Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$

Question

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.

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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.

Answer 1

$$\int_0^\infty (\sqrt{x^4+a^4}-\sqrt{x^4+b^4}) dx \implies $$

$$\int_0^\infty (\sqrt{x^4+a^4}-x^2-(\sqrt{x^4+b^4}-x^2)) dx$$

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.

Answer 2

Another way to do it, may be.

Assuming $a>0$ and $t>0$, we have

$$I_a=\int_0^t \sqrt{x^4+a^4}\, dx=a^2 t \, _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{t^4}{a^4}\right)$$ Expanding as Taylor series for infinitely large values of $t$, $$I_a=\frac{t^3}{3}+\frac{\sqrt{\frac{\pi }{2}} a^3 \Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}-\frac{a^4}{2 t}+O\left(\frac{1}{t^3}\right)$$ So, for an infinite value of $t$,
$$\lim_{t\rightarrow\infty}(I_a-I_b)=(a^3-b^3)\frac{\sqrt{\frac{\pi }{2}} \Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{7}{4}\right)}=(a^3-b^3)\frac{\Gamma \left(\frac{1}{4}\right)^2}{6 \sqrt{\pi }}$$ and hence the result.

Answer 3

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.