# Homework Help: Improper integral

1. Jan 6, 2005

### Chen

I need to find a function f(x), if one exists, such that:
lim (x->inf) x^2*f(x) = 0
And the improper integral of f(x) from 1 to infinity doesn't exist.

I'm thinking that no function can satisfy these requirements, but apparently I'm wrong... help anyone?

2. Jan 6, 2005

### arildno

Let f(x) on the interval [1,2] be as follows:
0 for x rational, 1 for x irrational.

From [2,..) let f(x)=0 for all x
Since f is not integrable on [1,2], the improper integral doesn't exist either..
(I'm thinking of Riemann integrability here..)
(Perhaps you where looking for an f such that the integral didn't exist through divergence?)

3. Jan 6, 2005

### Chen

Yes, I'm sorry, the improper integral needs to diverge. Also I need to be able to write down the function as a formula, without these "cases".

Thank you!

4. Jan 6, 2005

### dextercioby

Take the continuous function:
$$f(x):(+1,+\infty)\rightarrow R$$ (1)

$$f(x)=\left\{\begin{array}{cc} \frac{1}{x-1}-1,&\mbox{ if } x\in (1,2]\\ 0,&\mbox{ if } x>2 \end{array}\right$$ (2)

It is easily checkable that the function is continuous at every point in its domain and
$$\lim_{x\rightarrow +\infty} x^{2}f(x)=0$$ (3)

Its integral
$$\int_{1}^{+\infty} f(x) dx=\int_{1}^{2} f(x)dx+\int_{2}^{+\infty} f(x)dx=\lim_{a\searrow 1}\int_{a}^{2} f(x) dx=\lim_{a\searrow 1}[\ln(2-1)-\ln(a-1)-1]$$
$$=-1-\lim_{a\searrow 1}\ln(a-1)=+\infty$$ (4)

Daniel.

Last edited: Jan 6, 2005
5. Jan 6, 2005

### Chen

Daniel, like I said above the function must be pronounced with a single formula, without different cases for different domains of X.

6. Jan 6, 2005

### dextercioby

My friend,that function is an good as any function.It is continuous and infinitely times differentiable on its entire domain of definition.It's as good as e^{x}.If u want a "better" looking one,please be my guest and find it.

Daniel.

7. Jan 6, 2005

### Chen

Our homework is submitted via an online software that is uncapable of accepting answers in the form you posted. Believe me it annoys me much more than it does you, but this is the world we live in. So what I'm after is a function that can be described by a single formula.

Chen

8. Jan 6, 2005

### dextercioby

What do you mean,"a software that is uncapable of accepting answers in the form you posted"?? What kind of f***** up software (actually the bonehead who created it) is that?? :grumpy: Can't u just write:
f(x) =\frac{1}{x-1}-1,if 1<x<2,0,if x>=1 ???

It's basically text editing.You could write what i've written even in "Notepad"...

As for the "function which is described by a single formula",i'll say again:"Be my guest and find it".

Daniel.

9. Jan 6, 2005

### arildno

What about an f defined on a punctuated line:
$$f(x)=\frac{1}{(x-a)^{3}}$$
For some "a" greater than 1.

10. Jan 6, 2005

### dextercioby

$$\int_{1}^{+\infty} \frac{dx}{(x-a)^{3}}dx=\frac{1}{2}[\lim_{x\searrow a}\frac{1}{(x-a)^{2}}-\lim_{x\nearrow a}\frac{1}{(x-a)^{2}}]+\frac{1}{2}\frac{1}{(1-a)^{2}}$$

Is the result $+\infty$ ??I'd say "no".

For the function
$$f(x)=\frac{1}{(x-a)^{4}}$$
,for some "a" greater than 1,the same integral yields:
$$\int_{1}^{+\infty} \frac{dx}{(x-a)^{4}}dx=\frac{1}{3}[\lim_{x\searrow a}\frac{1}{(x-a)^{3}}-\lim_{x\nearrow a}\frac{1}{(x-a)^{3}}]+\frac{1}{3}\frac{1}{(1-a)^{3}}$$

Is the result $+\infty$ ??I'd say "yes".

Daniel.

PS.I like my example more.

11. Jan 6, 2005

### arildno

Well, whatever.

The "single formula"-constraint is rather silly, anyway.
Besides, I still think my first function is more than good enough.

12. Jan 6, 2005

### Chen

The software is used not only for submitting the answers but also for checking them. It simply cannot handle expressions such as the one you posted. If you want to change, be my guest.

At any rate, the function you last posted will work.

And for christ's sake, if you got nothing helpful to post - don't post at all.

13. Jan 6, 2005

### learningphysics

Doesn't the function f(x)=1/(x-1)^3 satisfy the requirements of the question?

14. Jan 6, 2005

### dextercioby

I've come up with an answer.If you didn't like it,that's your problem.I use to think that my answers are helpful,but you're free to think otherwise.
You didn't request a specific answer.
I quote:
How would i know that your stupid software won't accept my solution??The way you formulated,u were looking for a solution,not a very particular solution.

Daniel.

15. Jan 6, 2005

### dextercioby

It is very good,i cannot imagine any simpler solution which can be accepted by that stupid software.
Congratulations!!
Don't delete messages,when you're not sure of them.This one was very good,and,had you seen it,you wiuldn't have deleted.

Daniel.

PS.I may write stupid things,but i leave them there to remind me i'm wrong,from times to times... :tongue2:

16. Jan 6, 2005

### arildno

I agree:learningphysics' answer is the best and simplest choice here!

17. Jan 6, 2005

### Chen

You would know because I said so, and I quote:
Now do you really wish to continue this worthless argument? Don't you have some more questions in need of answer, or a kitten stuck on a tree that needs to be saved?

Chen.

18. Jan 6, 2005

### Chen

I would definitely say so. Thanks learningphysics and arildno.

19. Jan 6, 2005

### learningphysics

Thanks. You're right about the posting. Won't delete in the future.

20. Jan 6, 2005

### dextercioby

You do that!!Now,excuse me,i'm a little busy,i gotta go and help some poor kitten stuck on a tree.I love kittens,they're adorable... :!!) I have three at home.I nicknamed my girlfriend 'my little delicious kitten'... :!!)

Daniel.