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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 ?

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2 Answers 2

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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, $$\int_0^\infty \frac{dx}{\sqrt{1 + x^{2 + a}}} = \frac{\Gamma\left(\frac{a}{4 + 2 a}\right) \Gamma\left(\frac{1}{2 + a}\right)}{\sqrt{\pi} (a + 2)} .$$

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  • $\begingroup$ Wow thanks ! Can you tell how the general integral in the last is evaluated ? $\endgroup$ Aug 7, 2023 at 9:04
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    $\begingroup$ 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/…. $\endgroup$ Aug 7, 2023 at 9:13
  • $\begingroup$ Thanks once more for the hinted substitution. Yes, I'm aware of the latter one which is related Wallis integrals. $\endgroup$ Aug 7, 2023 at 9:29
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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 $$ I(3)=\frac{2}{3 \sqrt{\pi}} \Gamma\left(\frac{1}{6}\right) \Gamma\left(\frac{1}{3}\right)=2I(6) $$

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

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