Recently I have asked about the evaluation of an integral involving a Trilogarithm $($which can be found here$)$. Pisco provided a quite elegant approach starting with a functional equation of the Trilogarithm. However, his attempt made use of the fact that

$$\int_0^{\infty}\frac{\operatorname{Li}_3(-x)}{1+x}x^{s-1}\mathrm dx=\frac{\pi}{\sin \pi s}[\zeta(3)-\zeta(3,1-s)]~~~~~\text{for }0<s<1$$

Which can be seen as the Mellin Transform of the function $f(x)=\frac{\operatorname{Li}_3(-x)}{1+x}$. He also supplied a proof of his claim by using the Mellin Inversion Theorem. Anyway, I thought about applying Ramanujan's Master Theorem but for this purpose it is conclusive to show that the given function, lets call it $f(x)$, has a power series of the form


With the help of WolframAlpha I got that

$$f(x)=\frac{\operatorname{Li}_3(-x)}{1+x}=\frac12\sum_{n=1}^{\infty}[\psi^{(2)}(1+n)-\psi^{(2)}(1)](-x)^n~~\text{for }|x|<1$$

Using this result combined with Ramanujan's Master Theorem one can easily verify the given relation by setting $\phi(n)=\frac{\Gamma(1+n)}2[\psi^{(2)}(1+n)-\psi^{(2)}(1)]$ and applying the theorem

$$\begin{align} \int_0^{\infty}\frac{\operatorname{Li}_3(-x)}{1+x}x^{s-1}\mathrm dx=\int_0^{\infty}f(x)x^{s-1}\mathrm dx&=\Gamma(s)\phi(-s)\\ &=\Gamma(s)\frac{\Gamma(1-s)}2[\psi^{(2)}(1-s)-\psi^{(2)}(1)]\\ &=\frac{\pi}{\sin \pi s}\frac12[(-2\zeta(3,1-s))-(-2\zeta(3))]\\ &=\frac{\pi}{\sin \pi s}[\zeta(3)-\zeta(3,1-s)] \end{align}$$

Where Euler's Reflection Formula and some basic properties of the Polygamma Functions were used. This attempt affirms Pisco's proof. Up to now this is not a question but more a proof verification hence I am not sure whether all my steps are legitimate like this or not.

Since the power series given by WolframAlpha worked out fairly good I was curious how to derive such a series expansion. First of all I asked for a proof here to see how to approach to $f(x)$. Hence I was not able to identify a pattern within the coefficients of the MacLaurin expansion I thought about using a Cauchy Product here since both functions out of which $f(x)$ is composited have a well-known series expansion therefore this seems to be worth a try. Further the cited proof made use of it in some way aswell. Thus, I tried to compute


But I struggled to get started. I worried about the fact that the first index $i$ starts at $0$ while the second one $j$ at $1$. This would not be a big deal since I could just apply an index shift to one of the two series but then the powers of $(-x)$ would not longer match. Further I am not totally sure how the author of refered proof dogded or accomplished this step.

My question is splitted up into two parts. $(1)$ Is my approach to the equality from above - beside the part where I "cheated" by asking WolframAlpha to expand $f(x)$ as a series - valid i.e. the usage of Ramanujan's Master Theorem here, the simplification of the RHS, etc.? $(2)$ How can one derive the series expansion of $f(x)$? Is it possible - without knowing the explicit power series - to deduce the pattern given by the MacLaurin expansion as values of the Polygamma Function $\psi^{(2)}$? Can we apply the Cauchy Prodcut here; if yes how can we take care of the unsuitable indices? I would be interested in a whole derivation of the series expansion for $f(x)$ as well.

Thanks in advance!

  • 1
    $\begingroup$ Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] \left( \sum_{k = 0}^\infty a_k x^k \right) \sum_{k = 0}^\infty b_k x^k = \sum_{k = 0}^n a_{n - k} b_k = \sum_{k = 1}^n \frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$ $\endgroup$ – Maxim Oct 9 '18 at 19:24
  • $\begingroup$ Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$? $\endgroup$ – mrtaurho Oct 9 '18 at 19:27
  • $\begingroup$ The second series is $0 x^0 + b_1 x^1 + \dots\,$, why isn't it well-defined? $\endgroup$ – Maxim Oct 9 '18 at 19:32
  • $\begingroup$ The second series is $\frac{1}{0}-x+\frac{x^2}8\mp\dots$. $\endgroup$ – mrtaurho Oct 9 '18 at 19:33
  • $\begingroup$ I defined $b_0 = 0$, not $b_0 = (-1)^0/0^3$. $\endgroup$ – Maxim Oct 9 '18 at 19:35

Combining Maxim's suggestion to define the $0^{th}$ coefficient of the Trilogarithm series to be zero and ysharifi's given proof here on AoPS it is rather simple to obtain the given expansion. Using the Cauchy Product leads to

$$\begin{align} \operatorname{Li}_3(-x)\cdot\frac1{1+x}&=\left(\sum_{n=1}^{\infty}\frac{(-x)^n}{n^3}\right)\cdot\left(\sum_{n=0}^{\infty}(-x)^n\right)\\ &=\sum_{n=1}^{\infty}\sum_{k=1}^n\frac1{k^3}(-x)^n\\ &=\sum_{n=1}^{\infty}\left[\sum_{k=1}^{\infty}\frac1{k^3}-\sum_{k=n+1}^{\infty}\frac1{k^3}\right](-x)^n\\ &=\sum_{n=1}^{\infty}\left[\sum_{k=0}^{\infty}\frac1{(1+k)^3}-\sum_{k=0}^{\infty}\frac1{(1+n+k)^3}\right](-x)^n\\ &=\sum_{n=1}^{\infty}\left[-\frac12\psi^{(2)}(1)+\frac12\psi^{(2)}(1+n)\right](-x)^n\\ \end{align}$$



Using the fact that $$\sum_{n=1}^\infty H_n^{(p)}x^n=\frac{\operatorname{Li}_p(x)}{1-x}$$

and by setting $p=3$ and replacing $x$ with $-x$, we get

$$\sum_{n=1}^\infty (-1)^nH_n^{(3)}x^n=\frac{\operatorname{Li}_3(-x)}{1+x}$$

Proof: Using $$\left(\sum_{n=1}^\infty a_nx^n\right)\left(\sum_{n=1}^\infty b_nx^n\right)=\sum_{n=1}^\infty x^{n+1}\left(\sum_{k=1}^na_k\ b_{n-k+1}\right)$$

Then \begin{align} \frac{\operatorname{Li}_p(x)}{1-x}&=\frac1{1-x}*\operatorname{Li}_p(x)\\ &=\sum_{n=1}^\infty x^{n-1}*\sum_{n=1}^\infty\frac{x^n}{n^p}\\ &=\frac1x\sum_{n=1}^\infty x^{n+1}\left(\sum_{k=1}^n\frac{1}{k^p}\right)\\ &=\sum_{n=1}^\infty H_n^{(p)}x^n \end{align}

  • $\begingroup$ This is indeed an interesting consequence of those generating functions. (+1) $\endgroup$ – mrtaurho Aug 14 at 21:57
  • $\begingroup$ Thank you @mrtaurho $\endgroup$ – Ali Shather Aug 14 at 22:07

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