Give an example of $f$ that is $C^\infty$, $\int_0^\infty f(t)dt$ converges but $f$ does not converge.

Question

Today, my teacher asked about a real function that was integrable, but such that $\displaystyle\lim_{t\to \infty}f(t)\neq 0$ ($f$ need not have a limit).

It's easy to forge an example: a function that has tightening spikes each with areas $1/n^2$ does the trick.

He then asked for a real $C^\infty$ function that was integrable, but such that $\displaystyle\lim_{t\to \infty}f(t)\neq 0$ ($f$ need not have a limit).

One heuristic argument is to take the previous function and "smoothen" the spikes. However, he referred to a nicer function that can be defined explicitely.

Do you have an idea ? I've looked for something involving $\sin,\cos,\tan$

Are you familiar with standard bump functions? See http://en.wikipedia.org/wiki/Bump_function — guest196883, Jun 10 '14 at 18:33
@DanielFischer Riemann I guess ($\int_0^\infty f(t)$ converges) — Gabriel Romon, Jun 10 '14 at 18:42

cnick · Accepted Answer · 2014-06-10T21:07:55.997

2

why not something like $f(x) = \sin(x^2)$? It has to oscillate faster as you go towards infinity so that it adds/subtracts less and less area each time in order for the integral to converge.

edited Jun 10 '14 at 21:07

answered Jun 10 '14 at 18:39

cnick

1,037

Right. You should add this reference in your answer http://math.stackexchange.com/questions/279540/how-to-show-that-int-0-infty-sinx2-dx-converges?rq=1 – Gabriel Romon Jun 10 '14 at 18:44

score 1 · Answer 2 · answered Jun 10 '14 at 18:35

HINT: Try to construct "smooth mountains" by using the Cauchy function

$$h:x\mapsto \begin{cases} e^{-\frac{1}{x}} &\text{if }x>0\\ 0&\text{if }x\leq 0 \end{cases}$$

which is very useful for constructing $C^\infty$ functions with that kind of properties.

Give an example of $f$ that is $C^\infty$, $\int_0^\infty f(t)dt$ converges but $f$ does not converge.

2 Answers2