Evaluate given Series in Closed Form

Question

$$f_c(x)=\sum_{r=1}^\infty \frac{r}{r^2-c^2}x^r$$ I have tried developing relations between differentials but have been unable to totally reduce to a closed form solution. Any help appreciated.

Answer 1

Notice that

$$\frac{2n}{n^2-c^2}=\frac1{n-c}-\frac1{n+c}$$

Thus, we are left with:

$$2f_c(x)=\sum_{n=1}^\infty\left[\frac{x^n}{n-c}-\frac{x^n}{n+c}\right]$$

Now notice that

$$\sum_{n=1}^\infty\frac{x^n}{n-c}=x\sum_{n=0}^\infty\frac{x^n}{n+1-c}=x\Phi^*(x,1,1-c)$$

Likewise,

$$\sum_{n=1}^\infty\frac{x^n}{n+c}=x\Phi^*(x,1,1+c)$$

where we use the LerchPhi function.

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Another approach could be to apply integration. Notice that:

$$\begin{align}\sum_{n=1}^\infty\frac{x^n}{n-c}&=x^c\sum_{n=1}^\infty\frac{x^{n-c}}{n-c}\\&=x^c\int_0^x\sum_{n=1}^\infty t^{n-c-1}\ dt\\&=x^c\int_0^x\frac{t^{-c}}{1-t}\ dt\end{align}$$

Likewise,

$$\sum_{n=1}^\infty\frac{x^n}{n+c}=x^{-c}\int_0^x\frac{t^c}{1-t}\ dt$$

Which gives

$$2f_c(x)=\int_0^x\frac{(x/t)^c-(t/x)^c}{1-t}\ dt$$

Answer 2

Hint:

Note that $\quad\dfrac r{r^2-c^2}=\dfrac12\dfrac1{r-c}+\dfrac12\dfrac1{r+c}$, so you have to find closed forms for $$\sum_{r\ge 1}\dfrac{x^r}{r-c}=x^c\sum_{r\ge 1}\dfrac{x^{r-c}}{r-c}\quad\text{and}\quad\sum_{r\ge 1}\dfrac{x^r}{r+c}=x^{-c}\sum_{r\ge 1}\dfrac{x^{r+c}}{r+c}.$$ Now $\;\dfrac{x^{r+c}}{r+c}$ is the antiderivative of $\;x^{r+c-1}$. Thus the last sum is the antiderivative of $$ \sum_{r\ge 1}x^{r+c-1}=x^{c-1}\sum_{r\ge 1}x^{r}=x^{c}\frac1{1-x}, $$ and similarly for the other sum.