\[\begin{align*} E(X) &= \int_{-\infty}^\infty x \cdot \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(\frac{(x-\mu)^2}{2\sigma^2}\right) dx\\ &=\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \frac{x}{\sigma} \exp\left(\frac{(x-\mu)^2}{2\sigma^2}\right) dx\\ \end{align*}\]
then, substitute \(t = \frac{x-\mu}{\sigma}\),
\(\Longrightarrow: x= t\sigma + \mu\), and
\(\Longrightarrow: \frac{dt}{dx} = \frac{1}{\sigma}\)
\(\therefore dx = \sigma dt\)
\[\begin{align*} E(X) &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \frac{t\sigma + \mu}{\sigma} \exp\left(-\frac{t^2}{2}\right)\sigma dt\\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \frac{t\sigma + \mu}{\sigma} \exp\left(-\frac{t^2}{2}\right)\sigma dt\\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} (t\sigma + \mu) \exp\left(-\frac{t^2}{2}\right) dt\\ &= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} t\sigma \exp\left(-\frac{t^2}{2}\right) dt + \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \mu \exp\left(-\frac{t^2}{2}\right) dt\\ &= \sigma\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} t \exp\left(-\frac{t^2}{2}\right) dt + \mu\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\left(-\frac{t^2}{2}\right) dt \end{align*}\]
Evaluating these two integrals we see that the second integral is a standard normal which integrates to 1:
\[\begin{align*} &= \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\left(-\frac{t^2}{2}\right) dt \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left(-\frac{t^2}{2}\right) dt \\ &= \frac{1}{\sqrt{2\pi}} \sqrt{2\pi} \\ &= 1 \end{align*}\]
while the first integral goes to 0 (integration by parts):
\[\begin{align*} \sigma\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} t \exp\left(-\frac{t^2}{2}\right) dt &= \frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} t \exp\left(-\frac{t^2}{2}\right) dt \\ &= \frac{\sigma}{\sqrt{2\pi}} \left[ \left(\sqrt{2\pi}t \right) -\int_{-\infty}^\infty \sqrt{2\pi}dt \right] \\ &= \frac{\sigma}{\sqrt{2\pi}} \left[ \left(\sqrt{2\pi}t \right) - \left(\sqrt{2\pi}t \right) \right] \\ &= 0 \end{align*}\]
Substituting back into the original expression:
\[\begin{align*} E(X) &= 0 + \mu(1)\\ &= \mu \end{align*}\]
\[\begin{align*} E(X^2) &= \int_{-\infty}^\infty x^2 \cdot \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(\frac{(x-\mu)^2}{2\sigma^2}\right) dx\\ &= \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{\infty} x^2 \exp\left(\left(\frac{x-\mu}{\sqrt{2}\sigma}\right)^2 \right)dx \end{align*}\]
then, substitute \(t = \frac{x-\mu}{\sqrt{2}\sigma}\),
\(\Longrightarrow: x= \sqrt{2}\sigma t + \mu\), and
\(\Longrightarrow: \frac{dx}{dt} = \sqrt{2}\sigma\)
\(\therefore dx = \sqrt{2}\sigma dt\)
\[\begin{align*} E(X^2) &= \frac{\sqrt{2}\sigma}{\sigma\sqrt{2}\sqrt{\pi}} \int_{-\infty}^{\infty} (\sqrt{2}\sigma t + \mu)^2 \exp(-t^2) dt\\ &= \frac{\sqrt{2}\sigma}{\sigma\sqrt{2}\sqrt{\pi}} \int_{-\infty}^{\infty} (2\sigma^2t^2 + 2\sqrt{2}\sigma\mu t +\mu^2)\exp(-t^2)dt\\ &= \frac{1}{\sqrt{\pi}} \left[ 2\sigma^2\int_{-\infty}^{\infty} t^2\exp(-t^2)dt + 2\sqrt{2}\sigma\mu\int_{-\infty}^{\infty} t\exp(-t^2)dt + \mu^2\int_{-\infty}^{\infty} \exp(-t^2)dt \right] \\ &= \frac{1}{\sqrt{\pi}} \left[ 