Inverse Cotangent

The original challenge was to integrate the following: $I = \int_{0}^{\infty}(12tan^{-1}x + 5cot^{-1})(cot^{-1}x)^3 dx$ I made the substitution $w = cot^{-1}x$ and noted that $cot(w) = tan(\frac{\pi}{2}-w)$ . Thus $x = cot(w) = tan(\frac{\pi}{2}-w) \text{ and } dx = -csc^2(w)dw$

The limits are related as $0 \leq x < \infty\ \quad \text{to}\quad \frac{\pi}{2} \geq w \geq 0$ . After substitution, $I = \int_{0}^{\frac{\pi}{2}}\big [12\big (\frac{\pi}{2}-w\big ) + 5w\big ] w^3 csc^2(w)dw$

$I = \int_{0}^{\frac{\pi}{2}}\big [\big (6\pi w^3-7w^4\big )\big ] csc^2(w)dw$

There are now two integrals to solve: $I_3 = \int_{0}^{\frac{\pi}{2}}w^3 csc^2(w)dw\ \textrm{ and }\ I_4 = \int_{0}^{\frac{\pi}{2}}w^4 csc^2(w)dw$

The online Wolfram integrator will solve both of these integrals. $I_3 = \frac{3}{8}\big (\pi^2 ln(4)-7\zeta(3)\big )$ $I_4 = \frac{1}{4}\big (\pi^3 ln(4)-9\zeta(3)\big )$ Here $\zeta(s)$ is the Reimann zeta function.

Our result before simplification is: $I=6\pi I_3 - 7I_4$ or $I = \frac{18}{8}\pi^3ln(4) - \frac{126}{8}\zeta(3) -\frac{7}{4}\pi^3ln(4) + \frac{63}{4}\zeta(3)$ The zeta functions cancel and the two remaining terms sum to $I=\frac{1}{2}\pi^3ln(4)=\pi^3ln(2)$

For anyone who wants to dig deeper, I did a bit and was able to duplicate the Wolfram results, please see the following paper by H.M. Srivastava et al. SOME DEFINITE INTEGRALS ASSOCIATED WITH THE RIEMANN ZETA FUNCTION, which can be found here.

In the paper they discuss, among others things, the integration of $\int_{0}^{\frac{\pi}{\omega}} t^s csc^2(t) dt ,\qquad s>1,\ \omega>1$

One last item. I tried solving this problem using a series expansion for $csc^2(x)$ that is given here. The final result was a messy function of the positive integer $k = 1 \rightarrow \infty$ . Numerically I could show that it was converging to the correct value, but I could not figure out how to combine/cancel terms to end up with a recognizable result. If anyone is interested in taking a go at it, perhaps I could submit it as a problem.