Tough quadrilogarithm integral Solve the follolwing definite integral 
$$\int \frac{\operatorname{Li}_4(z)}{1-z}\, dz$$
It is easy for lower powers!

Wolfram's integrator can't find it. But it's a relatively simple matter to get a solution in terms of the Kampé de Fériet hypergeometric function:
http://en.wikipedia.org/wiki/Kamp%C3%A9_de_F%C3%A9riet_function
as
$$\int \frac{\mathrm{Li}_4(z)}{1 - z} dz = \frac{z^2}{2}\ ^{1+5}f_{1+4} \left(\begin{matrix}2 : 1, 1; 1, 1; 1, 1; 1, 1; 1, 1; \\ 3 : 2, 1; 2, 1; 2, 1; 2, 1;\end{matrix} z, z\right) + C$$.
To get this solution, first write
$$\frac{\mathrm{Li}_4(z)}{1-z} = \sum_{m=1}^{\infty} \frac{z^m}{m^4 (1-z)}$$
Now, expand
$$\frac{z^m}{1-z} = z^{m-1} \frac{z}{1-z} = z^{m-1} \sum_{n=1}^{\infty} z^n = \sum_{n=1}^{\infty} z^{n+m-1}$$.
So,
$$\sum_{m=1}^{\infty} \frac{z^m}{m^4 (1-z)} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{z^{n+m-1}}{m^4}$$.
Then integrate term-by-term and rewrite with the lower summation indices as 0:
$$\begin{align}\int \left(\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{z^{n+m-1}}{m^4}\right) dz &= C + \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{z^{n+m}}{m^4 (n+m)} \\ &= C + z^2 \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{z^n z^m}{(m+1)^4 (n+m+2)}\end{align} \tag{*}$$.
Now, from the definition of the hypergeometric function:
$$^{p+q} f_{r+s}\left(\begin{matrix}a_1, ..., a_p: b_1, b_1'; ...; ; b_q, b_q' ; \\ c_1, ..., c_r : d_1, d_1' ; ... ; d_s, d_s' ;\end{matrix} z, w\right) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{(a_1)_{m+n} ... (a_p)_{m+n}}{(c_1)_{m+n} ... (c_r)_{m+n}} \frac{(b_1)_m (b_1')_n ... (b_q)_m (b_q)_n}{(d_1)_m (d_1')_n ... (d_s)_m (d_s')_n} \frac{z^m}{m!} \frac{w^n}{n!}$$
the ratio of coefficients of $z^m z^n$ with respect to $m$ (including the factorials) is
$$R_{m,n} = \frac{(a_1+m+n)...(a_p+m+n)}{(c_1+m+n)...(c_r+m+n)} \frac{(b_1+m)...(b_q+m)}{(d_1+m)...(d_s+m)} \frac{1}{m+1}$$
and with respect to $n$,
$$S_{m,n} = \frac{(a_1+m+n)...(a_p+m+n)}{(c_1+m+n)...(c_r+m+n)} \frac{(b_1'+n)...(b_q'+n)}{(d_1'+n)...(d_s'+n)} \frac{1}{n+1}$$.
The ratio of the coefficients of the series marked (*) with respect to $m$ is
$$\frac{(m+1)^4 (n+m+2)}{(m+2)^4 (n+m+3)} = \frac{(m+n+2)}{(m+n+3)} \frac{(m+1)(m+1)(m+1)(m+1)}{(m+2)(m+2)(m+2)(m+2)}$$
and with respect to $n$,
$$\frac{(m+n+2)}{(n+m+3)}$$.
So we see that we should assign $a_1 = 2$, $c_1 = 3$, and $b_{1...5} = 1$, $d_{1...4} = 2$, with one extra factor on b to cancel the factor resulting from the factorial. Then set $b_{1...5}' = 1$ and $d_{1...4}' = 1$ to cancel the $b'$/$d'$ stuff and the other factor resulting from the factorial. This means $p = 1$, $r = 1$, $q = 5$, and $s = 4$. The normalization factor $\frac{1}{2}$ is because the setting of the $a$s, etc. only determines the ratio of terms, and not the initial value to which the multiplication by the ratio is applied, which is given by the term with index $m = n = 0$, which in series (*) is $\frac{1}{2}$. This then gives the desired result for the integral.