2\sigma^2\int_{-\infty}^{\infty} t^2\exp(-t^2)dt + 2\sqrt{2}\sigma\mu\left[-\frac{1}{2}\exp(-t^2) \right]_{-\infty}^{\infty} + \mu^2 \sqrt{\pi} \right] \\ &= \frac{1}{\sqrt{\pi}} \left[ 2\sigma^2\int_{-\infty}^{\infty} t^2\exp(-t^2)dt + 2\sqrt{2}\sigma\mu(0) + \mu^2 \sqrt{\pi} \right] \\ &= \frac{2\sigma^2}{\sqrt{\pi}} \int_{-\infty}^{\infty} t^2\exp(-t^2)dt + \mu^2 \\ &= \frac{2\sigma^2}{\sqrt{\pi}} \left[\left(-\frac{t}{2}\exp(t^2) \right)_{-\infty}^\infty +\frac{1}{2}\int_{-\infty}^{\infty}\exp(-t^2)dt \right] +\mu^2\\ &= \frac{2\sigma^2}{\sqrt{\pi}} \left[\frac{1}{2}\int_{-\infty}^{\infty}\exp(-t^2)dt \right] +\mu^2\\ &= \frac{2\sigma^2 \sqrt{\pi}}{2\sqrt{\pi}} + \mu^2 \\ &= \sigma^2 + \mu^2 \end{align*}\]
Then:
\[\begin{align*} Var(X) &= E(X^2) - [E(X)]^2\\ &= \sigma^2 + \mu^2 - \mu^2\\ &= \sigma^2 \end{align*}\]
\[\begin{align*} M_x(t) & = \int_{-\infty}^\infty e^{tx}\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{(x-\mu)^2}{2\sigma^2}\right) dx\\ &= \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^\infty \exp\left(tx\right)\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) dx\\ &= \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^\infty \exp\left(tx -\frac{(x-\mu)^2}{2\sigma^2} \right) dx \end{align*}\]
then, substitute \(u = \frac{x-\mu}{\sqrt{2}\sigma}\):
\(\Longrightarrow: x= u\sqrt{2}\sigma + \mu\), and
\(\Longrightarrow: \frac{du}{dx} = \frac{1}{\sqrt{2}\sigma}\)
\(\therefore dx = \sqrt{2}\sigma du\)
Substituting back in:
\[\begin{align*} M_x(t) & = \frac{\sqrt{2}\sigma}{\sigma\sqrt{2\pi}} \int_{-\infty}^\infty \exp ((u\sqrt{2}\sigma + \mu)t - u^2) du\\ &= \frac{\sqrt{2}\sigma}{\sigma\sqrt{2\pi}} \int_{-\infty}^\infty \exp ((u\sqrt{2}\sigma t + \mu t) - u^2) du\\ &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^\infty \exp (u\sqrt{2}\sigma t - u^2) du\\ &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^\infty \exp (-(u^2 - u\sqrt{2}\sigma t)) du \end{align*}\]
Now complete the square \((y=(x + \frac{b}{2})^2 + c - (\frac{b}{2})^2)\):
\[\begin{align*} M_x(t) &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^\infty \exp \left(-\left(u - \frac{\sqrt{2}\sigma t}{2}\right)^2 + \left(\frac{\sqrt{2}\sigma t}{2}\right)^2\right) du\\ &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^\infty \exp \left(-\left(u - \frac{\sqrt{2}\sigma t}{2}\right)^2 + \left(\frac{2\sigma^2 t^2}{4}\right)\right) du\\ &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^\infty \exp \left(-\left(u - \frac{\sqrt{2}\sigma t}{2}\right)^2 + \left(\frac{\sigma^2 t^2}{2}\right)\right) du\\ &= \frac{\exp \left(\mu t + \frac{\sigma^2 t^2}{2}\right)}{\sqrt{\pi}} \int_{-\infty}^\infty \exp \left(-\left(u - \frac{\sqrt{2}\sigma t}{2}\right)^2\right) du\\ \end{align*}\]
then, substitute \(v = u - \frac{\sqrt{2}\sigma t}{2}\):
\(\Longrightarrow: \frac{dv}{du} = 1\), and
\(\Longrightarrow: dv = du\)
\[\begin{align*} M_x(t) &= \frac{\exp(\mu t + \frac{\sigma^2 t^2}{2})}{\sqrt{\pi}} \int_{-\infty}^\infty \exp (-v^2) dv\\ \end{align*}\]
And from the (\(\int_{-\infty}^\infty exp(-v^2)dv = \sqrt{\pi}\))
\[\begin{align*} M_x(t) &= \frac{\sqrt{\pi}\exp \left(\mu t + \frac{\sigma^2 t^2}{2}\right)}{\sqrt{\pi}} \\ &= \exp \left(\mu t + \frac{\sigma^2 t^2}{2}\right) \end{align*}\]