# Need Help with Improper Integrals

1. Apr 21, 2007

### Asclepius

1. The problem statement, all variables and given/known data

This is not a homework problem per se, I'm just reviewing old material for my upcoming final examination in my calculus course and find improper integrals are a weak point of mine. I've picked a fairly simple example that I'm having problems grasping.

It is the integral from -infinity to 0 of $$\int(xe^-x^2) dx$$

(I'm sorry I don't really know how to use Latex properly.)

3. The attempt at a solution

Well, I undertand that after we evaluate the integral, we get the limit as t--> -infinity [-1/2 + 1/2e^(-t^2)]

Now, I already know the answer will come out to -1/2, therefore making the second limit convergent and equal to 0. Why is this? I think I'm getting confused trying to square -infinity, then take the negative of THAT, then raise e to that number. I'm very confused here, is there an easy way to look at these problems? I try to look at the graph in my head, but that always ends up giving me the wrong answer.

Any tips for how to attack these problems would be very much appreciated.

Last edited: Apr 21, 2007
2. Apr 21, 2007

### trajan22

Remeber what happens when you have a number raised to a negative power ex. x^(-2)=1/(x^2) so the limit as x approaches infinity ultimately brings this term to zero. This is basically what happens to the -t^2 in your problem. its like [-1/2+1/2(1/(e^t^2))]

3. Apr 21, 2007

### Asclepius

Right, that's a good point. So if it was -1/2*e^(t^2) as t---> -inf, this would be divergent, correct? Because we square -inf to get +inf, and therefore have e^inf, right? And I don't believe the graph of e^x ever hits an asymptote.

By the way, thanks trajan; sometimes its such a simple point one overlooks. I smacked my forehead when I read your post!

4. Apr 21, 2007

### trajan22

Well nearly correct you would have the limit approach -infinity because e^x is multiplied by the -1/2. But just a word of warning when speaking math one shouldnt refer to infinity as a number, it is more of an expression meaning that the term gets larger all the time. So on work try not to put things like 1+(infinity) or e^(infinity) the professor will most likely mark off points for things like that.

5. Apr 21, 2007

### Gib Z

You want to solve the definite integral
$$\int^0_{-\infty} xe^{-x^2} dx$$.

To do so, rewrite as such : $$\lim_{a\to-\infty} \int^0_a xe^{-x^2} dx$$.

Now let $u=-x^2$ and evaluate the indefinite integral. You have to work it out for yourself, but I'm going to call it F(x) for now.

Once You have the anti derivative, The Solution to your integral is just:
$$F(0) - \lim_{a\to-\infty} F(a)$$

6. Apr 21, 2007

### HallsofIvy

Staff Emeritus
Use the substution u=- x2. You are very fortunate to have that "x" multiplying $e^{x^2}$!

The proble with your LaTex is that you need { }.
e^{-x^2} will give
$$e^{-x^2}$$

7. Apr 21, 2007

### Asclepius

Thanks very much all.

Trajan, you were mentioning I should be wary of treating infinity as an actual number, and I've actually wondered about this. What properties does infinity have in common with "real," numbers; for example +inf * -1 is -inf, right? I know it's a stupid question, but I want to make sure I understand all the properties.

8. Apr 21, 2007

### Gib Z

There are certain things you have to be careful about.

1^a = 1 for any real a.

But if you had a limit question, 1^infinity is not always 1.

For no real number b can 1/b =0. But if b is infinity it is.

Infinity + negative infinity is not always zero.

infinity/infinity is not always 1.

9. Apr 22, 2007

### trajan22

The best explanation that I could find on the way to treat infinity is here
http://www.math.vanderbilt.edu/~schectex/commerrs/
just control f and type in infinity, it is in the other common calculus errors section. If you have studied limits with l`hospitals rules then you should understand how to treat the forms of (inf)/(inf) (notice that this is called a form and not an actual arithmetic operation) and also the form of (inf)^(inf) as well as a few others.