I read this paper on parameter estimation using Bayesian theory, and I came across the following integral that I did not know how to solve. We have the following joint posterior probability for the parameters $A$ and $\alpha$,
$P(A\alpha|D\rho I)=exp(-\frac{Q}{2\rho^2})$
where
$Q=\sum_{i=1}^{N}(d_i-A\exp(-\alpha t_i))^2$.
To obtain the posterior probability, they integrate with respect to the variable A
$P(\alpha|D\rho I)\propto\int dA exp(-\frac{Q}{2\rho^2})$
The integral over A is a Gaussian integral. It is evaluated by completing the square in Q, followed by a change of variables to transform the integral into a Gaussian.
to obtain
$P(\alpha|D\rho I) \propto exp(\dfrac{\bar{h}^2}{2\rho^2})$,
where
$\bar{h}^2= \dfrac{\{\sum_{i=1}^{N}(d_i\exp(-\alpha t_i))\}^2}{\sum_{i=1}^{N} \exp(-2\alpha t_i)}$.
I am assuming they use Laplace's method to evaluate the integral, but I don't understand how to complete the squares in $Q$ and what would be a good change of variables to obtain a Gaussian integral.
For inclined people interested in more details, see paper, Appendix A1.
Write $$Q=\sum_{i=1}^n (d_i- A \,a_i)^2 \qquad \text{with} \qquad a_i=e^{-\alpha t_i}$$
$$Q=A^2 \sum_{i=1}^n a_i^2-2A\sum_{i=1}^n a_id_i+\sum_{i=1}^n d_i^2=A^2 S_1-2A S_2+S_3$$ $$A^2 S_1-2A S_2+S_3=S_1\left(A^2-2 \frac{S_2}{S_1}A+\left(\frac{S_2}{S_1}\right)^2\right)-S_1\left(\frac{S_2}{S_1}\right)^2+S_3$$ $$A^2 S_1-2A S_2+S_3=S_1\left(A-\frac{S_2}{S_1}\right)^2+\frac{S_1 S_3-S_2^2}{S_1}$$