# Integrating normal from 0 to Inf and finding the third moment

1. Apr 10, 2013

### Hejdun

This is (perhaps) a tricky question regarding the moment of normal distribution.
Let f(x) be the pdf of normal distribution with mean (-σ^2/b) and variance σ^2, where b is just a constant. The goal is to solve the integral

∫ x^3 f(x) dx

integrating from 0 to ∞.

I am stuck. Any suggestions?

2. Apr 10, 2013

### Stephen Tashi

Use integration by parts to express the problem as one in lower powers of X. The expressions involving the lower powers will have something to do with lower order moments.

If it is a normal distribution, aren't you integrating from $-\infty$ to $\infty$?

3. Apr 10, 2013

### Hejdun

I will try integration by parts, but I am not sure if that would solve it.

No, unfortunately it is from 0 to ∞. Otherwise it would have been a standard result.

4. Apr 10, 2013

### Bacle2

Since the exponent will contain multiples of x^2 and x, h'about trying a substitution like:

x^2=u , then x^3=u^(3/2) , etc.

5. Apr 13, 2013

### Hejdun

It seems that there is no analyc answer to this integral since integration will result in an error function. But thanks anyway!

6. Apr 22, 2013

### Herick

But the Error function is analytic! It is an entire function.

7. Apr 22, 2013

### Bacle2

I think by analytic Hedjun means that the function has an antiderivative in a

"nice closed form" , not in the complex-analytic sense.

Notice that in this case f is assumed real-valued.

Last edited: Apr 22, 2013
8. Apr 22, 2013

### Herick

Well, is it $e^x$ a closed form? If by closed form he means a convergent infinite series, then Err(x) is also 'closed form'. Right?

9. Apr 25, 2013

### Bacle2

I assume s/he , means a 'nice' antiderivative ( a combination of sums, products of

known/common functions ) F that would allow you to calculate the area between any two

x,y , without having to use numerical methods , or tables. AFAIK there isn't any one.

Do you know of one? I assumed, since tables are used for the normal density, that

there isn't any "nice-enough" antiderivative. But maybe the OP can tell us what s/he

was looking for.

10. Apr 28, 2013

### Hejdun

Yes, it is this that I was looking for, which I believe does not exist for the integral that started this thread. Thanks for your replies.