Closed Form Solution to Difficult Integral

Question

I need help finding a closed form solution to the following integral:

$F_D(d) = \pi d^2 -4d^2\arccos{\frac{1}{2d}} + 2d\sin{\arccos{\frac{1}{2d}}}$

$\int_{\frac{1}{2}}^{\sqrt{\frac{1}{2}}} F_D(d)^2$

Mathematica can only express it in terms of Polylogarithims. I have a feeling that there is no simpler way to express the integral, but am interested to see if anyone else has any ideas.

Answer 1

This is too long for a comment.

I do not know how far you have been able to simplify with Mathematica. Waht I got by the end is $$\frac{17}{60}-\frac{1}{30 \sqrt{2}}-\frac{\pi ^2}{160}-\frac{C}{10}-\frac{1}{20} \Im\left(\text{Li}_2\left(-(-1)^{3/4}\right)\right)+\frac{1}{20} \Im\left(\text{Li}_2\left((-1)^{3/4}\right)\right)$$ where appaears Catalan number.

To me, the question is : what is taking us to polylogarithms ? Expanding the integrand, it becomes $$\left(-1+\pi ^2 d^4+4 d^2+4 \pi \sqrt{1-\frac{1}{4 d^2}} d^3\right)+16 d^4 \cos ^{-1}\left(\frac{1}{2 d}\right)^2-8 \pi d^4 \cos ^{-1}\left(\frac{1}{2 d}\right)-16 \sqrt{1-\frac{1}{4 d^2}} d^3 \cos ^{-1}\left(\frac{1}{2 d}\right)$$ The integration of the first term (inside parentheses) does not make any problem; same for the third term. For the second and the fourth term, the antiderivatives express in terms of polylogarithms. If you want an exact result, I am afraid you could not avoid them.

Answer 2

A possible closed-form of the integral is given by $$ \frac{17}{60} - \frac{1}{30\sqrt{2}} - \frac{C}{10} - \frac{\pi^2}{160} - \frac{\pi^2}{80\sqrt{2}} + \frac{1}{320\sqrt{2}}\left(\psi_1\left(\tfrac18\right) + \psi_1\left(\tfrac38\right)\right), $$ where $C$ is Catalan's constant and $\psi_1$ is the trigamma function. The numerical value of the integral is $$ 0.18176959561984121660888996745341730696110527282932662\dots $$