Integrating Exponential Function: Finding Error

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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 :(
 
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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?
 
quasar987 said:
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)
 
What is it supposed to give?
 
quasar987 said:
What is it supposed to give?
it gives me sqrt (2 pi)
after differentiating
 
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}
 
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