Need Help with Improper Integrals

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In summary, the conversation revolves around solving a definite integral from -infinity to 0 of xe^-x^2. The solution is found by rewriting the integral and using a substitution, and caution is given when treating infinity as a number. The conversation also touches on the properties of infinity and how to properly approach problems involving infinity in calculus.
  • #1
Asclepius
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Homework Statement



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 [tex]\int(xe^-x^2) dx[/tex]

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

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.
 
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  • #2
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
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
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 shouldn't 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
You want to solve the definite integral
[tex]\int^0_{-\infty} xe^{-x^2} dx[/tex].

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

Now let [itex]u=-x^2[/itex] 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:
[tex]F(0) - \lim_{a\to-\infty} F(a)[/tex]
 
  • #6
Use the substution u=- x2. You are very fortunate to have that "x" multiplying [itex]e^{x^2}[/itex]!

The proble with your LaTex is that you need { }.
e^{-x^2} will give
[tex]e^{-x^2}[/tex]
 
  • #7
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
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.

Your example is correct.

Infinity + negative infinity is not always zero.

infinity/infinity is not always 1.
 
  • #9
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.
 

1. What are improper integrals?

Improper integrals are integrals that cannot be evaluated using the standard methods because either the upper limit, lower limit, or both are infinite or the integrand function is not defined at certain points within the limits.

2. How do you evaluate improper integrals?

To evaluate improper integrals, you first need to determine if the integral is convergent or divergent. If it is convergent, you can use the limit definition of integration to evaluate it. If it is divergent, you can use comparison tests or integration by parts to evaluate it.

3. Can improper integrals have both infinite limits?

Yes, improper integrals can have both infinite limits. An example of this is the integral of 1/x from 1 to infinity, where both limits are infinite.

4. What is the difference between a type 1 and type 2 improper integral?

Type 1 improper integrals have infinite limits, while type 2 improper integrals have discontinuous integrands. Type 1 improper integrals can be evaluated using the limit definition, while type 2 improper integrals require other methods such as integration by parts or comparison tests.

5. Why are improper integrals important?

Improper integrals are important because they allow for the evaluation of integrals that would otherwise be impossible to solve using standard methods. They also have applications in various fields of science and engineering, such as calculating areas under curves and determining the convergence or divergence of infinite series.

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