Breathing Implies Alive

It is easy to show that there exists a $C^{\infty}$ function $\psi:[-1,1]\longrightarrow[0,1]$ with $\psi^{(k)}(-1)=\psi^{(k)}(1)=0$ for all $k\geq 0$ . For example $\psi(t)=\exp(-\frac{1}{1-t^{2}})$ . (Disclaimer: need to check this particular form, but there definitely exists such a function, as it is used in Functional Analysis.)

Let $C:=\int_{[-1,1]}\psi\in(0,1]$ . Set $f(t)=\psi(2^{n}(t-n))$ for $t\in[n-2^{-n},n+2^{-n}]$ and $n\in\mathbb{N}^{+}$ and elsewhere $f\equiv 0$ . It is not hard to check, that $f\in C^{\infty}([0,\infty))$ due to the boundary conditions on the derivatives of $\psi$ . Moreover $\int_{[n-2^{-n},n+2^{-n}]}f=2^{-n}\int_{[-1,1]}\psi= 2^{-n}C$ . Thus $\int_{[0,\infty)} f=\sum_{n\geq 1}2^{-n}C\in(0,\infty)$ . In particular, $f$ is Riemann-integrable.

On the other hand, $f(n)=1$ for all $n\geq 1$ , so that 2 is false. And since $f(n+0.5)=0$ for all $n\geq 1$ , there are two convergent subsequences with $t_{i}\longrightarrow\infty$ and different limits of $f(t_{i})$ . Hence 1 is false.

$\displaystyle \int_{0}^{\infty} f(x)dx=a$ For some $a\in \Re$ $\Rightarrow \displaystyle \int f(x)dx=F(x)+c$ Such that $\lim_{x\to \infty}F(x)-\lim_{x\to 0} F(x)=a$ And the last equation says nothing of each limit to exist apart from the other

Take $f(x)=\sin^2(x)$ , which is a smooth function (and hence, Riemann integrable). We have $\int_{0}^{\infty} f(x)= \infty,$ but $\lim_{x\to \infty} f(x)$ does not exist.

Counter example: $f\left(x\right)=\sin \left(x^2\right).$

Let $f (x)=sin (e^x)$ Then $f$ is a smooth riemann integrable function without a limit