Computing expectation value of x for a Gaussian distribution

[cscott]

Homework Statement

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)

The Attempt at a Solution

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

 

[HallsofIvy]

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

 

[rock.freak667]

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

not

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

 

[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?

 

[rock.freak667]

No you can't

 

[HallsofIvy]

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

not

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

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

 

[rock.freak667]

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

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

 

[dextercioby]

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

Why would you think that ?

 

[christianjb]

Why would you think that ?

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).

 

[rock.freak667]

ah oh my...stupid me...sorry...my bad