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.

AsukaMinato
  • 1,007
aga7689
  • 93

1 Answers1

0

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.

André Nicolas
  • 507,029
  • I don't understand what to do with the second term (the integral). That integral just seems as complicated as the first one. – aga7689 Apr 17 '16 at 17:35
  • We are not evaluating, we are proving convergence. Since $|\sin x|\le 1$, and $\int_1^\infty \frac{1}{x^{3/2}},dx$ converges, the integral $\int_1^\infty \frac{\sin x}{x^{3/2}},dx$ converges absolutely, and therefore converges. – André Nicolas Apr 17 '16 at 17:38
  • I get it now. Thanks a lot. – aga7689 Apr 17 '16 at 17:41
  • You are welcome. Integration by parts can be a useful tool for producing estimates, not just exact answers. – André Nicolas Apr 17 '16 at 17:55