,M,M)$ for some $M=M(a,b,c)$ which is the analogue for these
“Picard periods” of the AGM; and then
$$I(a,b,c) = I(M,M,M) = \frac{\pi \sqrt{4/3}}{M}.$$

For example, if we want to evaluate $g(3.1,4.1,5.9)$ then we write
$$g(3.1,4.1,5.9) = \frac1{3 \cdot 3.1 \cdot 4.1 \cdot 5.9} I\left(\frac1{3.1},\frac1{4.1},\frac1{5.9}\right).$$
Applying $\psi^3$ to $(1/3.1, 1/4.1, 1/5.9)$ yields
$M = 0.2453101037492669116299747362\ldots$, so
$I(1/3.1, 1/4.1, 1/5.9) = \pi \sqrt{4/3} / M = 14.78780805610937446473708974\ldots$,
and $g(3.1,4.1,5.9) = 0.0657332322345471756512603615-$
which agrees with the numerical computation of $g(a,b,c)$
(try telling

intnum(x=0,[+1],((x^3+3.1^3)*(x^3+4.1^3)*(x^3+5.9^3))^(-1/3))


to gp).

The reference where I found formulas equivalent to the above recipe
for $I(a,b,c)$ is

Keiji Matsumoto and Hironori Shiga:
A variant of
Jacobi type formula for Picard curves
,
J. Math. Soc. Japan 62 #1 (2010), 305$-$319.
(doi: 10.2969/jmsj/06210305)

This paper in turn attributes the formulas for $\psi$ and $M$ to
Theorem 2.2 on page 134 of

Kenji Koike and Hironori Shiga:
Isogeny formulas
for the Picard modular form and a three terms arithmetic geometric mean
,
J. Number Theory 124 (2007), 123$-$141.