<s>[INST] Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$ Turns out this integral has a very nice closed form:
$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$
I found it with Mathematica, but I can't figure out how to prove it.
The integral seems quite problematic to me. If the limits were finite, I would do this:
$$\frac{1}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{1}{a^4-b^4}(\sqrt{x^4+a^4}-\sqrt{x^4+b^4})$$
Then, for one of the integrals we will have:
$$\int_A^B \sqrt{x^4+a^4} dx=a^3 \int_{A/a}^{B/a} \sqrt{1+t^4} dt$$
This integral is complicated, but quite well known.
On the other hand $\int_0^\infty \sqrt{1+t^4}dt$ diverges, so I can't consider the two terms separately.
But the integral behaves like I can! If we look at the final expression, it seems like $\int_0^\infty \sqrt{1+t^4}dt=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}}$ even though it can't be correct.
I have to somehow arrive at Beta function, since we have a squared Gamma as an answer.

I'm interested in this integral, since it represents another kind of mean for two numbers $a$ and $b$. If we scale it appropriately:
$$I(a,b)=\frac{8 \sqrt{\pi}}{\Gamma(1/4)^2 } \int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}= \frac{4}{3} \frac{a^2+ab+b^2}{a^3+ab(a+b)+b^3}$$
So, $1/I(a,b)$ is a mean for the two numbers.
 [/INST] $$\int_0^\infty (\sqrt{x^4+a^4}-\sqrt{x^4+b^4}) dx \implies $$
$$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx$$
Because the $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx $ is convergent so the integration can be linearly seperated.
$$\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx - \int_0^\infty(\sqrt{x^4+b^4}-x^2) dx$$
Out of symmetry, the question becomes how to solve integration $\int_0^\infty (\sqrt{x^4+a^4}-x^2)dx$ in terms of $a$.
Let $a^2 \sqrt{t} = \sqrt{x^4+a^4}-x^2 \implies x^2 = a^2 \frac{1-t}{2\sqrt{t}}$ which brought us to an old problem: see the second answer in this post by votes sort.
After simplification the integration becomes $\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} a^3$.
Therefore, the original formula should be $$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx = \frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} (a^3-b^3)$$.
Hence the question get solved.
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