Evaluate $\int_0^\infty \frac{(1-x^2) \arctan x^2}{1+4x^2+x^4} \,{\rm d} x$

Question

Evaluate $$\int_0^\infty \frac{(1-x^2) \arctan x^2}{1+4x^2+x^4} \,{\rm d} x$$ using any analytic method taught to you.

Edit: This is not a homework problem, I simply want to learn how to evaluate such a kind of integral. I tried simplifying the denominator and used a trig substitution, but I failed. I tried subbing x = (tan(a))^1/2 but ended up with a messy trig integral that I could not evaluate.

Wolfram Alpha's answer on my try with the trig sub

Answer 1

According to wolfram alpha,

$$ \int \frac{1-x^2}{1+4x^2+x^4} dx = \frac{-(\sqrt{3}-3)\sqrt{2+\sqrt{3}} \tan^{-1}(\frac{x}{\sqrt{2-\sqrt{3}}}) - \sqrt{2-\sqrt{3}}(3+\sqrt{3})\tan^{-1}(\frac{x}{\sqrt{2+\sqrt{3}}})}{2\sqrt{3}},$$

which can be obtained by doing a partial fraction decomposition.

It should be clear that,

$$ \frac{d}{dx} \tan^{-1}(x^2) = \frac{2x}{1+x^4}$$

This means that the integral can be done by performing an integration by parts where $U=\tan^{-1}(x^2)$ and $dV=(1-x^2)/(1+4x^2+x^4)dx$.

The result is quite complicated and I will not type it all here; but it will have integrals of the form,

$$ \int_0^\infty \tan^{-1}(\frac{x}{\sqrt{2-\sqrt{3}}})\frac{x}{x^4+1} dx =\pi^2/12,$$

$$ \int_0^\infty \tan^{-1}(\frac{x}{\sqrt{2+\sqrt{3}}})\frac{x}{x^4+1} dx =\pi^2/24,$$

these final values I got from wolfram alpha/mathematica.

You should be able to compute the integral with the information I have provided here.

Answer 2

$$\begin{align*} I &= -\int_0^\infty \frac{x^2-1}{x^4+4x^2+1} \, \arctan\left(x^2\right) \, dx \\ &= - \int_0^\infty \frac{x^2-1}{x^4+4x^2+1} \left[\int_0^{x^2} \frac{dy}{1+y^2}\right] \, dx \tag1 \\ &= - \int_0^\infty \left[\int_{\sqrt y}^\infty \frac{x^2-1}{x^4+4x^2+1} \, dx\right] \, \frac{dy}{1+y^2} \\ &= \frac1{\sqrt2} \int_0^\infty \frac{\arctan\left(\sqrt{2-\sqrt3} \sqrt y\right)}{1+y^2} \, dy - \frac1{\sqrt2} \int_0^\infty \frac{\arctan\left(\sqrt{2+\sqrt3} \sqrt y\right)}{1+y^2} \, dy \tag2 \\ &= \frac{J\left(\sqrt{2-\sqrt3}\right) - J\left(\sqrt{2+\sqrt3}\right)}{\sqrt2} \\ &= \frac\pi{\sqrt2} \left[\arctan\left(1+\sqrt2\sqrt{2-\sqrt3}\right) - \arctan\left(1+\sqrt2\sqrt{2+\sqrt3}\right)\right] \tag{$\star$} \\ &= \frac\pi{\sqrt2} \left[\arctan\left(\sqrt3\right) - \arctan\left(2+\sqrt3\right)\right] \tag7 \\ &= \boxed{-\frac{\pi^2}{12\sqrt2}} \end{align*}$$

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Evaluating the parameterized integral:

$$\begin{align*} J(a) &= \int_0^\infty \frac{\arctan(a\sqrt y)}{y^2+1} \, dy \\[2ex] J'(a) &= \int_0^\infty \frac{\sqrt y}{\left(a^2y+1\right)\left(y^2+1\right)} \, dy \tag3 \\ &= \int_0^\infty \frac{2y^2}{\left(a^2y^2+1\right) \left(y^4+1\right)} \, dy \tag4 \\ &= \frac\pi{\sqrt2} \cdot \frac{1}{a^2+\sqrt2\,a+1} \tag5 \\[2ex] J(a) &= \frac\pi{\sqrt2} \int_0^a \frac{db}{b^2+\sqrt2\,b+1} \, db \tag6 \\ &= \pi \arctan\left(1+\sqrt2\,a\right) - \frac{\pi^2}4 \tag{$\star$} \end{align*}$$

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• $(1)$ : recall $\displaystyle \int_0^x \frac{dy}{1+y^2}=\arctan(x)$
• $(2)$ : consult the antiderivative given in Spencer's answer
• $(3)$ : differentiate under the integral sign
• $(4)$ : substitute $y\mapsto y^2$
• $(5)$ : partial fraction expansion (residue calculus also works)
• $(6)$ : fundamental theorem of calculus; note that $J(0)=0$
• $(7)$ : denest the square roots and simplify; note that $\arctan\left(2+\sqrt3\right)=\dfrac{5\pi}{12}$