A more straightforward and general way to calculate these kinds of integrals is by changing of variable: Suppose your normal distribution has mean $\mu$ and variance $\sigma^2$: $\mathcal{N(\mu, \sigma^2)}$ $$ E(x) = \frac{1}{\sigma\sqrt{2 \pi}} \int x \exp(-\frac{(x-\mu)^2}{2\sigma^2})dx $$ now by changing the variable $y = \frac{x-\mu}{\sigma}$ and $\frac{dy}{dx}=\frac{1}{\sigma} \rightarrow dx = \sigma dy$.

$$ E(x) = \frac{\sigma}{\sigma\sqrt{2 \pi}} \int (\sigma y + \mu) \exp(-\frac{y^2}{2})dy = \\ \frac{\sigma}{\sigma\sqrt{2 \pi}} \left[ \int \sigma y e^{-\frac{y^2}{2}} dy + \mu \int e^{-\frac{y^2}{2}} dy \right] $$ The first integral is an odd integral since $y$ is an odd function and $e^{-\frac{y^2}{2}}$ is an even function which results in an odd function with symmetric integral boundaries and is zero. The second integral itself has an answer of $2\pi$. $$ E(x) = \frac{\mu \sigma \sqrt{2 \pi}}{\sigma \sqrt{2 \pi}} = \mu $$ In your case since your distribution had a mean of zero by your definition the answer is $E(x) = 0$.

A more straightforward and general way to calculate these kinds of integrals is by changing of variable: Suppose your normal distribution has mean $\mu$ and variance $\sigma^2$: $\mathcal{N(\mu, \sigma^2)}$ $$ E(x) = \frac{1}{\sigma\sqrt{2 \pi}} \int x \exp(-\frac{(x-\mu)^2}{2\sigma^2})dx $$ now by changing the variable $y = \frac{x-\mu}{\sigma}$ and $\frac{dy}{dx}=\frac{1}{\sigma} \rightarrow dx = \sigma dy$.

$$ E(x) = \frac{\sigma}{\sigma\sqrt{2 \pi}} \int (\sigma y + \mu) \exp(-\frac{y^2}{2})dy = \\ \frac{\sigma}{\sigma\sqrt{2 \pi}} \left[ \int \sigma y e^{-\frac{y^2}{2}} dy + \mu \int e^{-\frac{y^2}{2}} dy \right] $$ The first integral is an odd integral since $y$ is an odd function and $e^{-\frac{y^2}{2}}$ is an even function which results in an odd function with symmetric integral boundaries and is zero. The second integral itself has an answer of $2\pi$. $$ E(x) = \frac{\mu \sigma \sqrt{2 \pi}}{\sigma \sqrt{2 \pi}} = \mu $$ In your case since your distribution has a mean of zero by your definition the answer is $E(x) = 0$.

A more straightforward and general way to calculate these kinds of integrals is by changing of variable: Suppose your normal distribution has mean $\mu$ and variance $\sigma^2$: $\mathcal{N(\mu, \sigma^2)}$ $$ E(x) = \frac{1}{\sigma\sqrt{2 \pi}} \int x \exp(-\frac{(x-\mu)^2}{2\sigma^2})dx $$ now by changing the variable $y = \frac{x-\mu}{\sigma}$ and $\frac{dy}{dx}=\frac{1}{\sigma} \rightarrow dx = \sigma dy$.

$$ E(x) = \frac{\sigma}{\sigma\sqrt{2 \pi}} \int (\sigma y + \mu) \exp(-\frac{y^2}{2})dy = \\ \frac{\sigma}{\sigma\sqrt{2 \pi}} \left[ \int \sigma y e^{-\frac{y^2}{2}} dy + \mu \int e^{-\frac{y^2}{2}} dy \right] $$ The first integral is an odd integral since $y$ is an odd function and $e^{-\frac{y^2}{2}}$ is an even function which results in an odd function with symmetric integral boundaries and is zero. The second integral itself has an answer of $2\pi$. $$ E(x) = \frac{\mu \sigma \sqrt{2 \pi}}{\sigma \sqrt{2 \pi}} = \mu $$ In your case since your distribution had a mean of zero by your definition the answer is $E(x) = 0$. You can use the same method to solve the expected value of higher moments which is very easy.

A more straightforward and general way to calculate these kinds of integrals is by changing of variable: Suppose your normal distribution has mean $\mu$ and variance $\sigma^2$: $\mathcal{N(\mu, \sigma^2)}$ $$ E(x) = \frac{1}{\sigma\sqrt{2 \pi}} \int x \exp(-\frac{(x-\mu)^2}{2\sigma^2})dx $$ now by changing the variable $y = \frac{x-\mu}{\sigma}$ and $\frac{dy}{dx}=\frac{1}{\sigma} \rightarrow dx = \sigma dy$.

$$ E(x) = \frac{\sigma}{\sigma\sqrt{2 \pi}} \int (\sigma y + \mu) \exp(-\frac{y^2}{2})dy = \\ \frac{\sigma}{\sigma\sqrt{2 \pi}} \left[ \int \sigma y e^{-\frac{y^2}{2}} dy + \mu \int e^{-\frac{y^2}{2}} dy \right] $$ The first integral is an odd integral since $y$ is an odd function and $e^{-\frac{y^2}{2}}$ is an even function which results in an odd function with symmetric integral boundaries and is zero. The second integral itself has an answer of $2\pi$. $$ E(x) = \frac{\mu \sigma \sqrt{2 \pi}}{\sigma \sqrt{2 \pi}} = \mu $$ In your case since your distribution had a mean of zero by your definition the answer is $E(x) = 0$.