Prove $\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} ( \text{agm} (\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} ))^{-1}$

Question

The following definite integral turns out to be expressible as the Arithmetic-Geometric Mean: $$I_4(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} \left( \text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)\right)^{-1}$$

$$I_4(1,1)=\frac{\pi}{2 \sqrt2}$$

I would like to remind you that:

$$I_2(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2} \left( \text{agm} \left(a,b \right)\right)^{-1}$$

$$I_2(1,1)=\frac{\pi}{2}$$

Which is why I have two questions:

How do we prove the identity for $I_4(a,b)$?

Is it possible to also express other integrals of this type using agm? Such as $I_8(a,b)$?

---

Because of the relation of the agm to elliptic integrals, we can also write:

$$I_4(a,b)=\frac{1}{a b \sqrt{a^2+b^2}} K \left( \frac{a-b}{\sqrt{2(a^2+b^2)}} \right)$$

Here the parameter convention is $$K(k)=\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}}$$

This seems to be the best way to prove the identity, but I don't know which substitution to use.

Another way to express this integral would be through the hypergeometric function:

$$I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right)$$

And for every integral of this type we have:

$$I_m(a,b)=\frac{I_m(1,1)}{a^{m-1}} {_2F_1} \left(\frac{1}{2},\frac{m-1}{m};1;1-\frac{b^m}{a^m} \right)$$

The outline for the proof can be found in this question for $m=3$ and is easily adapted to the general case.

Here we assume $a \geq b$.

As for the arithmetic geometric mean, I tried to get it into a simpler form, but the only transformation I was able to achieve is this:

$$\text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)=\text{agm} \left(\frac{a+b}{2}+i \frac{a-b}{2},\frac{a+b}{2}-i \frac{a-b}{2} \right)$$

Since we have complex conjugates, it's quite obvious, that they will give real numbers at first iteration, an it would give the left hand side.

Answer 1

Since $K(k)=\frac{\pi}{2} ~{}_2F_1(1/2,1/2;1;k^2)$ and it was shown in the question that $$ I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right), $$ the first question is equivalent to proving the equality

$$ \frac{\sqrt{2} }{b \sqrt{b^2+1}}{_2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{(1-b)^2}{2 \left(b^2+1\right)}\right)}={_2F_1\left(\frac{1}{2},\frac{3}{4};1;1-b^4\right)}\tag{1}. $$

$\it{Proof}$. The proof can be accomplished by combining 3 transformations for hypergeometric function.

By Pfaff's transformation $$ {_2F_1\left(\frac{1}{2},\frac{3}{4};1;1-b^4\right)}=\frac{1}{b^2}{~_2F_1\left(\frac{1}{2},\frac{1}{4};1;\frac{b^4-1}{b^4}\right)}. $$ By quadratic transformation 2.11(5) from Erdelyi, Higher transcendental functions (put $z=\frac{b^2-1}{b^2+1}$) $$ {~_2F_1\left(\frac{1}{2},\frac{1}{4};1;\frac{b^4-1}{b^4}\right)}=\sqrt{\frac{2b^2}{1+b^2}}{~_2F_1\left(\frac{1}{4},\frac{1}{4};1;\left(\frac{b^2-1}{b^2+1}\right)^2\right)}. $$ By transformation 2.11(2) (put $z=\frac{(1-b)^2}{2 \left(b^2+1\right)}$) $$ {~_2F_1\left(\frac{1}{4},\frac{1}{4};1;\left(\frac{b^2-1}{b^2+1}\right)^2\right)}={~_2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{(1-b)^2}{2 \left(b^2+1\right)}\right)}. $$

Answer 2

Substitition is sufficient.
Let $$\displaystyle z=x-\frac1x,w=x+\frac1x$$ then $$\displaystyle \frac{\mathrm dx}{\sqrt{x^8+p x^4+1}}=\frac12\left( \frac{\mathrm dz}{\sqrt{z^4+4z^2+2+p}}+\frac{\mathrm dw}{\sqrt{w^4-4w^2+2+p}}\right) $$ So $$f(p)=\displaystyle \int_0^\infty \frac{\mathrm dx}{\sqrt{x^8+p x^4+1}} =\frac12\int_{-\infty}^\infty \frac{\mathrm dx}{\sqrt{x^4+4x^2+2+p}} =\int_0^\infty \frac{\mathrm dx}{\sqrt{x^4+4x^2+2+p}} $$ let $$x^4+4x^2+2+p= (x^2+u^2)(x^2+v^2) $$ We know that $$\displaystyle \operatorname{agm}(u,v)= \frac{\pi}2\left( \int_0^\infty \frac{\mathrm dx}{\sqrt{(x^2+u^2)(x^2+v^2)}}\right)^{-1} =\frac{\pi}2 f(p)^{-1} $$ With the identity of agm, $$ \operatorname{agm}(u,v)=\operatorname{agm}\left(\frac{u+v}2,\sqrt{uv}\right) =\operatorname{agm}\left(\sqrt{\frac{\sqrt{p+2}}2+1},\sqrt[4]{p+2}\right) $$ So $$ \int_0^\infty \frac{\mathrm dx}{\sqrt{(x^4+a^4)(x^4+b^4)}} =\frac1{ab\sqrt{ab}}f\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right) =\frac\pi{2ab\sqrt{ab}}\left(\operatorname{agm}\left( \frac{a+b}{\sqrt{2ab}},\sqrt{\frac{a^2+b^2}{ab}} \right)\right)^{-1} =\frac\pi{2\sqrt2 ab}\left(\operatorname{agm}\left( \frac{a+b}2,\sqrt{\frac{a^2+b^2}2} \right)\right)^{-1} $$