I know this is a silly question, but I've tried to find an answer using my TI-89 calculator, Maple and wolframalpha but none of those could tell me whether

$$\int_0^\infty |\cos(x^2)| \mathrm dx$$

converges or diverges.

Thus, I'd be very happy if someone could help me out and tell me, whether the given integral converges or not (and why?). Thanks a lot.

  • $\begingroup$ It diverges. More specifically, $|cos(x)|$ has average value $2/\pi$ (over a single period, or any whole number of periods) and it seems reasonable that $|cos(x^2)|$ should have the same property since the squaring shouldn't perturb any single half-period "too badly". So I'd guess that $\int_0^t |cos(x^2)| \: dx \approx (2/\pi)t$, which agrees with numerical results. $\endgroup$ – Michael Lugo Dec 11 '10 at 4:34
  • $\begingroup$ @Michael Lugo: Your guess is correct. That is basically what my answer gives, except I wrote it in an ugly way. $\endgroup$ – Jonas Meyer Dec 11 '10 at 20:32

It diverges. You should be able to prove that $|\cos(x^2)|>0.1$ for most x. If you let x=$\sqrt{\pi}u$ it is easier to assess the range of u where the cosine is close to zero.


Another method: You could write it as the sum of the integrals on the intervals $\left[\sqrt{\frac{\pi}{2}+k\pi},\sqrt{\frac{\pi}{2}+(k+1)\pi}\right]$, and make a substitution $u=x^2$ to bound the integral on such an interval below by $\frac{1}{\sqrt{\frac{\pi}{2}+(k+1)\pi}}$. (I'm ignoring the interval $\left[0,\sqrt{\frac{\pi}{2}}\right]$.)

Using the inequality $\frac{\pi}{2}+(k+1)\pi\leq(k+2)\pi$ along with an integral comparison then leads to the estimate $$\int_0^\sqrt{\frac{\pi}{2}+N\pi}|\cos(x^2)|dx\geq\frac{2}{\sqrt{\pi}}(\sqrt{N}-\sqrt{2}).$$

  • $\begingroup$ A rough upper bound on that integral using the same methods is $\frac{2}{\sqrt \pi}\sqrt N +2$. $\endgroup$ – Jonas Meyer Dec 10 '10 at 22:53

There is a standard technique we learned for dealing with integrals of the form $(*) \int_{a}^{\infty}f(x)\cos (\alpha x)dx$ or $\int_{a}^{\infty}f(x)\sin (\alpha x)dx$ when $f(x)$ is continuous and positive in $[a, \infty)$.

  1. If $\int_{a}^{\infty}f(x)dx$ converges, then (*) converges absolutely by the comparison test because $|f(x)\sin (\alpha x)|\leq f(x)$ (or $|f(x)\cos (\alpha x)|\leq f(x)$).

  2. If $\int_{a}^{\infty}f(x)dx$ diverges while $f(x)$ is decreasing and $\lim_{x\to\infty}f(x)=0$ then the integrals (*) converge conditionally by Dirichlets test.

I'll illustrate the technique on your integral:

First, lets substitute $t=x^2$, then $(**) \int_{0}^{\infty} \cos(x^2) \mathrm dx=\int_{0}^{\infty} \frac{\cos(t)}{2 \sqrt t} \mathrm dt$.

Let $f(x)=\frac{1}{2\sqrt x}$ then $f(x)$ is decreasing and $\lim_{x\to\infty}f(x)=0$. Also $cos(x)$ has a bounded anti-derivative. Therefore (**) converges by Dirichlets test.

Now we observe that $|\frac{\cos(x)}{2 \sqrt x}|\geq \frac{\cos^2(x)}{2 \sqrt x}=\frac{1}{4\sqrt x}+\frac{\cos(2x)}{4\sqrt x}$. By the same arguments above $\int_{0}^{\infty}\frac{\cos(2x)}{4\sqrt x}dx$ converges. But then if we assume that $\int_{0}^{\infty}\frac{\cos^2(x)}{2 \sqrt x}$ converges, we get that $\int_{0}^{\infty}\frac{1}{4\sqrt x}$ converges and that's obviously not true.


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