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Prove that:

$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$ $(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$

What I do for $(1)$ is (something trival): $$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx$$ $$\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$ so it remains to prove that $$\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$ Thanks in advance!

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  • $\begingroup$ Well, the integrals are non elementary to begin with, so there has to be something to it, a trick I am currently overlooking $\endgroup$
    – imranfat
    May 10 '13 at 16:58
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Hint of 1. Substitute $x=\frac{3}{2}y$

But I have no idea about 2...

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  • $\begingroup$ ,Grateful!It seems so obvious.. $\endgroup$
    – lsr314
    May 10 '13 at 17:06
  • $\begingroup$ Maybe i should ask them as two diffirent problems. $\endgroup$
    – lsr314
    May 10 '13 at 17:25
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    $\begingroup$ @Hecke I add a proof of (2). $\endgroup$
    – Hanul Jeon
    May 10 '13 at 17:26
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    $\begingroup$ sorry,but something is wrong,$$\int_0^1 \frac{1}{\sqrt{8(x^2+1)}} dx+\int_1^{\infty } \frac{1}{\sqrt{8 (x^3+x^2)}} \ dx =\frac{\sinh ^{-1}(1)}{2 \sqrt{2}}+\frac{\sinh ^{-1}(1)}{\sqrt{2}}≈0.934$$ $\endgroup$
    – lsr314
    May 10 '13 at 17:41
  • $\begingroup$ @Hecke Oh... it is my mistake :( $\endgroup$
    – Hanul Jeon
    May 10 '13 at 17:53
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I modified the lower bound for the integral (2) though it is still not exactly 1. Still I'm giving my bounds.

\begin{equation*} \int_{0}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx= \int_{0}^1 \frac{1}{\sqrt{(8x^3+x+7)}}dx+\int_{1}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx=I_1+I_2 \end{equation*}

Now, for $x \in [0,1],\ x^2\geq x^3$, Hence it follows, \begin{equation*} I_1=\int_{0}^1 \frac{1}{\sqrt{(8x^3+x+7)}}dx > \int_{0}^1 \frac{1}{\sqrt{(8x^2+x+7)}}dx \end{equation*} which can be simplified, by standard methods, to $$\int_{0}^1 \frac{1}{\sqrt{(8x^2+x+7)}}dx=\frac{1}{2\sqrt{2}}\ln\left(\frac{17+16\sqrt{2}}{1+4\sqrt{14}}\right)\approx 0.321387 $$

Using the transformation $y=1/x$ $$I_2=\int_{0}^1 \frac{1}{\sqrt{x(7x^3+x^2+8)}}dx$$ and using the fact that for all $x \in [0,1],\ x^2\geq x^3$, we get, $$I_2>\int_{0}^1 \frac{1}{\sqrt{8x(x^2+1)}}dx$$ Using the transformation $x=\tan(\theta/2)$, we get $$I_2> 1/4\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\sin{\theta}}}=\frac{1}{8}\beta\left(\frac{1}{4},\frac{1}{2}\right)=\frac{\left(\Gamma\left(\frac{1}{4}\right)\right)^2}{8\sqrt{2\pi}}\approx 0.655514$$ Hence, my bound is $$\int_{0}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx>0.321387+0.655514=0.976901$$

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  • $\begingroup$ Estimation of $I_2$ is nice, but still long way to go to 1.00014... $\endgroup$ Jun 29 '13 at 0:10
  • $\begingroup$ @i707107 true, the actual value is so close to 1 that it hardly seems any closed form expression can achieve the bound 1 $\endgroup$ Jun 29 '13 at 5:45
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The second integral in closed form (by computer algebra) equals $$ \frac1{\sqrt{2(a_2-a_1)}} F\left(\text{asin}(\sqrt{1-a_2/a_1}), \sqrt{\frac{a_1-a_3}{a_1-a_2}}\right), $$ where $a_1$ is the real root and $a_{2,3}$ are the complex roots of $8x^3+x+7=0$, and $F(\phi,k)$ is the incomplete elliptic integral of the first kind. This evaluates to $$ 1.0001425023196181464480\ldots$$

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  • $\begingroup$ So, is everything inside square roots complex number? $\endgroup$ Jun 29 '13 at 6:40
  • $\begingroup$ @i707107 Yes. (Complex numbers are numbers too!) $\endgroup$
    – Kirill
    Jun 29 '13 at 16:01
  • $\begingroup$ Then, there are two square roots of complex number, which one of them? $\endgroup$ Jun 30 '13 at 3:42
  • $\begingroup$ There are two squares for every real number too. Pick a branch of the (multi-valued) square root by putting the branch cut on $(-\infty,0]$ (giving the principal branch), and then you just need to be consistent in using the same branch. So $\sqrt{e^{i\theta}} = e^{i\theta/2}$, $-\pi<\theta\leq\pi$. Note by the way that if you replace $\sqrt{x}\to-\sqrt{x}$ in the expression, the value remains the same. mathworld.wolfram.com/SquareRoot.html $\endgroup$
    – Kirill
    Jun 30 '13 at 6:25
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We could use Carleman's Inequality which states that, $$ \int_0^\infty f(x)dx \ge 1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx $$

Then, by substituting our function as f(x), we can prove the statement by proving that $$ 1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx \gt 1 $$ This is done by noting that for all x >0 $$ 9t^3 + 9 \gt 8t^3 + t + 7 $$

Simplifying the right hand integral might prove the case.

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