evaluate : $\int_0^{\infty}\frac{\arctan\left({\frac{\sqrt{4x^2 +2x\sqrt{4x^2+1}}}{\sqrt{4x^2+1}-x}}\right)}{1+x^2} dx=\frac{\pi^2}{6}$

Let $f(x)$ denote the arctangent term:

$f(x) = \arctan\left({\frac{\sqrt{4x^2 +2x\sqrt{4x^2+1}}}{\sqrt{4x^2+1}-x}}\right)$

Lemma. For $x \geq 0$, we have

$f(x) = \frac{1}{2} \left[ \arccos\left( \frac{1+ix}{1+3ix} \right) + \arccos\left( \frac{1-ix}{1-3ix} \right) \right].$

Assuming this lemma, it is easy to obtain $I = \frac{\pi^2}{6}$. Indeed, pugging the lemma into the integral,

Now, for $z = x-iy$ with $x \in \mathbb{R}$ and $y > 0$,

$\left| \frac{1+iz}{1+3iz} \right| = \sqrt{\frac{x^2+(1+y)^2}{9x^2 + (1+3y)^2}} < 1.$

Hence, the integrand of $\text{(1)}$ defines a meromorphic function on the lower-half plane which has pole at $z = -i$ and grows like $\frac{\log|z|}{|z|^2}$ as $|z| \to \infty$. So by invoking the standard trick involving semicircular contour and the residue theorem, we obtain

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Proof of Lemma. Let $x \geq 0$. By using the formula $\arctan t = \frac{1}{2}\arccos\left( \frac{1-t^2}{1+t^2} \right)$, valid for $t \geq 0$, we have

where $a = a(x) = \frac{1+ix}{1+3ix}$ and $b = b(x) = \frac{1-ix}{1-3ix}$, and $\sqrt{\cdot}$ is the principal square root. The last expression is suggestive of the cosine addition formula. However, directly inverting the cosines is subtle for complex arguments due to branch cuts. A more rigorous approach is to differentiate.

Let $c = ab - \sqrt{1 - a^2}\sqrt{1 - b^2}$, then

$\sqrt{1 - c^2} = a\sqrt{1-b^2} + b\sqrt{1-a^2}.$

From this, we obtain

Integrating both sides and noting that $a(0) = b(0) = c(0) = 1$, we obtain

$f(x) = \frac{\arccos a(x) + \arccos b(x)}{2},$

proving the desired claim.

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Addendum. A similar argument shows that, for $1 \leq a \leq b$,

this solution by : Sangchul Lee