Convergence Test for Improper Integral with e^-x^2 Function

[matadorqk]

Homework Statement

Test for convergence:
b. $$\int_0^\infty{e^{-x^2}}$$

Homework Equations

Any method you choose to approach the problem in order to test for convergence.

The Attempt at a Solution

First, I attempted to integrate the problem. However, I am not exactly sure how to approach such integration. I would appreciate some hints as to how to approach this. It would help me in continuing to solve this problem.

 

[w3390]

When doing improper integrals, replace the infinity with a variable, A for example. Then proceed through the integral while taking the limit as A approaches infinity. If any part of the process yields an answer of infinity, the integral is divergent.

 

[matadorqk]

When doing improper integrals, replace the infinity with a variable, A for example. Then proceed through the integral while taking the limit as A approaches infinity. If any part of the process yields an answer of infinity, the integral is divergent.

Yes, I realize what you say, so:

lim (b-> $$\infty$$) $$\int_0^b{e^{-x^2}}$$

I realize that $$e^\infty$$ is infinity, therefore, making it divergent. However, that's wrong because I have to integrate it first. Hence my question, how would I approach integrating $${e^{-x^2}}$$

 

[Dick]

When doing improper integrals, replace the infinity with a variable, A for example. Then proceed through the integral while taking the limit as A approaches infinity. If any part of the process yields an answer of infinity, the integral is divergent.

That's not very helpful because there no elementary antiderivative for e^(-x^2). You want to do a comparison test with something you can integrate. For example, if x>1 which is larger, e^(-x) or e^(-x^2)?

 

[matadorqk]

That's not very helpful because there no elementary antiderivative for e^(-x^2). You want to do a comparison test with something you can integrate. For example, if x>1 which is larger, e^(-x) or e^(-x^2)?

Hmm, a comparison test does seem interesting. So let's try this out.
Let $$f(x)=e^{-x^2}$$
and Let$$g(x)=e^{-x}$$
Where $$0\leq{f(x)}\leq{g(x)}$$

Given the comparison test, if g(x) is convergent, we can assume that f(x) is also convergent.

$$\stackrel{lim}{b\rightarrow\infty}\int_0^b{e^{-x}}$$

$$\stackrel{lim}{b\rightarrow\infty}[-e^{-x}]\stackrel{b}{0}$$

$$\stackrel{lim}{b\rightarrow\infty}[-e^{-b} + 1]$$

$$\stackrel{lim}{b\rightarrow\infty}[-e^{-\infty} + 1]$$

Errr. The answer I am getting is that g(x) is divergent. I can draw no conclusion from this.

Ohh no wait!

$$\stackrel{lim}{b\rightarrow\infty}[1/-e^{\infty} + 1]$$

$$\stackrel{lim}{b\rightarrow\infty}[1/{\infty} + 1]$$

$$\stackrel{lim}{b\rightarrow\infty}[0 + 1]$$

CONVERGENT!

:) Thanks Dick & W3390, for quick responses and help.

 

[Dick]

Err. e^(-infinity) is zero. Isn't it? I.e. limit b->infinity e^(-b)=0. g(x) is NOT divergent. And f(x)<=g(x) only for x>=1. Why don't you have to worry about 0<=x<1? It's good to think about these things.

 

[matadorqk]

Err. e^(-infinity) is zero. Isn't it? I.e. limit b->infinity e^(-b)=0. g(x) is NOT divergent. And f(x)<=g(x) only for x>=1. Why don't you have to worry about 0<=x<1? It's good to think about these things.

Yeah, It is, I figured it out as you can see haha. However, the f(x) <= g(x) is only for x>=1, we don't worry about x=0 -> x<1 because.. hmm, I'm honestly not sure.

 

[Dick]

You're pretty sharp. I'll give you two minutes to figure it out and then I'm off to bed. Hint: you know the integral of f(x) converges in [1,infinity) now. Why it converges over [0,1] is even easier.

 

[matadorqk]

You're pretty sharp. I'll give you two minutes to figure it out and then I'm off to bed. Hint: you know the integral of f(x) converges in [1,infinity) now. Why it converges over [0,1] is even easier.

Ok let's just solve this real quickly.

See if f(x) converges from 0->1.

Using the same, if g(x) converges, we'd be using the same steps, except from 0 to 1 now. This would give you

-1/e + 1. Given that they are all constants (its approaching a value), we know it converges.

I know this is probably not the theoretical answer you are seeking, I just seem to be missing something.

 

[Dick]

Nah, simpler than that. 0<=f(x)<=1 for x in [0,1]. For x>1 compare with g(x)=exp(-x). For x<1 just compare with g(x)=1. Sorry, I rushed you. I know you would have figured it out.

 

[matadorqk]

Nah, simpler than that. 0<=f(x)<=1 for x in [0,1]. For x>1 compare with g(x)=exp(-x). For x<1 just compare with g(x)=1. Sorry, I rushed you. I know you would have figured it out.

Not a problem. Anyways, thanks for the help.