Does
$$\int_{-\infty}^\infty \text{e}^{\ a\ (x+b)^2}\ \text dx=\int_{-\infty}^\infty \text{e}^{\ a\ x^2}\ \text dx\ \ \ \ \ ?$$
hold, even if the imaginary part of $b$ is nonzero?
What I really want to understand is what the phrase "By analogy with the previous integrals" means in that link. There, the expression $\frac{J}{a}$ is complex but they seem to imply the integral can be solved like above anyway.
The reusult tells us that the integral is really independend of $J$, which is assumed to be real here. I wonder if we can also generalize this integral to include complex $J$. In case that the shift above is possible, this should work out.
But even if the idea is here to perform that substitution, how to get rid of the complex $a$ to obtain the result. If everything is purely real or imaginary, then this solves the rest of the problem.
Answer
Let us write $b= r+it$. The real part of $b$ does not matter as you have already proven yourself. So wlog $r=0$.
For shifting along the imaginary axis, we have to employ the residue theorem. We have
$$
\begin{align}
\int_{-\infty}^\infty f(x+i t) \,dx&- \int_{-\infty}^\infty f(x)\, dx\\
&=\int_{-\infty-it}^{\infty-it} f(x) \,dx- \int_{-\infty}^\infty f(x)\, dx \\
&= 2\pi i \sum \text{Res}(f)+ \int_{\infty-it}^{\infty} f(x) \,dx
- \int_{-\infty-it}^{-\infty} f(x) \,dx
\end{align},$$
where $\sum \text{Res}(f)$ is the sum over the residues of $f$ in the area $z\in \mathbb{C}$ with $-t<\text{Im}\, z<0$.
So the two integrals are the same if there are no residues and if the two integral at $\pm \infty$ vanish (both of which is the case for your example as long as $\text{Re}\,a <0$).
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