MGF of quadratically transformed normal rv

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For Z~N(1,1/2), find the moment generating function of $W=Z^2$.

$$f_z(z)= \frac{1}{\sqrt{2\pi\sigma^2}}e^\frac{-(z-\mu)^2}{2\sigma^2}=\frac{1}{\sqrt\pi}e^{-(z-1)^2}$$ so $$M_W(s)=E[e^{sW}]=E[e^{sZ^2}]$$

but I can't get anywhere trying to evaluate this integral. What is the correct method for evaluating this integral?

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With $X \sim N(0,1)$ the MGF for $Y=X^2$ is

$$ \begin{align} M_Y(t) & = E[e^{tY}]=E[e^{tX^2}] \\ & \\ & = \int_{-\infty}^{\infty} e^{tx^2} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}\: \text{dx}\\ & \\ & = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} e^{-(1/2 -t)x^2} \: \text{dx}\\ \end{align} $$

Using

$$\int_{-\infty}^{\infty} e^{-ax^2} \text{dx} = \sqrt{\frac{\pi}{a}}$$

we obtain

$$M_Y(t) = \frac{1}{\sqrt{2 \pi}}\sqrt{\frac{\pi}{1/2 - t}} = \frac{1}{\sqrt{1 - 2t}}$$

Transforming to a general r.v. $N(\mu, \sigma^2)$

$$M_W(s) = e^{\mu s}M_Y(\sigma s) = \frac{e^{\mu s}}{\sqrt{1 - 2(\sigma s)}}$$