The lower cone begins at $\phi=\dfrac{3\pi}{4}$ and runs all the way down to $\phi=\pi$. If the starting point for $\phi$ isn't obvious, then you can find it as follows:
Let $x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi, z=r\cos\theta$. Substituting this into the cone equation, squaring, and rearranging, you obtain:
$$r^2\cos^2\theta = r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi\iff \tan^2\phi=1.$$
Since we're in the lower octants, this means $\tan\phi=-1$, i.e. $\phi=\dfrac{3\pi}{4}$.
Obviously, $0\le \theta\le 2\pi$.
$\rho$ will fun from the origin to the hemisphere. $0 \le \rho\le 2$.
First I want to define with the Stirling numbers of the first kind $\left[ \begin{array}{c} n \\ k \end{array} \right]$ a special generalization of the Riemann Zeta function :
$$\zeta_n(m):=\sum\limits_{k=1}^\infty \frac{1}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
and
$$\eta_n(m):=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
which are convergent for the integer values $\enspace m\geq 2$ .
For $\enspace n=0\enspace$ we have $\enspace\zeta_0(m)=\zeta(m)\enspace$ and $\enspace\eta_0(m)=\eta(m)\enspace$ .
Note: Obviously (because of the other results) these series can be expressed by sums of the polylogarithm function and modifications of that.
Please also see here, part Expansion by harmonic numbers, with $\enspace\displaystyle w(n,m):=\frac{m!}{(n-1)!}\left[ \begin{array}{c} n \\ {m+1} \end{array} \right]\enspace$ and it's recursion formula.
Secondly, an extension of an integral as a series, $n\in\mathbb{N}_0$ and $z\in\mathbb{R}\setminus \{2\mathbb{N}\}$ and $nz>-1$:
$ \displaystyle \int\limits_0^\pi x^n \left(2\sin\frac{x}{2}\right)^z dx=i^{-z} \int\limits_0^\pi x^n e^{i\frac{xz}{2}}(1- e^{-ix})^z dx= e^{-i\frac{\pi z}{2}} \int\limits_0^\pi x^n \sum\limits_{k=0}^\infty\binom{z}{k}(-1)^k e^{-ix(\frac{z}{2}-k)} dx$
$\displaystyle =\int\limits_0^\pi x^n e^{i(x-\pi)\frac{z}{2}} dx+ \sum\limits_{v=0}^n \frac{(-1)^v\pi^{n-v} n!}{i^{v+1}(n-v)!} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}} $
$\displaystyle \hspace{3.5cm} -i^{n-1}n!e^{-i\frac{\pi z}{2}} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{ (-1)^k}{(\frac{z}{2}-k)^{n+1}}$
using the main branch of the logarithm and therefore $\displaystyle i=e^{i\frac{\pi}{2}}$ .
The Stirling numbers of the first kind are usually defined by $\enspace \displaystyle \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) $ .
Because of $\enspace \displaystyle (\sum\limits_{v=0}^\infty x^v \frac{d^k}{dz^k}\binom{z}{v}) |_{z=0} =\frac{d^k}{dz^k}(1+x)^z |_{z=0} =(\ln(1+x))^k=k!\sum\limits_{v=k}^\infty (-1)^{v-k} \left[\begin{array}{c} v \\ k \end{array} \right] \frac{x^v}{v!}$
we get $\enspace \displaystyle \binom{z}{k}|_{z=0}=0^k\enspace$ , $\enspace \displaystyle \frac{d}{dz} \binom{z}{k} |_{z=0} = (-1)^{k-1} \left[\begin{array}{c} k \\ 1 \end{array} \right] \frac{1}{k!}= \frac{(-1)^{k-1}}{k} \enspace$ , $\enspace \displaystyle \frac{d^2}{dz^2} \binom{z}{k} |_{z=0} = (-1)^{k-2} \left[\begin{array}{c} k \\ 2 \end{array} \right] \frac{2!}{k!}= \frac{(-1)^k 2}{k}\sum\limits_{j=1}^{k-1}\frac{1}{j} \enspace$ and $\enspace \displaystyle \frac{d^3}{dz^3} \binom{z}{k} |_{z=0} = (-1)^{k-3} \left[\begin{array}{c} k \\ 3 \end{array} \right] \frac{3!}{k!}= \frac{(-1)^{k-1} 3}{k}( (\sum\limits_{j=1}^{k-1}\frac{1}{j})^2 - \sum\limits_{j=1}^{k-1}\frac{1}{j^2} ) $ .
