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I'm trying to show that the integral $$\int_1^\infty \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx \quad \text{is convergent}.$$

We know that $$\frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }} < \frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}{\text{ for}}\,\,{\text{all}}\,\,\,x \in \left\langle {1, + \infty } \right\rangle $$ Now this is reduced to prove that $$\int_1^\infty {\frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}}dx $$ is convergent, but the latter integral is difficult to calculate or prove to be convergent.

I know that this integral is convergent because when I calculate with Maple is $$\int_1^\infty {\frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}}dx = \frac{{{\pi ^{3/2}}\csc \left( {\frac{\pi }{8}} \right)}}{{4\Gamma \left( {\frac{7}{8}} \right)\Gamma \left( {\frac{5}{8}} \right)}} \approx {\text{2}}{\text{.327185143}}$$

So my question is:

Is there a simpler way of proving that $\int_1^\infty \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx$ converges?

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    $\begingroup$ $\arctan x$ is bounded, so we can forget about it. $\endgroup$ May 16, 2015 at 14:30
  • $\begingroup$ @AndréNicolas Yes, but it implies that $\mathop \smallint \limits_1^\infty \frac{{\frac{\pi }{2}}}{{\sqrt {{x^4} - 1} }}\;{\text{d}}x$ converges, which it is still difficult to prove $\endgroup$ May 16, 2015 at 14:34
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    $\begingroup$ Two potential problems, "at" $\infty$ and at $1$. At $\infty$, $1/\sqrt{x^4-1}$ behaves like $1/x^2$, no problem. For behaviour near $1$, use $\sqrt{x^4-1}\gt $\sqrt{x-1}$. $\endgroup$ May 16, 2015 at 14:40
  • $\begingroup$ $$I ~=~ \int_1^\infty\frac{\tan^{-1}x}{\sqrt{x^4-1}}~dx ~=~ \int_0^1\frac{\cot^{-1}x}{\sqrt{1-x^4}}~dx ~=~ \pi\sqrt\pi~\frac{\Gamma\bigg(\dfrac54\bigg)}{\Gamma\bigg(-\dfrac14\bigg)} ~+~ \sqrt{2\pi}\cdot\Gamma^2\bigg(\dfrac54\bigg)~+$$ $+~_3F_2\bigg(\dfrac34~,~1,~1~;~\dfrac32~,~\dfrac74~;~1\bigg),~$ where the numerical value of the last term is very close to that of the positive global minimum $a_\min$ of $~F(a) ~=~ \displaystyle\int_0^\infty\exp\big(-x^a\big)~dx ~=~ \Gamma\bigg(1+\frac1a\bigg).~$ See OEIS A$030169$ for more information. $\endgroup$
    – Lucian
    May 16, 2015 at 23:19

3 Answers 3

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On $(1,+\infty)$ the function $\arctan x $ is positive and bounded by $\frac{\pi}{2}$, while: $$ \int_{1}^{+\infty}\frac{dx}{\sqrt{x^4-1}}=\int_{0}^{1}\frac{dy}{\sqrt{1-y^4}}\leq\int_{0}^{1}\frac{dy}{\sqrt{1-y^2}}=\frac{\pi}{2}.$$

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    $\begingroup$ D'Aurizo good idea thanks! $\endgroup$ May 16, 2015 at 14:36
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    $\begingroup$ @D'Aurizo Yes, that is!!. The substitution you used the substitution is: $x=\frac{1}{y}$ and the inequality $\frac{1}{{\sqrt {1 - {y^4}} }} < \frac{1}{{\sqrt {1 - {y^2}} }}$ is easy to prove when $0\leq y <1$ $\endgroup$ May 16, 2015 at 15:03
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Note that your integral has infinities in two points: in $1$ because of the denominator, and in $\infty$. Split your integral in two pieces $\int \limits _1 ^2 + \int \limits _2 ^\infty$ and let us study each of them individually.

For the second one, let us apply the ratio test: call your function to integrate $f$, and let $g = \frac 1 {x^2}$. Note that $\lim \limits _{x \to \infty} \frac {f(x)} {g(x)} = \frac \pi 2 \in (0, \infty)$ which shows that $\int _2 ^\infty f \mathbb d x$ and $\int _2 ^\infty g \mathbb d x$ will behave similarly. But the second one is clearly convergent.

For the first one we shall use the ratio test again: consider the function $g(x)=\frac 1 {\sqrt {x-1}}$. Then $\lim \limits _{x \to 1^+} \frac {f(x)} {g(x)}= {\frac \pi 8} \in (0,\infty)$, which again shows that $\int _1 ^2 f \mathbb d x$ and $\int _1 ^2 g \mathbb d x$ will behave similarly. Again, the second integral is convergent.

So, the originl integral is convergent. The proof is quite short, it seems long because I have written down plenty of words and details.

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The integral is improper at $x=1$ and $x=\infty$. We first split the integral at a point, say, $x=2$.

For $\displaystyle\int_2^\infty\frac{dx}{\sqrt{x^4-1}}$, apply the limit comparison test to the integrand $\frac{1}{x^2}$, whose integral over $[2, \infty)$ is convergent by p-test.

For $\displaystyle\int_1^2\frac{dx}{\sqrt{x^4-1}}$, apply the substitution $x=1+y$, and we have the integral $\displaystyle\int_0^1\frac{dy}{\sqrt{4y+6y^2+4y^3+y^4}}$. Apply limit comparison test to the integrand $\frac{1}{\sqrt{y}}$, whose integral over $(1, 2)$ is convergent by p-test.

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