Let $I_{n,k} = \frac{(n+k)!}{k!(k-1)!(n-k)!}$. This is a sort of generalization of the Apéry's numbers, with $I_{n,n} =$ the $n$-th Apéry number. I am studying integrals of the form: $$f_u(x)=\int \prod_{j=1}^u \frac{x+j}{j-x}~dx.$$ Where $u$ is a natural number. For $u>2$, I have shown that $$\tag{1}f_u(x) = C+ (-1)^ux+\sum_{w=1}^{u}(-1)^w \log(x-w)I_{u,w}.$$ For example, $$f_3(x)=-x-60\log(x-3)+60\log(x-2)-12\log(x-1).$$ I am looking for clarification on two things:

  1. My eq. $(1)$ does not work for $u=1,2$. It generates a function that is almost what the integral evaluates to; whereas $f_1(x) = -x-2\log(1-x)$, eq. $(1)$ gives me $-x+2\log(x-1)$. Is there any amendment I can make to $(1)$ to ensure that it holds for all natural $u$?
  2. My eq. $(1)$ surely appears like it should be written as one sum, but I have not been able to manipulate the summand to include the $(-1)^u x$ term. Is there a way to rewrite $(1)$ as a single sum, barring $C~$?
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    $\begingroup$ What if the arguments of your logs are negative? Shouldn't you have log|▪︎| everywhere? $\endgroup$ – Wolfgang Feb 1 at 21:28
  • $\begingroup$ notice that $\log(x-w)$ and $\log(w-x)$ only differ by an (imaginary) constant, so this can be absorbed in the $C$ in $f_u(x)$; the only difference between $u=1,2$ and $u>2$ that matters is that the sign in front of the logarithm is different from $(-1)^w$; if I am allowed to change the sign of the definition of $I_{u,w}$ for $u=1,2$, I'm done. $\endgroup$ – Carlo Beenakker Feb 1 at 21:29
  • $\begingroup$ are you sure your equation (1) is correct? I think $-\sum_{w=1}^u$ should be $+\sum_{w=1}^u$ $\endgroup$ – Carlo Beenakker Feb 1 at 21:38
  • $\begingroup$ @CarloBeenakker you are correct, fixed. $\endgroup$ – Descartes Before the Horse Feb 1 at 21:40

I may be mistaken, but I get $$f_u(x)=\int \prod_{j=1}^u \frac{x+j}{j-x}~dx= C+ (-1)^ux+\sum_{w=1}^{u}(-1)^w \log(x-w)I_{u,w}$$ and this is correct for all $u=1,2,3,...$, so it seems issue 1 is resolved.

  • $\begingroup$ Thank you. If no answer addressing the second part of my question is posted in some time, I will accept this. $\endgroup$ – Descartes Before the Horse Feb 1 at 21:42

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