How to Evaluate $\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$?

Question

How to find $$\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$$

In fact, \begin{align*} &\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} \\ =& \int_0^1 {\frac{{u\arcsin u}}{{\left( {{u^2} - \sqrt {2\sqrt {13} - 2} u + \sqrt {13} } \right)\left( {{u^2} + \sqrt {2\sqrt {13} - 2} u + \sqrt {13} } \right)}}du} \\ = &\frac{1}{{2\sqrt {2\sqrt {13} - 2} }}\left[ {\int_0^1 {\frac{{\arcsin u}}{{{u^2} - \sqrt {2\sqrt {13} - 2} u + \sqrt {13} }}du} - \int_0^1 {\frac{{\arcsin u}}{{{u^2} + \sqrt {2\sqrt {13} - 2} u + \sqrt {13} }}du} } \right]. \end{align*}

But how can we continue?

Answer 1

Rewrite $\arcsin u = \arctan \dfrac u{\sqrt{1-u^2}} = \arctan v$:

$$\begin{align*} I &= \int_0^1 \frac{u \arcsin u}{u^4 + 2u^2 + 13} \, du \\ &= \int_0^1 \frac{u \arctan \frac u{\sqrt{1-u^2}}}{u^4 + 2u^2 + 13} \, du \\ &= \int_0^\infty \frac{\frac v{\sqrt{1+v^2}} \arctan v}{\frac{v^4}{\left(1+v^2\right)^2} + \frac{2v^2}{1+v^2} + 13} \, \frac{dv}{\left(1+v^2\right)^{3/2}} \\ &= \int_0^\infty \frac{v \arctan v}{16v^4 + 28v^2 + 13} \, dv \\ &= \frac12 \int_{-\infty}^\infty \frac{v \arctan v}{16v^4 + 28v^2 + 13} \, dv \end{align*}$$

Now integrate the complex function,

$$f(z) = \frac{z \arctan z}{16z^4 + 28z^2 + 13} = -\frac i2 \frac{z}{16z^4 + 28z^2 + 13} \left(\log\left\lvert\frac{i-z}{i+z}\right\rvert + i \arg\left(\frac{i-z}{i+z}\right)\right)$$

along an indented semicircular contour $C$ (a modified reproduction of the contour provided in @Accelerator's answer here), avoiding a branch cut taken along $i[1,\infty)$ and enclosing the poles $\omega_1=\color{red}-\dfrac{\sqrt{-7\color{red}-i\sqrt3}}{2\sqrt2}$ and $\omega_2=\color{red}+\dfrac{\sqrt{-7\color{red}+i\sqrt3}}{2\sqrt2}$, indicated approximately by the green X marks. (I use the branch of $\sqrt z$ with $\arg z\in(-\pi,\pi)$.)

NB: Rewriting the integrand in terms of $\arctan$ isn't strictly necessary; just personal preference. With $\arcsin$, one can just as easily take a branch cut along $[-1,1]$ and integrate along a dogbone contour, as demonstrated here.

By the residue theorem, and taking a shortcut with Mathematica to simplify the RHS as much as possible,

$$\oint_C f(z) \, dz = i2\pi \sum_{\omega_1,\omega_2} \operatorname{Res} f(z) = \frac{\pi^2}{8\sqrt3} - \frac\pi{8\sqrt3} \arctan \left(\frac4{79} \sqrt{21+114\sqrt{13}}\right)$$

The integrals along the circular arcs $\Gamma$ and $\gamma$ will vanish as their respective radii get arbitrarily large/small. As the paths $\lambda_1,\lambda_2$ approach the cut from either side, we have

$$\begin{align*} \int_{\lambda_1} f(z) \, dz &= \int_R^{1+\varepsilon} f(\varepsilon+iy) \cdot i\,dy \\ &\to -\frac i2 \int_1^\infty \frac{y}{16y^4 - 28y^2 + 13} \left(\log\left(\frac{y-1}{y+1}\right) + i \pi\right) \, dy \\[2ex] \int_{\lambda_2} f(z) \, dz &= \int_{1+\varepsilon}^R f(-\varepsilon+iy) \cdot i\,dy \\ &\to \frac i2 \int_1^\infty \frac{y}{16y^4 - 28y^2 + 13} \left(\log\left(\frac{y-1}{y+1}\right) - i \pi\right) \, dy \\[2ex] \implies \int_{\lambda_1\cup\lambda_2} f(z)\,dz &= \pi \int_1^\infty \frac{y}{16y^4 - 28y^2 + 13} \, dy \\ &= \pi \int_1^\infty \frac{y}{16 \left(y^2 - \frac78\right)^2 + \frac{3}{4}} \, dy = \frac{\pi^2}{12\sqrt3} \end{align*}$$

It follows that

$$\int_{-\infty}^\infty f(v) \, dv = \frac{\pi^2}{24\sqrt3} - \frac\pi{8\sqrt3} \arctan \left(\frac4{79} \sqrt{21+114\sqrt{13}}\right) \\ \implies I = \boxed{\frac{\pi^2}{48\sqrt3} - \frac\pi{16\sqrt3} \arctan \left(\frac4{79} \sqrt{21+114\sqrt{13}}\right)}$$

Answer 2

This is probably not an answer

Being too lazy to attack such monsters, I should use Taylor series $$\sin^{-1}(u)=\sum_ 0^\infty\frac{(2n)!}{4^n(n!)^2(2n+1)}u^{2n+1}\qquad (|u|\leq 1)$$ and perform the long division by the denominator. This would lead to $$\frac{{u\sin^{-1} u}}{{{u^4} + 2{u^2} + 13}}=\frac{u^2}{13}+\frac{u^4}{1014}-\frac{79 u^6}{263640}+O\left(u^7\right)$$ If you are brave enough, take more terms !