# $\mu_{2r+2}=\sigma^3\frac{d\mu_{2r}}{d\sigma}+\sigma^2\mu_{2r}$

How do I go about proving this?

If $\mu_x$ be the $x'th$ order central moment of the distribution, then for a normal distribution with mean $0$ and standard deviation $\sigma$,$$\mu_{2r+2}=\sigma^3\frac{d\mu_{2r}}{d\sigma}+\sigma^2\mu_{2r}$$

I have made little progress on this. The generalized normal distribution is $$f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-t)^2}{2\sigma^2}}$$ where $t$ is the mean and $\sigma$ is the standard deviation.Here, $t=0$ and so $f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}$. $\mu_{2r+2}=\frac{1}{\sigma\sqrt{2\pi}}\int^{\infty}_{-\infty}x^{2r+2}e^{-\frac{x^2}{2\sigma^2}}dx$ and $$\mu_{2r+2}-\sigma^2\mu_{2r}=\frac{1}{\sigma\sqrt{2\pi}}\int^{\infty}_{-\infty}x^{2r}e^{\frac{-x^2}{2\sigma^2}}(x^2-\sigma^2)dx$$ After that I have no idea how to complete the problem. Can anyone point ourt an error if there is one or tell me how I can complete the proof?

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Hint $$\frac{d\mu_{2r}}{d\sigma} = \frac{d}{d\sigma} \left(\frac{1}{\sqrt{2\pi} \sigma} \right) \cdot \int_{-\infty}^{\infty} x^{2r} e^{-\frac{x^2}{2\sigma^2}} \, dx + \frac{1}{\sigma \sqrt{2\pi}} \cdot \int_{-\infty}^{\infty} \frac{d}{d\sigma} \left( x^{2r} e^{-\frac{x^2}{2\sigma^2}} \right) \, dx \\ = \frac{1}{\sqrt{2\pi}} \cdot \int_{-\infty}^{\infty} x^{2r} e^{-\frac{x^2}{2\sigma^2}} \, dx + \frac{1}{\sigma \sqrt{2\pi}} \cdot \int_{-\infty}^{\infty} x^{2r} \cdot e^{-\frac{x^2}{2\sigma^2}} \cdot \frac{x^2}{\sigma^3} \, dx$$