Question arising from the proof of the existence of $\mathbb E[X]$ when $X$ ~ $\mathcal{N}(\mu, \sigma ^2)$

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$\mathbb E[X]$ when $X$ ~ $\mathcal{N}(\mu, \sigma ^2)$

I know that $\mathbb E[X]$ exists iff $\mathbb E[|X|]<\infty$

$\mathbb E[|X|]=\int_{\mathbb R}|x|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx$

My Professor then goes into the next step saying: $\int_{\mathbb R}|x|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx\leq\int_{\mathbb R}|x-\mu|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx+\mu$

and this is where I get confused.

I recognize that he is trying to get to a substitution $y=x-\mu$, and $\mathbb E[|X|]=\mathbb E[|X-\mu+\mu|]\leq\mathbb E[|X-\mu|]+\mu$

But why is $\mathbb E[|X-\mu|]=\int_{\mathbb R}|x-\mu|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx$ and not equal $\int_{\mathbb R}|x-\mu|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-2\mu)}{2\sigma^2}}dx$. I mean surely the density function $\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-2\mu)}{2\sigma^2}}$ shifts to the left if $\mu$ is subtracted.

Additional Question out of interest: If I can prove $\mathbb E[X] \in \mathbb R$, can I then automatically assume $\mathbb E[X]$ exists? In other words, why is $\mathbb E[|X|]<\infty$ neccessary rather than $\mathbb E[X]$? I realize that it has something to do with the Lebesgue Integral that only looks at positive functions, but I am not sure.

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What your professor did is simply applying the triangle inequality (assuming $\mu\geq 0$) $$ |x|\leq|x-\mu|+\mu, $$ which implies that \begin{align} \int_{\mathbb R}|x|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx &\leq \int_{\mathbb R}|x-\mu|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx +\int_{\mathbb R}\mu\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx\\ &= \int_{\mathbb R}|x-\mu|\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(x-\mu)}{2\sigma^2}}dx +\mu. \end{align}

For the additional question: lots of authors define $EX$ only for absolutely integrable random variables (i.e., $E|X|<\infty$) and leave $EX$ undefined when $E|X|=\infty$. Whenever $EX$ is defined, it must be a real number (in the context of real random variables).