$I = \int_{1}^{\infty}$$\frac{(1-x+[x])(x^a-x^{1-a})}{x^2} dx$ where $[x]$ denotes Greatest integer function and $a\in \mathbb{C}$ is a constant.
Prove that there exists some $c \in (1,\infty)$ such that $$\int_{1}^{\infty}\frac{(1-x+[x])(x^a-x^{1-a})}{x^2}dx = (c^a-c^{1-a})\int_{1}^{\infty}\frac{(1-x+[x])}{x^2}dx.$$
My try:
$$0<1-x+[x]\leq 1 \Rightarrow 0< \frac{1-x+[x]}{x^2} \leq \frac{1}{x^2}$$
$$0<\int_{1}^{\infty}\frac{1-x+[x]}{x^2}dx \leq 1.$$
Consider, $$\frac{\int_{1}^{\infty}\frac{(1-x+[x])(x^a-x^{1-a})}{x^2}dx }{\int_{1}^{\infty}\frac{(1-x+[x])}{x^2}dx} = \lambda \Rightarrow \int_{1}^{\infty}\frac{(1-x+[x])[(x^a-x^{1-a})-\lambda]}{x^2}dx=0$$