Painful? Moment Generating Function

Question

Part 1

Let $X$ be a random variable with the p.d.f. $f(x)=\frac{1}{4\pi}e^{\frac{-x^2}{4}}$, compute the MGF of $X$.

So I know I want $\psi_X(t)=E(e^{tx})=\frac{1}{4\pi}\int^{\infty}_{-\infty}e^{tx}e^{\frac{-x^2}{4}}\,dx$, but I'm having a problem trying to figure out how to integrate this one, which is what I would really like a little help with. Maybe a substitution recommendation? I tried to change $e^{tx}e^{\frac{-x^2}{4}}$ to something like $e^{x(t-\frac{x}{4})}$, but that didn't seem to get me anywhere.

Part 2 Suppose $X_1$ and $X_2$ are indep N(0,1) and let $Y=X_1+X_2$. Compute the MGF of Y and compare to the MGF of X. What is the p.d.f. of $Y$?

So this part I got okay, with $\psi_Y(t)= \psi_X(t)^2= e^{t^2}$, since the MGF of $X$ is $e^t$.

So how can I finish the first integral to compare them?

Answer 1

Hint: $$\begin{eqnarray} \int_{-\infty}^\infty \exp\left(tx - \frac{x^2}{a}\right) \,dx &=& \int_{-\infty}^\infty \exp\left(\frac{atx - x^2}{a}\right) \,dx = \int_{-\infty}^\infty \exp\left(\frac{-(x - \frac{1}{2}at)^2 + \frac{1}{4}a^2 t^2}{a}\right) \,dx \\ &=& exp\left(\frac{\mu^2}{2\sigma^2}\right) \int_{-\infty}^\infty \exp\left(\frac{-(x - \mu)^2}{2\sigma^2}\right) \,dx \text{ where } \begin{matrix} \mu = \frac{1}{2}at \\ \sigma = \sqrt{\frac{1}{2}a} \end{matrix} \end{eqnarray}$$

Answer 2

So by drhab's comment, I obtained: $\psi_X(t)=E(e^{tx})=\frac{1}{4\pi}\int^{\infty}_{-\infty}e^{tx}e^{\frac{-x^2}{4}}\,dx=\frac{1}{4\pi}\int^{\infty}_{-\infty}e^{tx-\frac{x^2}{4}}\,dx=\frac{1}{4\pi}\int^{\infty}_{-\infty}e^{t^2-\frac{(x-2t)^2}{4}}\,dx=\frac{1}{4\pi}e^{t^2}\int^{\infty}_{-\infty}e^{-\frac{(x-2t)^2}{4}}\,dx=\frac{2\sqrt{\pi}}{4 \pi}e^{t^2}=\frac{1}{2\sqrt{\pi}}e^{t^2}$. Then $\frac{1}{2\sqrt{\pi}}\psi_Y(t)=\psi_X(t)$, i.e.$\psi_Y(t)=2\sqrt{\pi}\cdot\psi_X(t)$ and so the p.d.f.of $Y$ is $f_Y(y)=\frac{1}{2\sqrt{\pi}}e^{\frac{-x^2}{4}}$. Is this the correct way since $E(aX)=aE(X)$?