Without evaluating the integrals directly, how can I prove that : $$\int_0^\infty\frac{1}{\sqrt{x^3+1}}\mathrm dx = 2 \int_0^\infty\frac{1}{\sqrt{x^6+1}}\mathrm dx$$
For the second integral, I tried the change of variable $x^2 \to x$ to get :
$$\int_0^\infty\frac{1}{\sqrt{x^6+1}}\mathrm dx=\int_0^\infty\frac{1}{2\sqrt{x}\sqrt{x^3+1}}\mathrm dx$$
However, I cannot relate it with the first one as there is an extra $\sqrt{x}$ term in the denominator.
How can I proceed further or are there better methods to solve it ?
Best Answer
Hint It's not much more effort to look for a $1$-parameter family of such identities. Substituting $$x = \frac{1}{u^b}, \qquad dx = -\frac{b\,du}{u^{b + 1}}$$ in $$\int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + a}}}, \qquad a > 0 ,$$ gives $$\int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + a}}} = b \int_0^\infty \frac{du}{\sqrt{u^{2 - a b} + u^{2 + 2 b}}},$$ so choosing $b = \frac{2}{a}$ and renaming $u$ as $x$ on the r.h.s. gives $$\int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + a}}} = \frac{2}{a} \int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + \frac{4}{a}}}} .$$
Taking $a = 1$ (or $a = 4$) gives our case.
Incidentally, $$\int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + a}}} = \frac{\Gamma\left(\frac{a}{4 + 2 a}\right) \Gamma\left(\frac{1}{2 + a}\right)}{\sqrt{\pi} (a + 2)} .$$