# Show the integral equality $\int_0^\infty\frac{1}{\sqrt{x^3+1}}\mathrm dx = 2 \int_0^\infty\frac{1}{\sqrt{x^6+1}}\mathrm dx$ without evaluating

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 ?

Hint It's not much more effort to look for a $$1$$-parameter family of such identities. Substituting $$x = \frac{1}{u^b}, \qquad dx = -\frac{b\,du}{u^{1 + b}}$$ 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, we can evaluate the general integral explicitly in terms of the gamma function: $$\int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + a}}} = \frac{\Gamma\left(\frac{a}{2 (2 + a)}\right) \Gamma\left(\frac{1}{2 + a}\right)}{\sqrt{\pi} (2 + a)} ,$$ so $$\int_0^\infty \frac{dx}{\sqrt{1 + x^3}} = \frac{\Gamma\left(\frac16\right) \Gamma\left(\frac13\right)}{3 \sqrt{\pi}} = 2.8043642106\ldots .$$

• Wow thanks ! Can you tell how the general integral in the last is evaluated ? Commented Aug 7, 2023 at 9:04
• You're welcome. That question is perhaps worthy of its own post, but substituting $x^{2 + a} = \tan^2 \theta$ transforms the integral to $\frac{2}{2 + a} \int_0^\frac{\pi}{2} \sin^l \theta \cos^m \theta \,d\theta$ for some $l, m$, but the latter integral can immediately be expressed in terms of the beta function, hence the gamma function; see en.wikipedia.org/wiki/…. Commented Aug 7, 2023 at 9:13
• Thanks once more for the hinted substitution. Yes, I'm aware of the latter one which is related Wallis integrals. Commented Aug 7, 2023 at 9:29
• +1). Please see my solution. Commented Aug 1 at 15:02

Maybe you would like this solution. Note $$\begin{eqnarray} I&=:&\int_0^\infty\frac{1}{\sqrt{x^3+1}}\mathrm dx -2 \int_0^\infty\frac{1}{\sqrt{x^6+1}}\overset{x^2\to x}{\mathrm dx}\\ &=&\int_0^\infty\frac{\sqrt{x}-1}{\sqrt x\sqrt{x^3+1}}\mathrm dx\overset{x\to \frac1x}=\int_0^\infty\frac{1-\sqrt{x}}{\sqrt x\sqrt{x^3+1}}\mathrm dx\\ &=&-I\end{eqnarray}$$ and hence $$I=0$$.

Update: Generally, from Travis Willse's answer, one has $$\int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + a}}} = \frac{2}{a} \int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + \frac{4}{a}}}}$$ which can be shown by using the same method without evaluation of these two integrals.

Proof by Gamma function

We consider the general integral $$I(a)=\int_0^{\infty} \frac{1}{\sqrt{x^a+1}} d x$$ \begin{aligned} I(a) & =\int_0^1 y^{-\frac{3}{2}}\cdot\frac{1}{a}\left(\frac{1}{y}-1\right)^{\frac{1}{a}-1} d y \\ & =\frac{2}{a} \int_0^1 y^{-\frac{1}{2}-\frac{1}{a}}(1-y)^{\frac{1}{a}-1} d y \\ & =\frac{2}{a} B\left(\frac{1}{2}-\frac{1}{a}, \frac{1}{a}\right) \\ & =\frac{2}{a} \frac{\Gamma\left(\frac{1}{2}-\frac{1}{a}\right) \Gamma\left(\frac{1}{a}\right)}{\Gamma\left(\frac{1}{2}\right)} \\ & =\frac{2}{a \sqrt{\pi}} \Gamma\left(\frac{1}{2}-\frac{1}{a}\right) \Gamma\left(\frac{1}{a}\right) \end{aligned} Putting $$a=3$$ and $$a=6$$ respectively gives $$I(3)=\frac{2}{3 \sqrt{\pi}} \Gamma\left(\frac{1}{6}\right) \Gamma\left(\frac{1}{3}\right)=2I(6)$$

• Even though I didn't asked by directly evaluating the integral, thanks for it. Commented Aug 8, 2023 at 12:57