The term $\int_{-\infty}^{+\infty} e^{-\frac{1}{2}(x-a)^2}\text{d}x$ is easy to be calculated in the following procedures. \begin{align} &\left[\int_{-\infty}^{+\infty}e^{-\frac{1}{2}(x-a)^2}\text{d}x\right] \left[\int_{-\infty}^{+\infty}e^{-\frac{1}{2}(y-a)^2}\text{d}y\right]\\ &=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-\frac{1}{2}[(x-a)^2+(y-a)^2]}\text{d}x\text{d}y\\ &=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-\frac{1}{2}(z_1^2+z_2^2)}\text{d}z_1\text{d}z_2\\ &=\int_0^{2\pi}\int_{0}^{+\infty} e^{-\frac{1}{2}r^2}\text{d}r.r\\ &=2\pi \left[-e^{-\frac{1}{2}r^2}\right]_{0}^{+\infty}\\ &=2\pi \end{align} Thus, $\int_{-\infty}^{+\infty}e^{-\frac{1}{2}(x-a)^2}\text{d}x=\sqrt{2\pi}$. This term can also be calculated by Gaussian integral by regarding $a$ as the mean of Gaussian random variable $x$.
As I use Matlab to verify them, I find they are equal. But, how to analytically calculate the term $\int_{-\infty}^{+\infty} e^{-\frac{1}{2}(x-\mathbb{J}a)^2}\text{d}x$ with $\mathbb{J}^2=-1$? The method above can't be applied to calculate this term since $r$ is a real-vale number while it is a complex-valued number in the term involving $\mathbb{J}a$. Meanwhile, the Gaussian integral is also not suitable since a real-valued random variable does not has a complex-valued mean $\mathbb{J}a$.
