Recurrence relation in complex domain I am bumping in the following problem : the expansion of
$$x^{i/k} \operatorname{LerchPhi}[x,1,i/k]$$
leads, for each value of i, to a linear combination of k terms, each of them writing 
$$a(j) \log[1+b(j)x^{1/k}]$$
$a(j)$ and $b(j)$ being fractional powers of $-1$.
I found that $b(j)= (k+1)(j/k)-1$ but I am unable to find anything for $a(j)$.
Any help will greatly be appreciated
 A: From the definition of the Lerch transcendant and Mma you probably want :
\begin{align}
x^{j/k}\; \phi\left(x,1,\frac jk\right)&=x^{j/k}\;\sum_{n=0}^\infty \frac {x^n}{n+\frac jk}\\
&=\sum_{n=0}^\infty \frac {x^{n+j/k}}{n+\frac jk}\\
&=\int_0^x \sum_{n=0}^\infty u^{n+j/k-1}\;du\\
&=\int_0^x u^{j/k-1}\sum_{n=0}^\infty u^{n}\;du\\
&=\int_0^x \frac {u^{j/k-1}}{1-u}\;du\\
\\
\text{Setting $u=t^k\,$ gives} \ \;du=k&\,t^{k-1}dt\,\ \text{and :}\\
\\
&=k\int_0^{x^{1/k}} \frac {t^{j-k+k-1}}{1-t^k}\;dt\\
&=k\int_0^{x^{1/k}} \frac {t^{j-1}}{1-t^k}\;dt\\
\text{or using partial fractions : }\\
&=\,k\int_0^{x^{1/k}} t^{j-1}\sum_{n=0}^{k-1}\frac{(-1)^k}k\frac {e^{2\pi in/k}}{t-e^{2\pi in/k}}\;dt\\
&=(-1)^k\sum_{n=0}^{k-1}\,e^{2\pi in/k}\int_0^{x^{1/k}} \frac {t^{j-1}}{t-e^{2\pi in/k}}\;dt\\
\text{Setting $v:=\dfrac t{e^{2\pi in/k}}$ gives :}\\
&=(-1)^k\sum_{n=0}^{k-1}\,e^{2\pi in/k}\ e^{2\pi i(j-1)n/k}\;\int_0^{e^{-2\pi in/k}\;x^{1/k}} \frac {v^{j-1}}{v-1}\;dv\\
\end{align}
$\quad$and the final result for $0<j\le k$ :
$$\boxed{\displaystyle x^{j/k}\; \phi\left(x,1,\frac jk\right)=-\sum_{n=0}^{k-1}\,e^{2\pi ijn/k}\;\ln\left(1-x^{1/k}\;e^{-2\pi i\,n/k}\right)}$$
Hoping this helped,
