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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

Raymond Manzoni
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Claude Leibovici
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  • I attempted to improve your formatting. Please check to make sure I didn't mess anything up. It won't be easy for someone to help you, though. First, she has to know what a LerchPhi is; then, she has to figure out what kind of expansion is being done; then, she has to convince herself that said expansion leads to a linear combination of $k$ terms of the given form --- and all that, just to understand what is being asked, before even attempting to answer. She might get tired, and go do something else, instead. Why not make it easier for someone to help? – Gerry Myerson Jul 01 '13 at 10:56
  • The use of $i$ as something besides $\sqrt{-1}$ here is problematic. – graveolensa Jul 01 '13 at 11:43

1 Answers1

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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,

Raymond Manzoni
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  • This is a splendid answer. I thank you very much – Claude Leibovici Jul 02 '13 at 03:44
  • Glad you liked it @Claude ! – Raymond Manzoni Jul 02 '13 at 08:30
  • I spent all my time working with what you so nicely sent me. Again, this is a beautiful work which helps me a lot. Thanks again. If I may, another stupid question: how comes suddenly a definite integral and what is x^(1/p) ? – Claude Leibovici Jul 03 '13 at 05:21
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    @Claude: Merely a typo for $x^{1/k}$. I added the integrals bounds explicitly everywhere. Another non trivial transition is the passage from $\int_0^{e^{-2\pi in/k};x^{1/k}} \frac {v^{j-1}}{v-1};dv;$ to $;\ln\left(1-x^{1/k};e^{-2\pi i,n/k}\right)$. In fact additional $v$ polynomials terms appear during integration but disappear after considering $\sum_{n=0}^{k-1},e^{2\pi ijn/k}$(polynomials). – Raymond Manzoni Jul 03 '13 at 07:08
  • may I ask you how you found the track you followed to arrive to this beautiful result ? I plan to publish results of my research and I would like to provide your demonstration and aknowledge your contribution. Any problem with that ? By the way, what is your job ? Regards and thanks. – Claude Leibovici Jul 04 '13 at 04:40
  • I would of course be glad to be cited in your publication so no problem about that. The track is simply the one I exposed here : starting from the definition of Lerch $\phi$, getting the first and second integral and the partial fraction decomposition (using limits), creating all along pari/gp scripts to compare numerically new formulas with the first one, trying some explicit cases for $\frac jk$ with Wolfram Alpha (with the inconvenient that Alpha combined some log terms together...), the last part was more difficult (my $\log$ terms were written differently and the results... – Raymond Manzoni Jul 04 '13 at 10:36
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    of my last script were all unrelated to the wished one). It seemed that the additional polynomial terms (from $\int_0^{e^{-2\pi in/k};x^{1/k}} \frac {v^{j-1}}{v-1};dv$) didn't cancel... Anyway my last change of variables clarified things and gave the final answer (note that this part is not formally proved here, the exposition was laborious enough... :-)). At this point the neat and simple final result made me think to the polylogarithm, polygamma and Clausen product formulas I saw earlier and I thought that... the result was probably trivial ! :-) – Raymond Manzoni Jul 04 '13 at 10:36
  • A quick glance in the Wolfram function didn't prove this but returned this more general formula for $\phi(e^{2\pi it},s,p/q)$. $\Gamma$ and Hurwitz $\zeta$ are singular for $s=1$ and the limit of the product makes only sense for $\zeta(1-s,1/2)$ I think (giving $-\frac{\ln(2)}2$) so that the overlap is minimal... Concerning my job I am in software development for quite some years... Let's add my pari/gp scripts to be complete and... Cheers, – Raymond Manzoni Jul 04 '13 at 10:37
  • s(j,k,x)=x^(j/k)sumpos(n=0,x^n/(n+j/k));
    sj(j,k,x)=intnum(t=1e-20,x,t^(j/k-1)/(1-t));
    ss(j,k,x)=(-1)^k    sum(n=0,k-1,exp(2PiIn/k)intnum(t=1e-20,x^(1/k),t^(j-1)/(t-exp(2PiIn/k))));
    su(j,k,x)=-sum(n=0,k-1,exp(2    PiInj/k)log(1-x^(1/k)exp(-2PiIn/k)));     – Raymond Manzoni Jul 04 '13 at 10:38
  • @Raymond : I am finishing my paper. I acknowledge your significant contribution. Do you have a specific title (Dr or anything else) ? Do you want me to put your affiliation ? Thanks for answering this message (I do not know how to contact people in SE). – Claude Leibovici Aug 02 '13 at 19:41
  • @ClaudeLeibovici: Thank you very much for that ! I don't have a title nor affiliation (I am software developer for a private society). Just indicating my (real) name and StackExchange if you want would be very kind. Btw my public email is < raymman (at) free (point) fr >. Cheers, – Raymond Manzoni Aug 02 '13 at 21:10