\[\begin{align*} E(X) &= \int_{-\infty}^\infty x\frac{1}{2b} \exp\left(-\frac{|x-\mu|}{b}\right) dx \\ &= \int_{-\infty}^\mu x\frac{1}{2b} \exp\left(-\frac{\mu-x}{b}\right) dx + \int_{\mu}^\infty x\frac{1}{2b} \exp\left(-\frac{x-\mu}{b}\right) dx \\ &= \int_{-\infty}^\mu x\frac{1}{2b} \exp\left(\frac{x-\mu}{b}\right) dx + \int_{\mu}^\infty x\frac{1}{2b} \exp\left(-\frac{x-\mu}{b}\right) dx\\ \text{let: } u &= x-\mu \\ \Rightarrow du &= dx \\ \Rightarrow x &= u+\mu\\ E(X) &= \int_{-\infty}^0 (u+\mu)\frac{1}{2b} \exp\left(\frac{u}{b}\right) du + \int_0^\infty (u+\mu)\frac{1}{2b} \exp\left(-\frac{u}{b}\right) du \\ &= \frac{1}{2b}\left[\mu\int_{-\infty}^0 \exp\left(\frac{u}{b}\right) du + \int_{-\infty}^0 u\exp\left(\frac{u}{b}\right) du + \mu\int_0^\infty \exp\left(-\frac{u}{b}\right) du + \int_0^\infty u \exp\left(-\frac{u}{b}\right) du\right] \\ &= \frac{1}{2b}\left[\mu\int_{-\infty}^0 \exp\left(\frac{u}{b}\right) du + \mu\int_0^\infty \exp\left(-\frac{u}{b}\right) du + \int_{-\infty}^0 u\exp\left(\frac{u}{b}\right) du + \int_0^\infty u \exp\left(-\frac{u}{b}\right) du\right] \\ &= \frac{1}{2b}\biggl[\mu b\exp\left(\frac{u}{b}\right)\big|_{-\infty}^0 - \mu b\exp\left(-\frac{u}{b}\right)\big|_0^{\infty} + \left(ub\exp\left(\frac{u}{b}\right)\big|_{-\infty}^0 - \int_{-\infty}^0b\exp\left(\frac{u}{b}\right)du\right) \\ &\qquad + \left(-ub\exp\left(-\frac{u}{b}\right)\big|_0^\infty + \int_0^\infty b\exp\left(-\frac{u}{b}\right)du\right)\biggr] \\ &= \frac{1}{2b}\biggl[\mu b\exp\left(\frac{u}{b}\right)\big|_{-\infty}^0 - \mu b\exp\left(-\frac{u}{b}\right)\big|_0^{\infty} + \left(ub\exp\left(\frac{u}{b}\right)\big|_{-\infty}^0 - b^2\exp\left(\frac{u}{b}\right)\big|_{-\infty}^0\right) \\ &\qquad + \left(-ub\exp\left(-\frac{u}{b}\right)\big|_0^\infty - b^2\exp\left(-\frac{u}{b}\right)\big|_0^\infty\right)\biggr] \\ &= \frac{1}{2b} \left[\mu b - 0 - 0 +\mu b +0 - 0 - 0 + b^2 + 0 - b^2 + 0 - 0 \right] \\ &= \frac{1}{2b}\cdot 2b\mu \\ &= \mu \end{align*}\]
