If $X$ is a normal random variable, and $Z$ is the standard normal variable, so that $X =\mu+\sigma Z$, doesn't this mean that:
$$\int_{-\infty}^{\infty}x\frac1{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\ dx=E[\mu+\sigma Z]=\int_{-\infty}^{\infty}(\mu+\sigma x)\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}\ dx.$$
Put into word, if we consider the normal variable as a function of the standard normal variable and then use the "Law of the unconscious statistician", we get a strange result. Or am I missunderstanding something?
Really confused about this one, would be great with an answer!
What is strange about it? $X$ is a normal random variable with mean $\mu$ and variance $\sigma^2$, as shown by using the LoUS and a change of variables substitution.
$\begin{split} \mathsf E(X)&=\mathsf E(\mu+\sigma Z) \\&= \int_\Bbb R (\mu+\sigma z)\phi(z)\mathsf d z &: \phi(z)=\frac 1{\surd 2\pi}e^{-z^2/2}\\ &=\int_\Bbb R x\phi(\dfrac{x-\mu}{\sigma})\cdot\frac 1\sigma\mathsf d x &: z=\dfrac{x-\mu}{\sigma} \\ &= \int_\Bbb R \dfrac{x}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma ^2}\mathsf d x\end{split}$