# Homework Help: More integration

1. Oct 19, 2006

### stunner5000pt

More integration :)

$$\frac{1}{4 \pi \sigma^2} \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}}$$

we know that
$$\int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2}} = \sqrt{2 \pi \sigma^2}$$

and then differentiate both sides wrt sigma
$$\int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}} = \sigma^3 \sqrt{2 \pi}$$

sib the third into the first

$$\frac{1}{4 \pi \sigma^2} \sigma^3 \sqrt{2 \pi}$$

$$\frac{\sigma \sqrt{2 \pi}}{4 \pi}$$

something is wrong .. where did i go wrong ... pelase help :(

Last edited: Oct 19, 2006
2. Oct 19, 2006

### quasar987

You forgot a "-" sign...

$$\int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2}} = \sqrt{2 \pi \sigma^2}$$

Specifically, would you mind writing what you get after differentiating the integral that equals $\sqrt{2 \pi \sigma^2}$ wrt sigma?

3. Oct 19, 2006

### stunner5000pt

i got
sigma times sqrt(2 pi)

4. Oct 19, 2006

### quasar987

What is it supposed to give?

5. Oct 19, 2006

### stunner5000pt

it gives me sqrt (2 pi)
after differentiating

6. Oct 19, 2006

### quasar987

One things's for sure;

$$\int_{-\infty}^{\infty} \frac{\partial}{\partial \sigma}e^{-\frac{x^2}{2\sigma^2}}dx = = \frac{\partial}{\partial \sigma}\sqrt{2\pi}\sigma = 2 \pi$$

If I differentiate the exponential, I get

$$\frac{-x^2}{2}\frac{-2}{\sigma ^3} = \frac{x^2}{\sigma^3}$$

So

$$\int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}}dx = 2\pi \sigma^3$$

And

$$\frac{1}{4 \pi \sigma^2} \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2\sigma^2}} = \frac{\sigma}{2}$$

Last edited: Oct 19, 2006