For $(n;k):=(3;3)$ follows
$\displaystyle \int\limits_0^\pi x^3 \left(\ln\left(2\sin\frac{x}{2} \right)\right)^3 dx =$
$\hspace{2cm}\displaystyle =\frac{9\pi^2}{2}\left(\zeta(5)+3\eta(5)-4\eta_1(4)+2\eta_2(3)\right) $
$\hspace{2.5cm}\displaystyle - 90\left(\zeta(7)+\eta(7)\right) +72\left(\zeta_1(6)+\eta_1(6)\right) - 18\left(\zeta_2(5)+\eta_2(5)\right) $
Note:
For the calculations I have used $\enspace\displaystyle\int\limits_0^\pi x^n e^{iax}dx = \frac{(-1)^{n+1} n!}{(ia)^{n+1}}+e^{i\pi a}\sum\limits_{v=0}^n\frac{(-1)^v \pi^{n-v}n!}{(ia)^{v+1}(n-v)!}$
with $\enspace\displaystyle a=-(\frac{z}{2}-k)$ .
And it was necessary to calculate $\enspace\displaystyle\frac{d^m}{dz^m} \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}}|_{z=0}\enspace$ and $\enspace\displaystyle\frac{d^m}{dz^m} e^{-i\frac{\pi z}{2}}\binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{n+1}}|_{z=0}\enspace$ for $\enspace m\in\{0,1,2,3\}$ .
Best Answer
Define $ G(x) $ with derivative $g(x)$ such that (assume $\zeta\phi\rho=b$)
$$ G(x)= \begin{array}{cc} \left\{ \begin{array}{cc} 0 & x< 0 \\ 1-F(x) &x\ge0 \end{array} \right. \end{array} $$
$$ \implies g(x)= -f(x) $$ Now $\log_2(1+x)=\log_2(e)\cdot\log_e(1+x)$
Taking away $ \frac{-\log_2(e)}{2} $ the integral reduces to $$ I = \int_0^\infty \log (1+x) g(x) \,dx $$ We evaluate the improper integral, by taking the limit of the definite integral $$ I = \lim \limits_{y\to\infty} \left[ \int_0^y \log (1+x) g(x) \,dx \right] $$
Simplifying by parts
[EDIT: This is where the utility of introducing an auxiliary function is most useful. Due to the definition of $G(x)$, the indefinite integral of its derivative $g(x)$ is identical to it. Hence the integral evaluation by parts is elegant.]
$$ I = \lim \limits_{y\to\infty} \left[ G(y)\log(1+y)-G(0)log(1) - \int_0^y \frac{G(x)}{1+x} \,dx \right] $$ $$ I = \lim \limits_{y\to\infty} G(y)\log(1+y) - \int_0^\infty \frac{G(x)}{1+x} \,dx $$ By application of L'hospital rule, we can evaluate the limit to be zero.
Hence, $I$ reduces to $$ I =-\int_0^\infty \frac{G(x)}{1+x} \,dx = -\int_0^\infty \frac{e^{-\frac{x}{b}}}{(1+x)(1+ax)} \,dx $$ $$ = \frac{1}{a-1} \int_0^\infty \frac{(1-a)e^{-\frac{x}{b}}}{(1+x)(1+ax)} \,dx = \frac{1}{a-1} \int_0^\infty \left(\frac{1}{1+x}-\frac{a}{1+ax}\right) e^{-\frac{x}{b}} \,dx $$ Multiplying with $ \frac{-\log_2(e)}{2} $, we get result (2)