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.

Another way to split the integrals is to replace the square root by a power $p$:
$$\sqrt{x^4 + a^4}\longrightarrow \left(x^4 + a^4\right)^{p}$$
The integral of the separate terms will then converge for $p<-\frac{1}{4}$ and can be expressed in terms of the beta-function. You can substitute $p = \frac{1}{2}$ in the final answer, despite the individual integrals not converging by invoking analytic continuation.
To get to the beta-functions, you can substitute $x = a t$ to get $a$ out of the way, then $u = t^4 + 1$ and finally $u = \frac{1}{v}$ will yield the explicit beta-function form.
