Tough quadrilogarithm integral

Question

Solve the follolwing definite integral

$$\int \frac{\operatorname{Li}_4(z)}{1-z}\, dz$$

It is easy for lower powers!

Answer 1

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.

Answer 2

I think the best approach that can work on most ranges of $z$ should be as follows:

$\int\dfrac{\text{Li}_4(z)}{1-z}dz$

$=-\int\text{Li}_4(z)~d(\ln(1-z))$

$=-\ln(1-z)\text{Li}_4(z)+\int\ln(1-z)~d(\text{Li}_4(z))$

$=-\ln(1-z)\text{Li}_4(z)+\int\dfrac{\ln(1-z)\text{Li}_3(z)}{z}dz$

$=-\ln(1-z)\text{Li}_4(z)-\int\text{Li}_3(z)~d(\text{Li}_2(z))$

$=-\ln(1-z)\text{Li}_4(z)-\text{Li}_2(z)\text{Li}_3(z)+\int\text{Li}_2(z)~d(\text{Li}_3(z))$

$=-\ln(1-z)\text{Li}_4(z)-\text{Li}_2(z)\text{Li}_3(z)+\int\dfrac{(\text{Li}_2(z))^2}{z}dz$

Then substitute $\text{Li}_2(z)=\sum\limits_{n=1}^\infty\dfrac{z^n}{n^2}$ for $|z|\leq1$ , $\text{Li}_2(z)=\dfrac{\pi^2}{3}-\dfrac{(\ln z)^2}{2}-\sum\limits_{n=1}^\infty\dfrac{1}{n^2z^n}-i\pi\ln z$ for $z\geq1$ and $\text{Li}_2(z)=-\dfrac{\pi^2}{6}-\dfrac{(\ln(-z))^2}{2}-\sum\limits_{n=1}^\infty\dfrac{1}{n^2z^n}$ for $z\leq-1$ respectively for example according to Short calculation of the dilogarithm? for further calculation.