Improper convergence of $ \cos(x)/{x^{1/2}} $ I have to evaluate the convergence of the improper integral $ \int_1^\infty \frac {\cos(x)}{x^{1/2}}dx $.
As the function is continuous on every $ [1, M] $, I can tell that this function is Riemann integrable on every interval $ [1,M] $, M > 1.  So all I have to do is to evaluate the limite at the bounds :
$$ \lim_{b\to \infty}\int_1^b \frac {\cos(x)}{x^{1/2}}dx $$.  The problem is, I don't know how to evaluate this integral.  I've tried integrating by parts, but it doesn't work as the power of x isn't an integer.  Should I use the comparison theorem?  Or should I integrate this?  
Thank you for your help.

We prove convergence. Integrate by parts, letting $u=x^{-1/2}$ and $dv=\cos x\,dx$. Then $du=(-1/2)x^{-3/2}$ and we can take $v=\sin x$. So our integral from $1$ to $M$ is
$$\left. x^{-1/2}\sin x\Large\right|_1^M-\int_1^M (-1/2)x^{-3/2}\sin x\,dx.$$
Both parts behave nicely as $M\to\infty$, because $|\sin x|\le 1$.
Remark: By looking at our function from $\pi$ to $2\pi$, and $2\pi$ to $3\pi$, and so on, one can show that $\int_1^\infty x^{-1/2}|\cos x|\,dx$ doe not converge. 
