Wallis meets quantum mechanics

Consider the integral, change $x$ to be $x+2$ , we get $\begin{aligned} \int_{-\infty}^{\infty} \exp(2(x+2)(2-x)) \, dx &= \int_{-\infty}^{\infty} \exp(2(4-x^2)) \, dx\\[5pt] &=e^8 \int_{-\infty}^{\infty} \exp(-2x^2) \, dx\\[5pt] &=\frac{e^8}{\sqrt{2}} \int_{-\infty}^{\infty} \exp(-x^2) \, dx\\[5pt] &=\frac{e^8 \sqrt{\pi}}{\sqrt{2}}. \end{aligned}$ Note that we changed $x$ to be $\frac{x}{\sqrt{2}}$ meanwhile. Therefore, $e^8 \sqrt{\frac{\pi}{2}}=e^8 \prod_{c=1}^{\infty}\frac{2c}{\sqrt{\alpha c^2-\beta}}.$ We can express this equation as $\frac{\pi}{2} = \prod_{c=1}^{\infty} \frac{4c^2}{\alpha c^2-\beta},$ which is Wallis formula, where $\alpha=4$ and $\beta=1$ . The answer is $4+1=\boxed{5}$ .