# Integrate $$\int_{0}^{\infty} (\sqrt{x^{4}+a^{4}} + \sqrt{x^{4}+b^{4}})^{-1} \mathrm{d} x$$

How to evaluate

\int\limits_{0}^{\infty} \frac{1}{\sqrt{x^{4}+a^{4}} + \sqrt{x^{4}+b^{4}}} \mathrm{d} x
\label{eq:160824-1}
\tag{1}

was a question on Mathematics Stack Exchange. One of the answers outlined a way to convert the integral into beta functions but did not work out the solution. To do so requires a clever trick to ensure convergence of the integral, which is my reason for posting a full solution here.

We begin by multiplying the integrand by

\frac{\sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}}}{\sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}}}

so that we now have

\frac{1}{a^{4}-b^{4}} \int\limits_{0}^{\infty} \sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}} \, \mathrm{d} x
\label{eq:160824-2}
\tag{2}

The problem here is that

\int\limits_{0}^{\infty} \sqrt{x^{4}+a^{4}} \, \mathrm{d} x

does not converge. Here is where we invoke the clever trick I referred to. Let us consider

\int\limits_{0}^{\infty} (x^{4}+a^{4})^{p} \mathrm{d} x
\label{eq:160824-3}
\tag{3}

Making the successive substitutions $$x^{4} = a^{4}y^{4}$$ and $$y^{4} = z$$ yields
\begin{align}
\int\limits_{0}^{\infty} (x^{4}+a^{4})^{p} \mathrm{d} x & = \frac{a^{4p+1}}{4} \int\limits_{0}^{\infty} z^{-3/4} (z+1)^{p} \mathrm{d} z \\
& = \frac{a^{4p+1}}{4} \mathrm{B}\left(\frac{1}{4},\frac{-1}{4}-p\right)
\label{eq:160824-4}
\tag{4}
\end{align}

Putting it all together and letting $$p = 1/2$$, we have
\begin{align}
\int\limits_{0}^{\infty} \frac{1}{\sqrt{x^{4}+a^{4}} + \sqrt{x^{4}+b^{4}}} \mathrm{d} x & = \frac{1}{a^{4}-b^{4}} \int\limits_{0}^{\infty} \sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}} \mathrm{d} x \\
& = \frac{1}{4} \frac{a^{3}-b^{3}}{a^{4}-b^{4}} \mathrm{B}\left(\frac{1}{4},\frac{-3}{4}\right) \\
& \approx \frac{a^{3}-b^{3}}{a^{4}-b^{4}} 1.23605
\end{align}