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