On the series expansion of $\frac{\operatorname{Li}_3(-x)}{1+x}$ and its usage

Recently I have asked about the evaluation of an integral involving a Trilogarithm $$($$which can be found here$$)$$. Pisco provided a quite an 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

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

$$f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{\phi(n)}{n!}x^n$$

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 as well. Thus, I tried to compute

$$\frac1{1+x}\cdot\operatorname{Li}_3(-x)=\left(\sum_{i=0}^{\infty}(-x)^i\right)\left(\sum_{j=1}^{\infty}\frac{(-x)^j}{j^3}\right)=~?$$

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 referred proof dodged or accomplished this step.

My question is split 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 Product 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.

• Regarding the product of two series, just assume $b_0 = 0$: $$[x^n] \left( \sum_{k = 0}^\infty a_k x^k \right) \sum_{k = 0}^\infty b_k x^k = \sum_{k = 0}^n a_{n - k} b_k = \sum_{k = 1}^n \frac {(-1)^n} {k^3} = (-1)^n H_{n, 3}.$$ Oct 9 '18 at 19:24
• Is this not a little bit of cheating since obviously the second series will not be well defined for $j=0$? Oct 9 '18 at 19:27
• The second series is $0 x^0 + b_1 x^1 + \dots\,$, why isn't it well-defined? Oct 9 '18 at 19:32
• The second series is $\frac{1}{0}-x+\frac{x^2}8\mp\dots$. Oct 9 '18 at 19:33
• @AliShather No worries. You're not annoying to me, you can edit if you feel like its appropriate. But I might change it back if I feel like it's not ;) Dec 4 '19 at 5:58

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}

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

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}

• This is indeed an interesting consequence of those generating functions. (+1) Aug 14 '19 at 21:57
• Thank you @mrtaurho Aug 14 '19 at 22:07

Different approach using a simple powerful identity

We proved here that

$$\sum_{n=1}^\infty a_nx^n=\frac1{1-x}\sum_{n=1}^\infty (a_n-a_{n-1})x^n,\quad a_{0}=0$$

Set $$a_n=H_n^{(a)}$$ we have

$$\sum_{n=1}^\infty H_n^{(a)}x^n=\frac1{1-x}\sum_{n=1}^\infty (H_n^{(a)}-H_{n-1}^{(a)})x^n=\frac{1}{1-x}\sum_{n=1}^\infty \frac{x^n}{n^a}=\frac{\operatorname{Li}_a(x)}{1-x}$$

Set $$a=3$$ and replace $$x$$ with $$-x$$ to have

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

• Yet again, quite interesting! (+1) Feb 22 '20 at 21:25
• @mrtaurho thank you . Feb 22 '20 at 21:53