Evaluate $\int_{0}^{\infty}\frac{\ln^m(x)}{\prod_{k=0}^{n}(1+\alpha_k x)}\mathrm dx=G(m)$

Question

Given that,

$$\int_{0}^{\infty}\frac{\ln^m(x)}{\prod_{k=0}^{n}(1+\alpha_k x)}\mathrm dx=G(m)$$

Assume $\alpha_n$ are integers

I was able to see a pattern for $m=1$

$$\int_{0}^{\infty}\frac{\ln x}{\prod_{k=0}^{n}(1+\alpha_k x)}\mathrm dx=-\frac{1}{2^{n+1}n!}\sum_{j=0}^{n}(-1)^j{n\choose j}\alpha_j^{n-1}\ln^2(\alpha_j)$$

but for $m\ge2$, I was not able to work out the general closed form for $G(m)$

How can we evaluate for $G(m)$?

Answer 1

Assume $a_k>0$.

Denote $$P(c)=\prod^n_{g=0,g\ne c}(1-\frac{a_g}{a_c})$$

Then, by residue theorem, one can obtain the recursive relation $$\sum^{m-1}_{j=0}C^m_j (2\pi i)^{m-j}G(j)=-2\pi i\sum^n_{k=0}\frac{(\pi i-\ln a_k)^m}{P(k)}$$

I will elaborate later.

Note that $G(0)$ cannot be defined using this relationship. $G(0)$ has to be evaluated explicitly. That’s not very difficult.