# Improper Integration

1. Feb 20, 2008

### Quiggy

[SOLVED] Improper Integration

Hi, I was working on this problem in my calculus class today and kept getting the answer of 0, while my teacher was saying that the integral diverges. I just want to know if I'm wrong, he's wrong, or something way over my head is going on here and we're both wrong.

The problem was to find:
$$\int_{-\infty}^{\infty} x^3 dx$$

I broke this into:
$$\lim_{b\rightarrow\infty} \int_{-b}^{b} x^3 dx$$

This in turn worked out to:
$$\lim_{b\rightarrow\infty} [\frac{1}{4} x^4]_{-b}^{b} = \lim_{b\rightarrow\infty} [(\frac{1}{4} b^4) - (\frac{1}{4} b^4)] = 0$$

Meanwhile, my teacher broke the original integral into:
$$\int_{-\infty}^{0} x^3 dx + \int_{0}^{\infty} x^3 dx$$

This yielded him with indeterminate form $$\infty - \infty$$.

So who's right?

2. Feb 20, 2008

### sutupidmath

well you cannot breake the integral as you did. And your prof is right, taht integral diverges, and he does not end up with that intermediate form as you said but rather with infinity+infinity=infinity, which simply tells you that the integ diverges.

3. Feb 20, 2008

### sutupidmath

what you are actually saying is that the area under the curve of the function X^3 from minus in finity to plus infinity is zero, which is not actually true. Remember in those cases you have to switch the sign of the integral so they won't cancel each other out!

4. Feb 20, 2008

### Quiggy

Why can't I do that? $$\lim_{x\rightarrow\infty} -x = -\infty$$, so it's not like there's a problem (that I see, anyways) with representing the lower limit $$-\infty$$ as $$\lim_{x\rightarrow\infty} -x$$.

I guess the other thing that's confusing me is that $$\int_{-1}^{1} x^3 dx = \int_{-10}^{10} x^3 dx = \int_{-100}^{100} x^3 dx = \cdots = 0$$, so why doesn't $$\int_{-\infty}^{\infty} x^3 dx = 0$$?

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Edit: Ninja'd. I never said anything about area. I'm finding integrals, and since, for instance, $$\int_{-2}^{0} x dx = -2$$, I don't understand why that area's even being brought up at all. Negative integrals do in fact exist.

Last edited: Feb 20, 2008
5. Feb 20, 2008

### Gear300

Probably because you're dealing with infinite. Once 0 and $$\infty\$$ enter...things get wierd.

Last edited: Feb 20, 2008
6. Feb 20, 2008

### sutupidmath

First whether you like it or not, definite rieman integrals are all about the areas under a curve. THat is when you evaluate the integral of x^3 from say -10 to 10, you are doing nothing else but finding the area that is enclosed by the curve of that equation on the interval (-10, 10). Now somewhere in your textbook there should be a warning when you integrate a function from -a to a. you should always switch the sign of the integral on that part.In other words you should switch the sign of the function on that region that the function itself becomes negative, that is lies under the x-axis. So the curve of X^3 obviously lies under the x-axis when x<0. that is

integ of x^3 from -10, to 10= -integ of x^3 form -10 to 0 + integ of x^3 from 0 to 10. Other wise you will end up gettin zero answers all along.

Last edited: Feb 20, 2008
7. Feb 20, 2008

### sutupidmath

Who said that you cannot integrate a function from a negative lower bound to a positive upper one. or sth??????????????

Edit: YOu are right for one thing though, that whenever we integrate odd functions between -infinity to +infinity the answer is always zero. However if we are interested to evaluate the area under that curve on the interval(-infinity, +infinity) than we have to do like i did. It depends on what your interest is.
NOw going back to your question, both of you are right. Depends what we want!

Last edited: Feb 20, 2008
8. Feb 21, 2008

### Quiggy

I'm not finding area at all, I'm performing the integration $$\int_{-\infty}^{\infty} x^3 dx$$. I agree that I would have to split it up if I was finding area and would get $$\infty$$, but that's not what I'm doing. Why then did my teacher and I get different answers?

9. Feb 21, 2008

### <---

I think I see the problem here.
This is what you are trying to do.$$\int_a^bx^3dx$$
Now solve like you would always and you get,
$$\frac{1}{4}(\underbrace{lim}_{b ->\infty}b^4 - \underbrace{lim}_{a -> -\infty}a^4 )$$

See now here you are subtracting a negative, so adding leading to a final answer of $$+\infity$$ :)

*I know the lim formatting is weird, don't have time to figure it out right now.

Quick edit: Shi.. I see tyhat's wrong as we are dealing with a number raised to a even power... Sorry than I'm not sure what's wrong :(

Last edited: Feb 21, 2008
10. Feb 21, 2008

### VietDao29

$$\int_{-\infty} ^ {+\infty} x ^ 3 dx$$ does not equal to: $$\lim_{b \rightarrow \infty}\int_{-b} ^ {+b} x ^ 3 dx$$.

