Substitute $g(x)=Ae^{-\beta x^2}$ into $\theta(x,t)=\frac{1}{2\sqrt{\pi\alpha t}}\int_{-\infty}^{\infty} g(\eta)\exp(-(x-\eta)^2/4\alpha t) \ d\eta$ 
I am trying to show that by directly substituting $g(x)=Ae^{-\beta x^2}$ into $$\theta(x,t)=\frac{1}{2\sqrt{\pi\alpha t}}\int_{-\infty}^{\infty} g(\eta)\exp(-(x-\eta)^2/4\alpha t) \ d\eta,$$ we obtain $$\theta(x,t)=\frac{A}{\sqrt{1+4\alpha\beta t}}\exp(-\beta x^2/(1+4\alpha\beta t).$$

My attempt:
Let $g(\eta)=Ae^{-\beta\eta^2}$. Then $$\theta(x,t)=\frac{1}{2\sqrt{\pi\alpha t}}\int_{-\infty}^{\infty} \exp(-\beta\eta^2-(x-\eta)^2/4\alpha t) \ d\eta.$$ However, this doesn't give the desired result. Is someone able to please give me a hint on the correct path that I should take?
 A: Let us first compute the expression in exponential
\begin{align}
-\beta\eta^2 - \frac{1}{4\alpha t}(x-\eta)^2 &= -\beta\eta^2 - \frac{1}{4\alpha t}(x^2-2x\eta+\eta^2) \\
&=-\frac{x^2 }{4\alpha t}-\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta^2 - 2\frac{x}{4\alpha t\beta+1}\eta\right) \\
&=-\frac{x^2 }{4\alpha t}-\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta - \frac{x}{4\alpha t\beta+1}\right)^2 -\frac{x^2}{(4\alpha t\beta+1)4\alpha t} \\
&=-\frac{\beta x^2}{(4\alpha t\beta+1)} -\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta - \frac{x}{4\alpha t\beta+1}\right)^2
\end{align}
Replacing the previous expression in the initial calculation:
\begin{align*}
\frac{A}{2\sqrt{\pi\alpha t}}\exp\left(-\frac{\beta x^2}{(4\alpha t\beta+1)}\right)\int_{R}\exp\left(-\frac{4\alpha t\beta+1}{4\alpha t}\left(\eta - \frac{x}{4\alpha t\beta+1}\right)^2\right)d\eta = \\ \frac{A}{\sqrt{4\alpha t\beta+1}}\exp\left(-\frac{\beta x^2}{(4\alpha t\beta+1)}\right)
\end{align*}
In the last equation I used the Gauss integral computation : $\int_{R}\exp(-\mu(x-\gamma)^2)dx = \sqrt{\pi/\mu}$
