# Expectation value

1. Sep 23, 2007

### cscott

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

Can somebody help me integrate $$\int{x\cdot p(x)}$$ where $p(x)$ is the Gaussian distribution (from here http://hyperphysics.phy-astr.gsu.edu/hbase/math/gaufcn.html)

3. The attempt at a solution

I can't really get anywhere. It's true that $\int{e^{x^2}}$ has no analytical solution, right?

Last edited: Sep 23, 2007
2. Sep 23, 2007

### HallsofIvy

Staff Emeritus
Yes, it is true that you cannot find the indefinite integral
$$\int e^{-x^2}dx$$
in terms of elementary functions (though you can find the definite integral for some choices of upper and lower bound).

However, there is a very simple substitution that will give you
$$\int x e^{-x^2}dx$$

3. Sep 23, 2007

### rock.freak667

Is $$\int e^{-x^2} dx$$

not

$$\frac{e^{-x^2}}{-2x}$$ + K ?

4. Sep 23, 2007

### cscott

But I can't use that easy substitution for $$\int{x \cdot e^{-(x-x_0)^2} dx$$ for some constant $x_0$, can I?

Last edited: Sep 23, 2007
5. Sep 23, 2007

### rock.freak667

No you can't

6. Sep 24, 2007

### HallsofIvy

Staff Emeritus
No, it's not. Why in the world would you think it was?

7. Sep 24, 2007

### rock.freak667

because $$\frac{d}{dx}(\frac{e^{-x^2}}{-2x}) = e^{-x^2}$$

8. Sep 24, 2007

### dextercioby

Why would you think that ?

9. Sep 24, 2007

### christianjb

Oh, don't be so cutting! I can see why the poster would think that and so can you. It's clear that the poster forgot about the derivative of the denominator (in case you hadn't worked that out).

10. Sep 24, 2007