There's nothing guarantees that the upper bound, and lower bound of the integral both tend to infinity at the same rate..

What if I say that: $$\int_{-\infty} ^ {+\infty} x ^ 3 dx = \lim_{b \rightarrow + \infty}\int ^ {+b}_{-b \color{red}{- 1}} x ^ 3 dx$$ This integral is clearly divergent. Right?

So, in conclusion, it's better written as:

$$\int_{-\infty} ^ {+\infty} x ^ 3 dx = \lim_{\substack{a \rightarrow + \infty \\ b \rightarrow + \infty}} \int_{-a} ^ {b} x ^ 3 dx$$

Where a, and b are independent of each other.

This is why your integral diverges, instead of converging to 0, like you said. :)

One can prove that: Given a < b < c (a, and c can be infinity):

$$\int_{a} ^ {c} f(x) dx$$ converges if, and only if $$\int_{a} ^ {b} f(x) dx$$ and $$\int_{b} ^ {c} f(x) dx$$ both converge.

Last edited: Feb 21, 2008
11. Feb 21, 2008

### Quiggy

Huh. That's very strange, as I'd assume that $$\lim_{x\rightarrow\infty} \frac{1}{4} x^4$$ and $$\lim_{x\rightarrow-\infty} \frac{1}{4} x^4$$ would both approach $$\infty$$ at the same rate. I don't understand why, but I'll take your word for it. Thanks!

12. Feb 21, 2008

### sutupidmath

Well these go to $$\infty$$ at the same rate just because you assumed that the original bounds of the integral go to infinity at the same rate!!!
However Vietdao29's point was that nothing guarantees us that the original bounds of the integral go to infinity at the same rate, that's why you cannot let both the upper and the lowe bound be b, and -b respectively as b-->infinity! And he perfectly well illustrated it by taking the lower bound to be -b-1, so this obviously approaches negative infinity a little bit faster than b aproaches positive infinity.

Last edited: Feb 21, 2008
13. Feb 21, 2008

### <---

Sorry about my last post, shouldn't have posted not having enough time to really look at the problem.
I overlooked that divergence/convergence was being sought out.
As I understand it you always separate the positive from the negative.
The op's not wrong if you if you are looking for the value of the integral.
But that's not what he/she is looking for.

So to the op, the rates don't really matter because you are not looking for the value. You only need to know if on approaching + OR - $$\infty$$ (not at the same time) the the integral approaches a definite value.

Again convergence/divergence isn't necessarily if the value of the integral approaches a number, it's if the value in a specific direction(+/-) does.

Hope that's helpful, although I don't think I am explaining it that well.

14. Feb 21, 2008

### sutupidmath

I agree with this.

15. Feb 21, 2008

### Quiggy

Actually, that's a pretty good explanation. Neither my textbook nor my teacher did a very good job of explaining exactly what divergence is, so I suppose that that's the reason why it didn't work out. Thanks again.

16. Feb 22, 2008

### VietDao29

No, the OP is clearly wrong!!! And, divergent integrals do not return any specific values!!!

Err, what do you mean? The rate does matter..

Now, say, does the integral: $$\int_{-1} ^ {1} \frac{dx}{x}$$ converge, you think?

Huh?!?! Now that I am completely lost..

What does the 'specific direction' have anything to do here?

Are you sure you are not confusing the OP?..

Last edited: Feb 22, 2008
17. Feb 22, 2008

### Gib Z

It would probably be helpful for some people here to google "Cauchy Principal Value".

18. Feb 22, 2008

### <---

I told you I didn't think I was explaining very well. :P
And If I'm really just way off and have no Idea what I am talking about, than I apoligize...

And upon going over divergence/convergence again (haven't dealt with it myself for a bit.),
Yeah, you're right. Having the matter confused myself, I probably was just confusing the op.
I apologize.

What I think I was trying to explain (while mixing it with something completely unrelated myself) is the principal of evaluating for con/divergence by splitting the integral into two where you evaluate between 0 (or any other definite number) and +/- infinity separately.
Which is just what his professor did.

So in the end, yes I was probably looking at it in the same incorrect way as the op.
I myself had the very definition of con/divergence confused, and should have checked before replying.
So again, sorry, but keep in mind I am also here to learn.

To the op. He's right, ignore what I said before.

Edit: Thanks for your suggestion Gib Z, found this particularly relevant Link.

Last edited: Feb 22, 2008
19. Feb 22, 2008

### Quiggy

Y'know, somehow it's reassuring that this is gaining a good amount of disagreement, since it means I'm not missing something obvious :)

The whole bit with the integral split up differently and the link with the CPV discussion all help out quite a lot. I think I understand it enough now to know why the answer's not 0. Thanks once again.