I am trying to evaluate the integral as written below and I've tried the following:
$$\int\prod_{i=1}^{m}\dfrac{1}{x-i}\mathrm dx=\int\sum_{i=1}^{m}\dfrac{a_i}{x- i}\mathrm dx=\sum_{i=1}^{m}a_i\ln\left| x-i\right|+k, k\in\mathbb{R}$$ $$\begin{aligned}a_m=\dfrac{1}{(x-1)(x-2)\cdots\boxed{(x-m)}}=\dfrac{1}{(m-1)(m-2)\cdots1}&=\dfrac{1}{(m-1)!}\\a_1=\dfrac{1}{\boxed{(x-1)}(x-2)\cdots(x-m)}=\dfrac{1}{(-1)(-2)\cdots(1-m)}&=?\end{aligned}$$ We could also say that $a_i$ is simply $(x-i)$ divided by $\prod_{i=1}^{m}(x-i)$ at $x=i$. But that itself is not a good starting point to lead to anywhere.
I tried using the Heaviside Cover-Up method to find the $a_i$'s. But I am having a hard time generalizing the rule. Any hints for finding the general formula for $a_i$ are appreciated. Thanks
Edit 1:
The expression for $a_1$ can be written in the following form: $$a_1=\dfrac{(-1)^{m-1}}{(m-1)!}$$
Edit 2:
I believe a good guess to the general formula is $$a_i=\dfrac{(-1)^{m-i}}{(m-i)!}$$
We may write $$ a_i = \frac {1}{ \Pi _{1\le j\ne i \le m} (i-j)}$$