# Expectation value

## 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?

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

not

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

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?

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rock.freak667
Homework Helper
No you can't

HallsofIvy
Homework Helper
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
Homework Helper
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
Homework Helper
because $$\frac{d}{dx}(\frac{e^{-x^2}}{-2x}) = e^{-x^2}$$
Why would you think that ?

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
Homework Helper