Help Needed: Unbounded Function with Integral Exists

[Mathmos6]

Homework Statement

Pretty much explained in the title! I need to find an $f:[0,\infty) \to [0,\infty)$ such that f is unbounded but $\int_0^{\infty}f(x)dx$ exists - I've been pondering for about half an hour now but nothing which works has come to mind - could anyone help? I know it can't be periodic (trig etc) because then F(infinity) will be undefined, where F'=f, but other than that I'm feeling a bit clueless! Thanks very much, Mathmos6

 

[matt grime]

I presume f doesn't have to be continuous, have you thought about that?

In fact I don't think f can be continuous given the hypotheses: for the integral to exist, then f(x) tends to zero as x tends to infinity. In particular there is a T such that x>T implies |f(x)| <1, i.e. on the range [T,inf) f is bounded between 1 and -1. And on the range [0,T] f is bounded as f is continuous, thus f is bounded on the whole of [0,inf). Now, if the domain and range were (0,inf) that would be different.

 

[Mathmos6]

Ah, I apologize, I stupidly forgot to mention that yes, the function has to be a -continuous- unbounded function wherein the aforementioned integral exists.

 

[yyat]

I presume f doesn't have to be continuous, have you thought about that?

In fact I don't think f can be continuous given the hypotheses: for the integral to exist, then f(x) tends to zero as x tends to infinity. In particular there is a T such that x>T implies |f(x)| <1, i.e. on the range [T,inf) f is bounded between 1 and -1. And on the range [0,T] f is bounded as f is continuous, thus f is bounded on the whole of [0,inf). Now, if the domain and range were (0,inf) that would be different.

Why does f(x) have to go to 0 for x to infinity? Couldn't you have a function with "spikes" (scaled bump functions) that increase in height but decrease in area for large x?

 

[matt grime]

Yes, of course you can, yyat. How stupid of me not to see that. Clearly there will be many trivial to describe such examples: one springs to mind instantly with area in the region of pi^2/6 (possibly pi^2/6 - 1, if I'm exact). That's supposed to be a hint.

You could do something fancy, I suppose like x sin ((f(x)), for a suitable f, but I doubt it would be very easy to integrate to prove the result.

 

[Mathmos6]

Thanks very much, I'll have a look at that and see if i can ponder something out!

 

[mm08]

Hi, I'm stuck on the same question but haven't managed to figure it out yet with the help you've given. Could you give another clue please?

 

[Marksyb]

Hi, I'm stuck on the same question but haven't managed to figure it out yet with the help you've given. Could you give another clue please?

Well, yyat pretty much summed it up. You can have a continuous function that essentially looks like a set of spikes:
... /\
... /\.../...\
/\/...\/... \

(ignore the dots, they're there to preserve spacing.)

The height of these triangles is unbounded as you let $$x\rightarrow\infty$$. The trick here is to define them in such a way that the total area under them is finite even in the limit. A hint towards that end is to scale the triangles in some convenient way.