I need to prove that this integral diverges:
$$\int_{-\infty}^{\infty}\frac{\cos{x}}{(x^2-3x-4)x}dx$$
And I don't know why. Any help? Thanks!
I know that there are 3 real singularities:
$x=0$,
$x=-1$,
$x=4$
I need to prove that this integral diverges:
$$\int_{-\infty}^{\infty}\frac{\cos{x}}{(x^2-3x-4)x}dx$$
And I don't know why. Any help? Thanks!
I know that there are 3 real singularities:
$x=0$,
$x=-1$,
$x=4$
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Hint. Just because for $x$ near $0$, we have $$ \frac{\cos{x}}{(x^2-3x-4)x} \sim -\frac{1}{4x} \tag1 $$ then, for $\epsilon, \,b $ near $0^+$, we have $$ \int_{\large \epsilon}^b \frac{\cos{x}}{(x^2-3x-4)x} dx \sim -\frac{1}{4}\int_{\large \epsilon}^b\frac{1}{x}dx=-\frac{1}{4}(\log(b)-\log(\epsilon)) \tag2 $$ and, as $\epsilon \to 0^+$, the right hand side of $(2)$ is divergent with your initial integral.