Integral of $\ln\left(\sum_{k=0}^N a_kx^k\right)$

Question

Given the series $$\displaystyle S(x,N)=\sum_{k=0}^N a_kx^k$$ I would like to know if exists a closed formula to calculate the undefined integral of: $$G(x,N)=\ln\left(\sum_{k=0}^N a_kx^k\right)$$ It's easy to calculate: $$\int G(x,N)dx$$ for $N=2$ and also for $N=3$ even if in this last case the result is more complicated. What happens if $N\gt 3$? Thanks.

Answer 1

I think that in the case $N>3$ one would need to impose a certain number of conditions on the $a_{k}$. For example, a little sketching out on my notepad gave me a quick solution to the case $N=4$ subject to $a_{1} = a_{3}=0$; yielding the so-called "biquadratic case", namely, \begin{equation} S(x,4) = a_{0}+a_{2}x^{2}+a_{4}x^{4} \end{equation}

From which we make a simple substitution $s=x^{2}$, viz \begin{equation} S(s,4) =a_{0}+a_{2}s +a_{4}s^{2} \end{equation} From which we obtain, (wlog) the roots to this quadratic as \begin{equation} \Delta^{\pm} = \frac{a_{2} \pm \sqrt{a_{2}^{2} - 4a_{0}a_{4}}}{2a_{4}} \end{equation} This then brings about the four original roots of the biquadratic as \begin{eqnarray} x_{1} &=& - \sqrt{\Delta^{+}} \\ x_{2} &=& + \sqrt{\Delta^{+}} \\ x_{3} &=& - \sqrt{\Delta^{-}} \\ x_{4} &=& + \sqrt{\Delta^{-}} \\ \end{eqnarray} Thus (wlog) \begin{equation} S(x, 4) = (x_{1} + \sqrt{\Delta^{+}})(x_{2} - \sqrt{\Delta^{+}})(x_{3} + \sqrt{\Delta^{-}})(x_{4} - \sqrt{\Delta^{-}}). \end{equation} Now,by defintion of $G(x, N)$ we have; \begin{eqnarray} G(x, 4) &=& \ln S(x, 4) \\ &=& \ln \left( (x_{1} + \sqrt{\Delta^{+}})(x_{2} - \sqrt{\Delta^{+}})(x_{3} + \sqrt{\Delta^{-}})(x_{4} - \sqrt{\Delta^{-}})\right) \\ &=& \sum_{i=1}^{4} \ln \left(x_{i} +(-1)^{i+1} \Delta_{i} \right). \end{eqnarray} Where I have defined, \begin{equation} \left( \begin{array}{c} \Delta_{1} \\ \Delta_{2} \\ \Delta_{3} \\ \Delta_{4} \end{array} \right) = \left( \begin{array}{c} \sqrt{\Delta^{+}} \\ \sqrt{\Delta^{+}} \\ \sqrt{\Delta^{-}} \\ \sqrt{\Delta^{-}} \end{array} \right) \end{equation} Forming the integral of $\int G(x, 4)dx$ is now a trivial task. Of course, one can apply a "general" formula for the solution of a quartic in the case where all $a_{k}$ are non-zero, but the application of the "formula" is hugely cumbersome.