Solve the follolwing definite integral

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

It is easy for lower powers!

  • 3
    $\begingroup$ If it's a definite integral, you forgot the endpoints. $\endgroup$ – Robert Israel Aug 7 '13 at 5:04
  • 1
    $\begingroup$ It is easy to observe that the antiderivative can be found once we know how to evaluate $$ \int \frac{\mathrm{Li}_{2}^{2}(z)}{z} \, dz, $$ which I don't quite know how to manage. $\endgroup$ – Sangchul Lee Aug 8 '13 at 0:55
  • $\begingroup$ @sos look at the following math.stackexchange.com/questions/462389/… $\endgroup$ – Zaid Alyafeai Aug 8 '13 at 2:03
  • $\begingroup$ I think we can use the following generating function $$\sum_{n\geq 0} H^{(m)}_n z^n = \frac{\operatorname{Li}_m(z)}{1-z}$$ $\endgroup$ – Zaid Alyafeai Aug 8 '13 at 2:16
  • $\begingroup$ @Sangchul Lee; The integral in question is crucial if we want to find generating functions of quantities like $H_n^{(m)}/n^q$ where $m+q \ge 5$ and $n$ is a positive integer (see math.stackexchange.com/questions/1919402/…) . I wonder have you achieved any progress in tackling this integral ? $\endgroup$ – Przemo Sep 9 '16 at 11:25

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


$$\int \frac{\mathrm{Li}_4(z)}{1 - z} dz = \frac{z^2}{2}\ ^{1+5}f_{1+4} \left(\begin{matrix}2 : 1, 1; 1, 1; 1, 1; 1, 1; 1, 1; \\ 3 : 2, 1; 2, 1; 2, 1; 2, 1;\end{matrix} z, z\right) + C$$.

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

$$\frac{z^m}{1-z} = z^{m-1} \frac{z}{1-z} = z^{m-1} \sum_{n=1}^{\infty} z^n = \sum_{n=1}^{\infty} z^{n+m-1}$$.


$$\sum_{m=1}^{\infty} \frac{z^m}{m^4 (1-z)} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{z^{n+m-1}}{m^4}$$.

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 $n$,


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.

  • $\begingroup$ Your idea of linking the result in terms of Kampé de Fériet function is very good, but you don't really afraid that your procedure will have chance that only valid for $|z|<1$ ? If so, how about for $|z|\geq1$ ? $\endgroup$ – doraemonpaul Aug 7 '13 at 9:15
  • $\begingroup$ One should be able to analytically extend the function. This allows one to define the integral at real $z < -1$. In the complex plane, $z = 1$ is a branch point, so it is ambiguous for real $z > 1$ and not real-valued there. It can, of course, be extended to the cut plane, a Riemann surface, or as a multivalued "function". $\endgroup$ – The_Sympathizer Aug 8 '13 at 2:26

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








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.


I know that this is not the full answer but I think at least it is a step in the right direction. Firstly define: \begin{equation} J^{(p)}_q(x) := \int\limits_0^x [\log(\frac{x}{t})]^p \frac{Li_q(t)}{1-t} dt = p!\sum\limits_{n=1}^\infty \frac{H^{(q)}_{n-1}}{n^{p+1}} x^n \end{equation} As it has been noted in the previous answers integrating by parts and using the properties of the polylogarithms we easily get: \begin{equation} J^{(0)}_4(x)=\int\limits_0^x \frac{Li_4(t)}{1-t} dt = \sum\limits_{l=1}^2 Li_l(x) Li_{5-l}(x) (-1)^{l-1} + (-1)^{2} \int\limits_0^x \frac{[ Li_2(t)]^2}{t} dt \end{equation} We will tackle the second integral again by using integration by parts but in a slightly different way . We write $1/t = \log(t)^{'}$ and we integrate by parts three times . Our aim is to reduce the order of the polylogarithm from two to zero. At the first step we compute the antiderivative of $\log(t) \cdot \frac{Li_1(t)}{t}$ and at the second step we compute the antiderivative of the product of the later with $1/t$. As a result of that process we obtain the following: \begin{equation} J^{(0)}_4(x)=\int\limits_0^x \frac{Li_4(t)}{1-t} dt = -\frac{1}{3} Li_2(x) Li_3(x) + Li_1(x) Li_4(x) - \frac{2}{3} \underbrace{\int\limits_0^x \log(\frac{x}{t}) \frac{Li_3(t)}{1-t} dt}_{J^{(1)}_3(x)} \end{equation} In terms of Euler sums the result reads: \begin{eqnarray} J^{(0)}_4(x)=\sum\limits_{n=1}^\infty \frac{H^{(4)}_{n-1}}{n} x^n &=& -\frac{1}{3} Li_2(x) Li_3(x) + Li_1(x) Li_4(x) - \frac{2}{3}\sum\limits_{n=1}^\infty \frac{H^{(3)}_{n-1}}{n^2} x^n \\ &=& -Li_2(x) Li_3(x) + Li_1(x) Li_4(x) +2 \sum\limits_{n=1}^\infty \left(\frac{H_{n-1}^{(2)}}{n^3}+2\frac{H_{n-1}}{n^4}\right) x^n \\ &=&-Li_2(x) Li_3(x) + Li_1(x) Li_4(x) + J^{(2)}_2(x) + \frac{2}{3} J^{(3)}_1(x) \\ &=&-Li_2(x) Li_3(x) + Li_1(x) Li_4(x) + J^{(2)}_2(x) + 4 S_{3,2}(x) \end{eqnarray} In the second line we used the series expansion of $Li_2 (x) Li_3(x)$ to eliminate the appropriate Euler sum and in the last line we used A family of generating functions related to generalized harmonic numbers and power functions. . Here $S_{3,2}(x)$ is the generalized Nielsen polylogarithm. Now the only thing we need to compute if $J^{(2)}_2$.


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