F is integrable but f^2 is not integrable

[stevekho]

Give an example of a function f : R -> R, such that f is integrable but f^2 is not
integrable. Prove your result.


Seriously I don't know where to start, please help me guys

 

[Office_Shredder]

The two basic ways that a function can fail to be integrable is either it's too big as x goes to infinity, or there's a singularity around which f can't be integrated.

Which of these is going to be the case for f² here? (this isn't rigorous but will give you an intuitive idea of what kind of functions to look at)

 

[stevekho]

Well in that case, we are in R so f^2 will be positive so could be too big as x goes to inf. :/

 

[Whitishcube]

Really, I think either option Office_Shredder gave will work in this case, but the singularity option might be a tad easier to think about.

 

[D H]

Well in that case, we are in R so f^2 will be positive so could be too big as x goes to inf. :/

What are the conditions regarding the behavior of some function as x goes to plus or minus infinity?

Hint: Suppose a is a real number. Then a²>|a| if |a|>1 but a²<|a| if |a|<1.

 

[stevekho]

"Suppose a is a real number. Then a2>|a| if |a|>1 but a2<|a| if |a|<1."
...

sin(1/x)/x is integrable in R I saw
(sin(1/x)/x)^2 is not integrable in R

Could someone confirm me?

 

[D H]

You picked one ugly f(x), but yep, it works. This is your homework, so you are the one who needs to show that my "yep, it works" is correct.

Do you need a hint?

 

[stevekho]

(If you have a nicer f(x) could I also have a hint?)

 

[stevekho]

Also I know how to prove it but could you give me a better f(x)

 

[D H]

It's easy to show that (sin(1/x)/x)^2 is not integrable over R because its antiderivative has a simple representation that blows up at a certain point. Hint:

$$(\sin a)^2 = \frac{1-\cos 2a}2$$It is showing that sin(1/x)/x is integrable over R that is a bit tougher because this function is not integrable in the elementary functions. It's integral is closely related to a well-known special function. Hint:

$$\text{Si}(x) \equiv \int_0^x \frac{\sin t}t\,dt$$

Si(x), the sine integral, is not quite what you want, but it is very close.

 

[pic_beginner]

It's easy to show that (sin(1/x)/x)^2 is not integrable over R because its antiderivative has a simple representation that blows up at a certain point. Hint:

$$(\sin a)^2 = \frac{1-\cos 2a}2$$


It is showing that sin(1/x)/x is integrable over R that is a bit tougher because this function is not integrable in the elementary functions. It's integral is closely related to a well-known special function. Hint:

$$\text{Si}(x) \equiv \int_0^x \frac{\sin t}t\,dt$$

Si(x), the sine integral, is not quite what you want, but it is very close.

Hi,
I have the same qn as my tutorial qn. So far I was able to follow part of the discussion. Could you give a better idea on a simpler f(x) than the one mentioned?

 

[D H]

A function that is an appropriate power of x will do quite nicely.

 

[Dickfore]

Give an example of a function f : R -> R, such that f is integrable but f^2 is not
integrable. Prove your result.


Seriously I don't know where to start, please help me guys

what is the range of integration?

 

[pic_beginner]

what is the range of integration?

I believe the range doesn't matter, so long as f is integrable over the range while f^2 is not

 

[Dickfore]

I believe the range doesn't matter, so long as f is integrable over the range while f^2 is not

Well, the particular example does. Some functions are integrable on $[0, \infty)$, but are not on $(-\infty,\infty)$. Since you asked for a particular example, I would think it matters. Although, now I see that you have the provision $f: \mathbb{R} \rightarrow \mathbb{R}$, which would mean it is defined on the whole real axis.

 

[pic_beginner]

what if we define the range to be [0,1]? still need a clearer hint on a simple f(x).

 

[Dickfore]

what if we define the range to be [0,1]? still need a clearer hint on a simple f(x).

$$f(x) = \frac{1}{\sqrt{x}}$$

is integrable on $[0.1]$, but $f^{2}(x) = 1/x$ is not.