# Integral Problem

1. Jun 27, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
Hey, just stuck on an integral. I actually gave up after a while and typed it in wolfram and got something that is beyond me. Just looking for some clarity.

2. Relevant equations

$I = \int e^{-x^2} \, dx$

3. The attempt at a solution
The only way I know to tackle e is to make a substitution for the variable being raised to a power. I tried to make u = x^2 but that doesn't go anywhere. The problem clearly lies in the fact that the exponent on e is being raised to a power it's self. I don't know how to deal with this. If you pop that sucker in on wolfram you get something I have never seen ....called erf(x) ...not a clue. So I'm looking for an explanation of this. Please and thanks.

2. Jun 27, 2013

### Ray Vickson

The function $e^{-x^2}$ does not have an "elementary" anti-derivative; that means that it is impossible to write a *finite* formula for $I$ in terms of things like sums, products, reciprocals, powers, roots, exponentials, logarithms, trig functions, etc. Even if you allow formulas 10 million pages in length, you still cannot do it! BTW: it is NOT just that nobody has been smart enough to figure out how to write such a formula; rather, it has been proven to be impossible.

That is why functions like $\text{erf}(x)$have been invented; they allow us to express such integrals in a compact form, and numerical computations using them are not much more complicated than with $\log(x)$ or $\cos(x)$, etc.

3. Jun 27, 2013

### Jbreezy

Whoa. Cool. So how do you do integrals when you see something like I have written above. Can you elaborate on how to deal with them? Thanks amig

4. Jun 27, 2013

### SteamKing

Staff Emeritus
That's the point. These integrals are evaluated numerically or they are left in their original form. You can't do anything else with them.

5. Jun 27, 2013

### Curious3141

Just to add to Ray's post, $\text{erf}$ is called the "error function".

You need to invoke $\text{erf}(x)$ when doing the indefinite integral. However, if you wanted to evaluate the definite integral between certain special bounds, say between 0 and ∞, there is a way to do so exactly without needing $\text{erf}$, but the mathematics is slightly involved.

6. Jun 27, 2013

### Jbreezy

Well the question was actually asking me to state if the limit converged or not between - infinity and infinity.
So how does that work? If you don't mind or maybe someone else knows.
Thank you.

7. Jun 27, 2013

### LCKurtz

The integrand is an even function and $\int_{1}^{\infty}e^{-x^2}\, dx \le \int_{1}^{\infty}e^{-x}\, dx$

8. Jun 28, 2013

### Curious3141

The method I alluded to allows you to actually calculate that integral. You don't need to do that just to prove that the integral is finite. LCKurtz's observation works fine.

9. Jun 28, 2013

### Jbreezy

Lckrutz, What is it about it being even? I graphed them, I see but when you first looked what did you think?
P.S. I have a vector question. In another thread in a moment.

10. Jun 28, 2013

### LCKurtz

For an even function $f(x)$, the graph is symmetric about the y axis. So if the part where $x>0$ converges, so does the part where $x<0$, and$$\int_{-\infty}^{\infty} f(x)\, dx = 2\int_0^\infty f(x)\, dx$$

11. Jun 28, 2013

OK thanks