Lebesgue integral of $x \to \cos(x)$

Question

why we have $\cos(x) \notin L^1(\mathbb{R})$? i.e, why $$ \displaystyle\int_{-\infty}^{+\infty} |\cos(x)| dx $$ is not finite?

Thank you in advance to the help

Answer 1

I assume you want a formal proof. Note that $I_k = \int_{\pi k-\pi/2}^{\pi k+\pi/2} \lvert \cos(x)\rvert\,\mathrm{d}x = 2$ for any given $k$. Your integral is therefore bounded below by $\sum_{k=0}^n I_k = 2(n+1)$ for any $n$. Then let $n$ go to infinity to see that the integral cannot be finite.

Answer 2

$$ \displaystyle\int_{-\pi/2}^{\pi/2} |\cos(x)| dx =2 $$ $$ \displaystyle\int_{\pi/2}^{3\pi/2} |\cos(x)| dx =2 $$ $$ \displaystyle\int_{0}^{N} |\cos(x)| dx \geq \lfloor 2N/\pi \rfloor $$

so your integral from $-\infty$ to $\infty$ doesn't converge.