Evaluate $\int_0^1 \arctan^3 x\,dx$ I don't want to use the Fourier series. My work \begin{align}J&=\int_0^1 \arctan^3 x\,dx\\
 &=[x\arctan^3 x]_0^1 -3\int_0^1
 \frac{x\arctan^2 x}{1+x^2}\,dx\\
 &=\frac{\pi^3 }{64}-3\int_0^1 \frac{x\arctan^2
 x}{1+x^2}\,dx\\
 &= \frac{\pi^3
 }{64}-\frac{3}{2}\left[\ln(1+x^2)\arctan^2
 x\right]_0^1 +3\int_0^1 \frac{\ln(1+x^2)\arctan
 x}{1+x^2}\,dx\\
 &=\frac{\pi^3 }{64}-\frac{3\pi^2\ln
2}{32}+3\int_0^1 \frac{\ln(1+x^2)\arctan
 x}{1+x^2}\,dx\\
\end{align}
How to continue?
 A: I am not sure if this counts as a solution using Fourier series, but let me present my solution anyway: By the substitution $x=\tan\theta$, we get
$$ J = \int_{0}^{\frac{\pi}{4}} \theta^3 \sec^2\theta \, \mathrm{d}\theta. $$
In order to compute this integral, we will utilize the following regularized expansion:
$$ \sec^2\theta = \frac{4e^{2it}}{(1+e^{2it})^2} = \lim_{r \uparrow 1} \frac{4r e^{2it}}{(1+r e^{2it})^2} = 4 \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} n r^n e^{2in\theta} $$
Plugging this back to $J$, we can interchange the order of limit and integration by the uniform convergence, whence we get
\begin{align*}
J
&= 4 \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} n r^n \int_{0}^{\frac{\pi}{4}} \theta^3 e^{2in\theta} \, \mathrm{d}\theta.
\end{align*}
Now by the integration by parts,
$$ \int_{0}^{\frac{\pi}{4}} \theta^3 e^{2in\theta} \, \mathrm{d}\theta
= -\frac{3i^n}{8n^4} + \frac{3}{8n^4} + \frac{3\pi i^{n+1}}{16n^3} + \frac{3\pi^2 i^n}{64n^2} - \frac{\pi^3 i^{n+1}}{128n}. $$
Plugging this and taking limit,
\begin{align*}
J
&= \lim_{r \uparrow 1} \sum_{n=1}^{\infty} (-1)^{n-1} r^n \left( -\frac{3i^n}{2n^3} + \frac{3}{2n^3} + \frac{3\pi i^{n+1}}{4n^2} + \frac{3\pi^2 i^n}{16n} - \frac{\pi^3 i^{n+1}}{32} \right) \\
&= \sum_{n=1}^{\infty} (-1)^{n-1} \left( -\frac{3i^n}{2n^3} + \frac{3}{2n^3} + \frac{3\pi i^{n+1}}{4n^2} + \frac{3\pi^2 i^n}{16n} \right) + \frac{\pi^3}{64}(1-i) \\
&= - \frac{3}{2} \left( \frac{1}{2^3} - \frac{1}{4^3} + \frac{1}{6^3} - \dots \right)
+ \frac{3}{2} \left( \frac{1}{1^3} - \frac{1}{2^3} + \frac{1}{3^3} - \dots \right) \\
&\quad - \frac{3\pi}{4} \left( \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \dots \right)
+ \frac{3\pi^2}{16} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \dots \right)
+ \frac{\pi^3}{64} \\
&\quad + \underbrace{\text{[imaginary term]}}_{=0}.
\end{align*}
Simplifying this, we get
$$ J = -\frac{3\pi G}{4} + \frac{63\zeta(3)}{64} + \frac{\pi^3}{64} + \frac{3 \pi^2 \log 2}{32}, $$
where $G$ is the Catalan's constant.
A: Not an answer, but Wolfram Alpha finds the amazing $$\int (\arctan(x))^3\mathrm{d}x$$ $$=\frac{3}{2}\operatorname{Li}_3(-e^{2i\arctan(x)})-3i\arctan(x)\operatorname{Li}_2(-e^{2i\arctan(x)})+(\arctan(x))^2(x\arctan(x)-i\arctan(x)+3\ln(1+e^{2i\arctan(x)}))+c$$
And the even more beautiful
$$\int_0^1 (\arctan(x))^3\mathrm{d}x=\frac{1}{64}(\pi^2(\pi+\ln(64))+63\zeta(3)-48\pi C)$$
With Catalan's constant $C$ and the Riemann zeta function $\zeta$.
But, continuing from your last line,
$$\int_0^1 \frac{\ln(1+x^2)\arctan(x)}{1+x^2}\mathrm{d}x$$
We can make the substitution $x=\tan\theta, \mathrm{d}x=\sec^2(\theta)\mathrm{d}\theta$ to get
$$\int_0^{\arctan(1)} \theta\ln(\sec^2(\theta))\mathrm{d}\theta$$
Although knowledge of the polylogarithm will still be needed.
