Integrating Exponential Function: Finding Error

[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 :(

 

[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?

 

[stunner5000pt]

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?

i got
sigma times sqrt(2 pi)

 

[quasar987]

What is it supposed to give?

 

[stunner5000pt]

What is it supposed to give?

it gives me sqrt (2 pi)
after differentiating

 

[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

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