Absolute Convergence of the Improper integral $\int_{-\infty}^{-4-\epsilon}\frac{\cos(x)}{(x^2+3x-4)x}dx$

Question

I need to prove that the following improper integral converges absolutely, and I don't know how:

$$\int_{-\infty}^{-4-\epsilon}\frac{\cos(x)}{(x^2+3x-4)x}dx$$

where $\epsilon$ is a small number (because the function diverges in x=-4)

Any help?

Answer 1

$$\int_{-\infty}^{-4-\varepsilon}\frac{\cos x\,dx}{x(x-1)(x+4)}\,dx =\int_{-\infty}^{-\varepsilon}\frac{\cos (x-4)\,dx}{x(x-4)(x-5)}=-\int_{\varepsilon}^{+\infty}\frac{\cos (x+4)\,dx}{x(x+4)(x+5)}$$ Now if $x\geq\varepsilon$ we have $\left|\cos x\right|\leq 1$ and $0\leq \frac{1}{x(x+4)(x+5)}\leq \frac{1}{x^3}$, where: $$ \int_{\varepsilon}^{+\infty}\frac{dx}{x^3}=\frac{1}{2\varepsilon^2}.$$