# Solve Improper Integration: Who's Right?

• Quiggy
In summary: You want to know if there's a way to integrate a function from a negative lower bound to a positive upper one. In summary, the problem was to find:- an improper integration problem in calculus class- when dividing an integral by a negative number, the answer is always +infinity- when integrating a function between -infinity to +infinity, the answer is always zero.
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?

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.

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!

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 $$\int_{-\infty}^{\infty} x^3 dx = 0$$?

-----

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

Last edited:
Quiggy said:
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 $$\int_{-\infty}^{\infty} x^3 dx = 0$$?

-----

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.

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:
Quiggy said:
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.
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:
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?

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:
$$\int_{-\infty} ^ {+\infty} x ^ 3 dx$$ 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:
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!

Quiggy said:
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!

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:
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.

<--- said:
Hope that's helpful, although I don't think I am explaining it that well.
I agree with this.

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.

<--- said:
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.

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

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.

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

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

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.

Huh?? Now that I am completely lost..

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

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

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

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

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:
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.

## 1. What is improper integration?

Improper integration refers to the process of evaluating an integral when one or both of the limits of integration are infinite or when the integrand is unbounded at one or more points within the interval of integration.

## 2. How do I know if an integral is improper?

An integral is improper if one or both of the limits of integration are infinite or if the integrand is unbounded at one or more points within the interval of integration. Additionally, if the integrand contains a singularity (such as a vertical asymptote), the integral is also considered improper.

## 3. Why is it important to know how to solve improper integrals?

Improper integrals are important in many areas of mathematics and science, including physics, engineering, and statistics. They allow us to find the exact value of an integral in cases where traditional integration methods may fail. Additionally, improper integrals often arise in real-world applications, so being able to solve them is crucial for solving practical problems.

## 4. What are the two types of improper integrals?

The two types of improper integrals are Type 1 and Type 2. Type 1 improper integrals have an infinite limit of integration, while Type 2 improper integrals have an unbounded integrand within the interval of integration. Both types require different techniques for solving.

## 5. How do I solve improper integrals?

The approach to solving improper integrals depends on the type of improper integral. For Type 1 improper integrals, you can use the limit comparison test, the comparison test, or the Cauchy criterion. For Type 2 improper integrals, you can use the limit comparison test, the comparison test, or the Cauchy criterion, along with splitting the integral into smaller, more manageable integrals. In some cases, you may need to use integration by parts or other integration techniques. It is important to carefully analyze the integrand and limits of integration to determine the best approach for solving an improper integral.

• Calculus and Beyond Homework Help
Replies
8
Views
853
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
13
Views
665
• Calculus and Beyond Homework Help
Replies
17
Views
877
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
47
Views
3K