\[\begin{align*} E(X^2) &= \int_{-\infty}^\infty x^2\frac{1}{2b} \exp\left(-\frac{|x-\mu|}{b}\right) dx \\ &= \int_{-\infty}^\mu x^2\frac{1}{2b} \exp\left(-\frac{\mu-x}{b}\right) dx + \int_{\mu}^\infty x^2\frac{1}{2b} \exp\left(-\frac{x-\mu}{b}\right) dx \\ &= \int_{-\infty}^\mu x^2\frac{1}{2b} \exp\left(\frac{x-\mu}{b}\right) dx + \int_{\mu}^\infty x^2\frac{1}{2b} \exp\left(-\frac{x-\mu}{b}\right) dx\\ \text{let: } u &= x-\mu \\ \Rightarrow du&= dx \\ \Rightarrow x &= u+\mu \\ \therefore E(X^2) &= \frac{1}{2b}\left[ \int_{-\infty}^0 (u+\mu)^2 \exp\left(\frac{u}{b}\right)du + \int_0^{\infty} (u+\mu)^2 \exp\left(-\frac{u}{b} \right) du \right] \\ &= \frac{1}{2b}\left[ \int_{-\infty}^0 (u^2 + 2\mu u +\mu^2) \exp\left(\frac{u}{b}\right)du + \int_0^{\infty} (u^2 + 2\mu u +\mu^2) \exp\left(-\frac{u}{b} \right) du \right] \\ &= \frac{1}{2b}\Bigg[\int_{-\infty}^0 u^2 \exp(\frac{u}{b})du + 2\mu\int_{-\infty}^0 u \exp(\frac{u}{b})du + \mu^2\int_{-\infty}^0 \exp(\frac{u}{b})du\\ &\qquad \int_{-\infty}^0 u^2 \exp(-\frac{u}{b})du + 2\mu\int_{-\infty}^0 u \exp(-\frac{u}{b})du + \mu^2\int_{-\infty}^0 \exp(-\frac{u}{b})du \Bigg] \\ \end{align*}\]
Now:
\[\begin{align*} \int_{-\infty}^0 u^2 \exp(\frac{u}{b})du &= \left[u^2 \cdot be^{u/b}\right]_{-\infty}^0 - \int_{-\infty}^0 2u\cdot b\exp(\frac{u}{b})du \\ &= 0 - 2b\left(\left[ube^{u/b}\right]_{-\infty}^0 - \int_{-\infty}^0 be^{u/b}\,du\right) \\ &= -2b\left(0 - b\left[be^{u/b}\right]_{-\infty}^0\right) \\ &= 2b^3 \\ \int_{-\infty}^0 u \exp(\frac{u}{b})du &= \left[ub\exp(\frac{u}{b})\right]_{-\infty}^0 - \int_{-\infty}^0 b\exp(\frac{u}{b})du \\ &= 0 - b\left[b\exp(\frac{u}{b})\right]_{-\infty}^0 \\ &= -b^2 \\ \int_{-\infty}^0 \exp(\frac{u}{b})du &= \left[be^{u/b}\right]_{-\infty}^0 \\ &= b \\ \int_0^\infty u^2 \exp(-\frac{u}{b})du &= \left[-bu^2\exp(-\frac{u}{b})\right]_0^\infty + \int_0^\infty 2ub\exp(-\frac{u}{b})du \\ &= 0 + 2b\left(\left[-ub\exp(-\frac{u}{b})\right]_0^\infty + \int_0^\infty b\exp(-\frac{u}{b})du\right) \\ &= 2b\left(0 + b\left[-b\exp(-\frac{u}{b})\right]_0^\infty\right) \\ &= 2b^3 \\ \int_0^\infty u\exp(-\frac{u}{b})du &= \left[-ub\exp(-\frac{u}{b})\right]_0^\infty + \int_0^\infty b\exp(-\frac{u}{b})du \\ &= 0 + b\left[-b\exp(-\frac{u}{b})\right]_0^\infty \\ &= b^2 \\ \int_0^\infty \exp(-\frac{u}{b})du &= \left[-b\exp(-\frac{u}{b})\right]_0^\infty \\ &= b \\ &= \frac{1}{2b}\Big[ (2b^3) + 2\mu(-b^2) + \mu^2(b) + (2b^3) + 2\mu(b^2) + \mu^2(b) \Big] \\ &= \frac{1}{2b}\Big[4b^3 + 2\mu^2 b\Big] \\ &= 2b^2 + \mu^2 \\ \mathrm{Var}(X) &= 2b^2 + \mu^2 - \mu^2 \\ &= 2b^2 \end{align*}\]
