I need to prove that the following integral converges:
$$\int_{-\infty}^{\infty}\frac{\cos(\frac{\pi}{2}x)}{(x^2+1)(x-1)}dx$$
From Dirichlet-Abel, I know that:
If $\alpha:[a, +\infty]$ and $U:[a, +\infty]$ and:
(1): $\alpha(x) > 0$ for all $x>0$
(2): $\alpha$ is decresing
(3): $\lim\limits_{x\rightarrow +\infty}\alpha(x)=0$
(4): Exists real $k$ and, for all $\lambda>a$: $|\int_{a}^{\lambda}U(x)dx|<k$
So,
$\int_{a}^{\infty}\alpha(x).U(x)dx$ converges
But I can't put $a=-\infty$ (my professor told me that), so I don't know how to use it here. In other situation, I would say that the function is even or odd (so I multiply the integral for 2 and say that $a=0$), but this is not the case, because $x-1$ is not even nor odd.
What can I do?
Thanks
Hint: $\displaystyle\left|\frac{\cos(\frac\pi2x)}{x-1}\right|=\left|\frac{\sin(\frac\pi2(1-x))}{x-1}\right|\lt\frac\pi2$. Therefore, the integrand is less than $\dfrac{\pi/2}{x^2+1}$.