\[\begin{align*} M_X(t) &= \int_{-\infty}^\infty \exp(tx) \cdot \frac{\exp(-\frac{|x-\mu|}{b})}{2b} dx\\ &= \frac{1}{2b} \int_{-\infty}^\infty \exp(tx) \cdot \exp(-\frac{|x-\mu|}{b})dx\\&= \frac{1}{2b} \left[\int_{-\infty}^\mu \exp(tx) \cdot \exp(-\frac{\mu-x}{b})dx + \int_\mu^\infty \exp(tx) \cdot \exp(-\frac{x-\mu}{b})dx\right]\\ &= \frac{1}{2b} \left[\int_{-\infty}^\mu \exp\left(tx -\frac{\mu-x}{b}\right)dx + \int_\mu^\infty \exp\left(tx-\frac{x-\mu}{b}\right)dx\right]\\ &= \frac{1}{2b} \left[\int_{-\infty}^\mu \exp\left(\frac{btx +x-\mu}{b}\right)dx + \int_\mu^\infty \exp\left(\frac{btx-x+\mu}{b}\right)dx\right]\\ &= \frac{1}{2b} \left[\exp\left(-\frac{\mu}{b}\right)\int_{-\infty}^\mu \exp\left(x(t+\frac{1}{b}) \right) dx + \exp\left(\frac{\mu}{b}\right) \int_{-\infty}^\mu \exp\left(x(t-\frac{1}{b}) \right) dx \right]\\ \text{for convergence:} |t| &< \frac{1}{b}: \\ &= \frac{1}{2b} \left[\exp\left(-\frac{\mu}{b}\right)\exp\left(x(t +\frac{1}{b})\right)\cdot \frac{1}{t+\frac{1}{b}} \biggr\rvert_{-\infty}^\mu + \exp\left(\frac{\mu}{b}\right)\exp\left(x(t -\frac{1}{b})\right)\cdot \frac{1}{t-\frac{1}{b}} \biggr\rvert_\mu^\infty \right]\\ &= \frac{1}{2b} \left[\exp\left(-\frac{\mu}{b}\right)\exp\left(\mu(t +\frac{1}{b})\right)\cdot \frac{1}{t+\frac{1}{b}} - \exp\left(\frac{\mu}{b}\right)\exp\left(\mu(t -\frac{1}{b})\right)\cdot \frac{1}{t-\frac{1}{b}} \right]\\ &= \frac{1}{2b} \left[\exp\left(-\frac{\mu}{b}\right)\exp(\mu t)\exp\left(\frac{\mu}{b}\right)\cdot \frac{1}{t+\frac{1}{b}} - \exp\left(\frac{\mu}{b}\right)\exp(\mu t)\exp\left(-\frac{\mu}{b}\right)\cdot \frac{1}{t-\frac{1}{b}} \right]\\ &= \frac{\exp(\mu t)}{2b} \left[\frac{1}{t+\frac{1}{b}} - \frac{1}{t-\frac{1}{b}} \right]\\ &= \frac{\exp(\mu t)}{2b} \left[\frac{b}{tb+1} - \frac{b}{tb-1} \right]\\ &= \frac{\exp(\mu t)}{2b} \left[b\cdot \frac{(tb-1) - (tb+1)}{(tb+1)(tb-1)}\right]\\ &= \frac{\exp(\mu t)}{2b} \left[\cdot \frac{-2b}{t^2b^2-1}\right]\\ &= -\frac{\exp(\mu t)}{t^2b^2-1}\\ &= \frac{\exp(\mu t)}{1-b^2t^2} \quad \text{for } |t| < \frac{1}{b} \end{align*}\]